AP Chemistry Unit 7: Equilibrium — Free Review Games.

At dynamic equilibrium, the forward and reverse reactions continue at equal rates. Concentrations remain constant but are not necessarily equal.

Kc = products over reactants, each raised to their stoichiometric coefficients: Kc = [NH3]^2 / ([N2][H2]^3).

When K >> 1, the numerator (products) is much larger than the denominator (reactants), meaning equilibrium lies far to the right, favoring products.

Adding reactant increases its concentration. The system shifts toward products to partially counteract the change. K does not change.

The concentration of a pure solid or liquid is constant since their density does not change. This constant is folded into the equilibrium constant.

For an exothermic reaction, heat is a product. Increasing temperature adds heat, shifting equilibrium to the left and decreasing K.

When Q < K, there are too few products relative to equilibrium. The reaction shifts right (toward products) until Q = K.

Kp = Kc(RT)^delta_n, where delta_n = moles of gaseous products minus moles of gaseous reactants.

Adding an inert gas at constant volume increases total pressure but does not change the partial pressures or concentrations of reacting gases. Equilibrium is unaffected.

A catalyst speeds up both forward and reverse reactions equally. It helps reach equilibrium faster but does not change K or equilibrium concentrations.

ICE table: \([A] = 1.0-x\), \([B] = 2x\). \(K_c = (2x)^2/(1.0-x) = 4.0\). Solving: \(4x^2 + 4x - 4 = 0\), \(x = (-1+\sqrt{5})/2 = 0.618\). \([B] = 2(0.618) = 1.24\) M.

AgCl \(\to\) Ag+ + Cl-. If solubility = s, then \(K_{sp} = s^2 = 1.8 \times 10^{-10}\). \(s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}\) M.

Decreasing volume increases pressure. The system shifts toward fewer moles of gas (1 mole N2O4 vs 2 moles NO2) to reduce pressure.

When reactions are added, their equilibrium constants are multiplied. K = K1 * K2 = 2.0 * 3.0 = 6.0.

PbCl2 \(\to\) Pb2+ + 2Cl-. With 0.10 M NaCl providing Cl-, \(K_{sp} = [Pb^{2+}][Cl^-]^2\). Since s is small: \(1.7 \times 10^{-5} = s(0.10)^2 = 0.01s\). \(s = 1.7 \times 10^{-3}\) M. The common ion effect reduces solubility.

At equilibrium, the forward and reverse reactions continue to occur, but at equal rates, so there is no net change in concentrations. This is why equilibrium is described as dynamic rather than static. Choice A describes conditions before equilibrium is reached when Q < K. Choice C is a common misconception — equilibrium is not a static state; molecules are still reacting in both directions.

A very small K (K << 1) means product concentrations in the numerator of the equilibrium expression are much smaller than reactant concentrations in the denominator, so reactants are heavily favored. Choice A describes a K >> 1 situation. Choice D is wrong — the reaction does occur but barely produces products before reaching equilibrium. Choice C would correspond to K near 1.

The equilibrium expression places product concentrations in the numerator and reactant concentrations in the denominator, each raised to the power of their stoichiometric coefficients. All coefficients here are 1, giving Kc = [PCl3][Cl2]/[PCl5]. Choice B is the inverse and represents the equilibrium constant for the reverse reaction (1/Kc). Choices C and D omit either the reactant or product term entirely.

Removing a product lowers its concentration, making Q < K. To restore equilibrium the reaction shifts in the forward direction, producing more product. Choice B is incorrect — shifting toward reactants would further decrease product concentration, moving the system even further from equilibrium. Choice C is a common misconception: K stays constant, but the equilibrium position (the actual concentrations) does shift in response to the stress.

For an endothermic reaction, heat can be treated as a reactant. Adding heat shifts equilibrium toward the products. Unlike concentration or pressure changes, temperature changes alter the value of K itself — K increases for endothermic reactions when temperature rises. Choice A describes what happens to an exothermic reaction when temperature increases. Choice C is a common error: any equilibrium shift caused by a temperature change is always accompanied by a change in K.

At equilibrium, concentrations are constant because the forward and reverse rates are equal — not because the reaction has stopped. Choice A is wrong because there is no requirement for reactant and product concentrations to be equal; that would only occur if K = 1. Choice C describes a complete (irreversible) reaction, not an equilibrium. Choice D is wrong because equilibrium is dynamic — both the forward and reverse reactions continue at equal rates.

Ksp is simply the equilibrium constant for the dissolution equilibrium of a sparingly soluble salt, with the pure solid omitted from the expression. For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), Ksp = [Ag⁺][Cl⁻]. Choice A confuses Ksp with a rate constant. Choice B is a common misconception — Ksp can be used to calculate maximum solubility, but it is defined as an equilibrium constant, not a concentration limit. Choice D is not a thermodynamic definition.

In the ICE table, let x = moles per liter of H2 (and I2) consumed. At equilibrium: [H2] = [I2] = 1.00 - x and [HI] = 2x. Kc = (2x)²/(1.00 - x)² = 50.0. Taking the square root of both sides: 2x/(1.00 - x) = 7.07. Solving: 9.07x = 7.07, so x = 0.780. Therefore [HI] = 2(0.780) = 1.56 M. Choice A (0.78 M) is the value of x — a common error of forgetting the stoichiometric coefficient of 2 for HI.

When Q > K, there are too many products relative to the equilibrium state. The reaction proceeds in the reverse direction, consuming products and producing reactants, which decreases Q until Q = K. Choice A is backwards — shifting toward products would increase Q further, moving the system farther from equilibrium. Choice D is wrong because K is a constant at a given temperature; it is Q that must change to match K, not the other way around.

Pressure changes only shift equilibrium when there is a net difference in moles of gas between products and reactants (Δn ≠ 0). Here the reactant side has 1 + 1 = 2 moles of gas and the product side also has 2 moles, so Δn = 0. Changing pressure has no effect on the equilibrium position. Choice D is wrong because Kc depends only on temperature — not on pressure or concentration changes.

The relationship is Kp = Kc(RT)^Δn, where Δn = moles of gaseous products minus moles of gaseous reactants = 2 - 1 = 1. Therefore Kp = (4.6 × 10⁻³)(0.0821 × 298)^1 = (4.6 × 10⁻³)(24.5) ≈ 0.11. Choice A is incorrect because Kp = Kc only when Δn = 0. Choice C would result from dividing by RT rather than multiplying, which reverses the formula.

The reactant side has 1 + 3 = 4 moles of gas; the product side has 1 + 1 = 2 moles. Decreasing volume increases pressure, and Le Chatelier's principle predicts a shift toward the side with fewer moles of gas — the products — to partially relieve the increased pressure. Choice A has the direction backwards. Choice D is incorrect because Kc is constant at constant temperature and does not depend on pressure.

Adding a common ion raises the concentration of one product ion in the dissolution equilibrium, making Q > Ksp. The system responds by shifting in the reverse direction (toward solid), which decreases the amount of dissolved salt. Choice A is incorrect — adding a common ion shifts the equilibrium backward, reducing solubility. Choice C is wrong because Ksp is a constant at a given temperature and cannot be altered by concentration changes.

Pure solids have constant activity equal to 1 and are excluded from equilibrium expressions. Since both CaCO3 and CaO are pure solids, only the gaseous CO2 appears, giving Kc = [CO2]. Choice A incorrectly includes pure solid concentrations in the expression. Choice D is the equilibrium constant for the reverse reaction (CO2 + CaO → CaCO3), which equals 1/Kc.

At pH 3.0, [H⁺] = 10⁻³ = 0.0010 M. Percent dissociation = ([H⁺] / [HA]₀) × 100 = (0.0010 / 0.10) × 100 = 1.0%. Choice A (10%) would result from incorrectly dividing [H⁺] by 0.010 instead of 0.10. Choice C (3.0%) confuses the pH value itself with the percent dissociation, which is a category error. Only 1% dissociation is consistent with typical weak acid behavior.

The correct thermodynamic relationship is ΔG° = -RT ln K. A spontaneous forward reaction (ΔG° < 0) corresponds to K > 1, which is consistent because ln(K > 1) > 0, making -RT ln K negative. Choice A lacks the negative sign, producing the wrong sign convention — it would predict a negative ΔG° for K < 1, which is backwards. Choice D omits temperature T, which is required for proper units and physical meaning.

For HA ⇌ H⁺ + A⁻, let x = [H⁺] at equilibrium. Ka = x²/(0.50 - x) ≈ x²/0.50 under the small-x approximation. Solving: x² = Ka × 0.50, so x = √(Ka × 0.50) = √(1.8 × 10⁻⁵ × 0.50) ≈ 3.0 × 10⁻³ M. Choice B multiplies Ka by concentration, which would apply to a strong acid scaled incorrectly. Choice D inverts the concentration inside the square root, which has no physical basis in the equilibrium expression.

Two manipulations are required. First, reverse the reaction: K_rev = 1/K = 1/(9.1 × 10⁻⁴¹) = 1.1 × 10⁴⁰. This gives 2H2 + O2 → 2H2O. Second, multiply the entire equation by ½ to obtain H2 + ½O2 → H2O. When a reaction is multiplied by a factor n, the new K = (K_old)^n. So K_final = (1.1 × 10⁴⁰)^(1/2) ≈ 1.05 × 10²⁰. Choice B is K for the reversed but unhalved reaction. The very large K confirms that water formation from H2 and O2 is highly thermodynamically favored.

Let x = amount of A that reacts. At equilibrium: Kc = x²/(0.50 - x) = 0.010. Rearranging: x² + 0.010x - 0.0050 = 0. Applying the quadratic formula: x = (-0.010 + √(0.0001 + 0.020))/2 = (-0.010 + 0.1418)/2 ≈ 0.066 M. Therefore [A] = 0.50 - 0.066 = 0.43 M. Choice C is x itself (the change), not the equilibrium concentration of A. The small-x approximation gives x ≈ 0.071, which is 14% of 0.50 — above the 5% threshold — confirming the quadratic is required.

Two factors must be optimized. (1) Temperature: the reaction is exothermic, so decreasing temperature shifts equilibrium toward products and increases K. (2) Pressure: the reactant side has 2 + 1 = 3 moles of gas and the product side has 2 moles, so increasing pressure shifts equilibrium toward the fewer-mole side — the products. Both changes in choice B favor SO3 production. Choice A applies the wrong direction for both factors. Choice C is incorrect because while a catalyst speeds attainment of equilibrium, it never shifts the equilibrium position, and increasing temperature for this exothermic reaction actually decreases SO3 yield.

After mixing equal volumes, all concentrations are halved: [Pb²⁺] = 5.0 × 10⁻⁵ M and [Cl⁻] = 5.0 × 10⁻⁵ M. The ion product Q = [Pb²⁺][Cl⁻]² = (5.0 × 10⁻⁵)(5.0 × 10⁻⁵)² = (5.0 × 10⁻⁵)(2.5 × 10⁻⁹) = 1.25 × 10⁻¹³. Since Q = 1.25 × 10⁻¹³ is far less than Ksp = 1.7 × 10⁻⁵, the solution is unsaturated and no precipitate forms. Choice A is a common misconception — the mere presence of both ions is not sufficient to cause precipitation; Q must exceed Ksp.

Using an ICE table: [PCl3]₀ = [Cl2]₀ = 1.00 M, [PCl5]₀ = 0. Since 0.82 mol/L of PCl5 forms, [PCl3] = [Cl2] = 1.00 - 0.82 = 0.18 M and [PCl5] = 0.82 M at equilibrium. Kc = [PCl5]/([PCl3][Cl2]) = 0.82/(0.18 × 0.18) = 0.82/0.0324 ≈ 25. Choice A (0.82) is just the equilibrium concentration of PCl5 — not the equilibrium constant. Choice D (0.040) is approximately 1/Kc, the equilibrium constant for the reverse dissociation reaction.

Ca(OH)2 dissolves as: Ca(OH)2(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq). If molar solubility = s, then [Ca²⁺] = s and [OH⁻] = 2s. Ksp = s(2s)² = 4s³ = 4.68 × 10⁻⁶. Solving: s³ = 1.17 × 10⁻⁶, so s = (1.17 × 10⁻⁶)^(1/3) ≈ 1.05 × 10⁻² M. Choice D (2.10 × 10⁻²) is [OH⁻] = 2s, not the molar solubility s. Choice B results from incorrectly using Ksp = s² as if the salt were a 1:1 electrolyte like AgCl, ignoring the stoichiometric coefficient of 2 for OH⁻.

Doubling the volume halves all concentrations: [A] = 0.50 M and [B] = 0.25 M. Q = [B]²/[A] = (0.25)²/(0.50) = 0.0625/0.50 = 0.125. Since Q = 0.125 < Kc = 0.25, the reaction shifts toward the products to increase Q back to 0.25. This also follows from Le Chatelier's principle: decreasing pressure by doubling volume favors the side with more moles of gas — the product side has 2 moles versus 1 for the reactants. Choice D has the correct Q but the wrong shift direction; Q < K always means a shift toward products.

In the presence of 0.010 M SO4²⁻ from Na2SO4, set up the ICE table: [Ba²⁺] = s and [SO4²⁻] = 0.010 + s ≈ 0.010 M (since s will be very small). Ksp = s(0.010) = 1.1 × 10⁻¹⁰. Solving: s = 1.1 × 10⁻¹⁰ / 0.010 = 1.1 × 10⁻⁸ M. Compared to pure water solubility of √(1.1 × 10⁻¹⁰) ≈ 1.05 × 10⁻⁵ M, the common ion SO4²⁻ reduces solubility by a factor of about 1000. Choice A is the solubility in pure water, not in the Na2SO4 solution. Choice C equals Ksp itself, not the solubility.

The equilibrium constant expression is products over reactants, with each concentration raised to the power of its stoichiometric coefficient. NH3 has a coefficient of 2, so [NH3] is squared; H2 has a coefficient of 3, so [H2] is cubed. Choice B is the inverse (reactants over products). Choice C uses an exponent of 1 on [NH3] instead of 2. Choice D incorrectly places stoichiometric coefficients as multipliers in the numerator and denominator rather than as exponents.

Pure solids have constant concentrations and are excluded from the equilibrium expression. Both CaO and CaSO3 are solids, so neither appears in the expression. The only species included is SO2(g), which is a reactant, making Kc = 1/[SO2]. Choices B and C incorrectly include solid concentrations. Choice D is the inverse of the correct expression, placing [SO2] in the numerator rather than the denominator.

A catalyst lowers the activation energy of both the forward and reverse reactions by the same amount, so both rates increase equally. Because the ratio of rates — and thus the ratio of concentrations at equilibrium — remains unchanged, neither the equilibrium position nor the value of Kc changes. A catalyst only allows equilibrium to be reached more quickly. Kc depends only on temperature, not on activation energy, making Choice D incorrect. Choices A and B incorrectly imply a selective effect on one direction of reaction.

When a reaction is reversed, the new equilibrium constant is the reciprocal of the original: K_reverse = 1 / K_forward = 1 / 0.040 = 25. Choice A incorrectly uses the same K value. Choice B equals (0.040)², which would apply if the equation were doubled in the same direction, not reversed. Choice D equals the square root of 0.040, which has no thermodynamic justification for a simple reversal.

In a rigid container at constant volume, adding an inert gas increases total pressure but does not change the partial pressures or molar concentrations of any reacting species. Because the concentrations are unchanged, Q still equals Kc, and no shift occurs. Kc depends only on temperature, which has not changed. If instead the container were flexible (constant pressure), adding helium would dilute the partial pressures of reactants and products and could cause a shift, but that scenario is not described here.

For an endothermic reaction, heat acts as a reactant. Increasing temperature provides more energy, shifting equilibrium to the right to absorb it, increasing the concentrations of N2 and I2. This also increases the value of Kc. Choice A describes the behavior of an exothermic reaction. Choice C is incorrect: although NI3 is a solid and excluded from the Kc expression, it still participates in the equilibrium and can shift. Choice D is internally inconsistent — for an endothermic reaction, increasing temperature increases, not decreases, Kc.

Calculate Q: Q = [SO2]²[O2] / [SO3]² = (0.10)²(0.10) / (0.50)² = (0.010)(0.10) / 0.25 = 0.0010 / 0.25 = 0.0040. Since Q (0.0040) is less than Kc (0.25), the reaction has not yet reached equilibrium and must proceed forward to produce more SO2 and O2. Choice B is wrong because Q < Kc, not Q > Kc. Choice C is wrong because 0.0040 is far from 0.25. Choice D is wrong because Kc does not change — only the concentrations shift until Q equals the fixed Kc.

The relationship between Kp and Kc is Kp = Kc(RT)^Δn, where Δn is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants). Kp equals Kc only when Δn = 0. For H2 + I2 ⇌ 2HI: Δn = 2 − 2 = 0, so Kp = Kc. For Choice A: Δn = 2 − 4 = −2. For Choice C: Δn = 2 − 1 = +1. For Choice D: Δn = 2 − 3 = −1. In all other cases Kp ≠ Kc.

Kc = [HF]² / ([H2][F2]) = (0.90)² / ((0.30)(0.15)) = 0.81 / 0.045 = 18. Choice A (20) results from forgetting to square [HF] and computing [HF] / ([H2][F2]) = 0.90 / 0.045 = 20. Choice C (0.056) is the reciprocal of 18, which would be Kc for the reverse reaction. Choice D (0.050) results from computing ([H2][F2]) / [HF]², inverting the entire expression.

The relationship is Kp = Kc(RT)^Δn. Here Δn = 1 − 2 = −1. Solving for Kc: Kc = Kp / (RT)^Δn = Kp × (RT)^1 = 8.8 × (0.08206 × 298) = 8.8 × 24.45 ≈ 215. Choice A incorrectly assumes Kp = Kc. Choice C (0.36) results from dividing instead of multiplying: Kp / RT = 8.8 / 24.45 ≈ 0.36, which is what you get if you misapply the exponent sign. Choice D is far too small and has no basis in the calculation.

Decreasing volume increases pressure (concentrations). By Le Chatelier's principle, the system shifts to reduce pressure by decreasing the total moles of gas. The reactant side has 2 + 1 = 3 moles of gas and the product side has 2 moles, so the equilibrium shifts right. Kc depends only on temperature, which has not changed, so it remains constant. Choice A correctly identifies the shift direction but wrongly states Kc increases. Choice B misapplies Le Chatelier's logic. Choice D has the shift direction reversed.

Setting up the ICE table with x as the decrease in [H2] and [I2]: at equilibrium, [H2] = [I2] = 1.00 − x and [HI] = 2x. Kc = (2x)² / (1.00 − x)² = 55.3. Taking the square root of both sides: 2x / (1.00 − x) = 7.44. Solving: 2x = 7.44 − 7.44x, so 9.44x = 7.44, giving x = 0.788 M. Therefore [HI] = 2(0.788) = 1.58 M. Choice A (0.21 M) is the equilibrium concentration of H2 (1.00 − 0.79). Choice B (0.79 M) is the value of x, not [HI]. Choice D neglects the stoichiometric factor of 2.

Calculate Q: Q = [PCl3][Cl2] / [PCl5] = (0.10)(0.25) / 0.50 = 0.025 / 0.50 = 0.050. Since Q (0.050) > Kc (0.042), the reaction will shift in the reverse direction, consuming PCl3 and Cl2 and forming more PCl5, until Q decreases to equal Kc. Choice A is wrong because Q > Kc, not Q < Kc. Choice C is incorrect because even a small difference between Q and Kc means the system is not at equilibrium. Choice D uses faulty reasoning — the concentrations of reactants versus products alone do not determine direction without comparing Q to Kc.

When two reactions are added together, their equilibrium constants are multiplied. A → B → C combines the two steps, so K_overall = K1 × K2 = 0.10 × 0.30 = 0.030. Choice A (0.40) incorrectly adds the two K values — equilibrium constants are multiplied, not added, when reactions are summed. Choice B (3.3) is 1/(K1 × K2), the reciprocal of the correct answer. Choice D (0.0030) might arise from a place-value error or from incorrectly squaring one of the constants.

Using the Henderson–Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). First, pKa = −log(1.8 × 10⁻⁵) = 4.74. Then, log(0.30/0.20) = log(1.5) = 0.18. So pH = 4.74 + 0.18 = 4.92. Choice B (4.74) is just the pKa, forgetting to include the log ratio term. Choice A (4.56) results from subtracting the log ratio instead of adding it (4.74 − 0.18), which would apply if [HA] > [A⁻]. Choice D (5.10) is a calculation error with an incorrect log value.

Adding N2 increases its concentration, causing Q < Kc. The system responds by shifting right (forward) to consume some of the added N2 and produce more NH3, restoring Q = Kc. Importantly, Kc depends only on temperature, which has not changed, so Kc remains constant. Choice A incorrectly states Kc changes. Choice C incorrectly states Kc decreases and misstates what equilibrium restores — the equilibrium restores the ratio defined by Kc, not the exact original concentration of N2. Choice D applies Le Chatelier's principle incorrectly; adding a reactant shifts equilibrium right, not left.

In 0.10 M NaCl, the Cl⁻ concentration is already 0.10 M (the common ion). AgCl dissolves by: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). If s is the molar solubility, then [Ag⁺] = s and [Cl⁻] ≈ 0.10 M (since s is tiny). Setting up Ksp = [Ag⁺][Cl⁻] = s × 0.10 = 1.8 × 10⁻¹⁰, gives s = 1.8 × 10⁻⁹ M. Choice A (1.3 × 10⁻⁵ M) is the solubility in pure water (√Ksp), ignoring the common ion. Choice C equals Ksp itself, which incorrectly treats Ksp as the solubility. Choice D results from an additional erroneous division step.

For the equilibrium NH3 + H2O ⇌ NH4⁺ + OH⁻, Kb = x² / (0.50 − x) ≈ x² / 0.50 = 1.8 × 10⁻⁵. Solving: x² = 9.0 × 10⁻⁶, so x = [OH⁻] = 3.0 × 10⁻³ M. The approximation check: (3.0 × 10⁻³ / 0.50) × 100% = 0.6%, which is well under 5% and valid. Choice A (9.0 × 10⁻⁶) is x² itself, forgetting to take the square root. Choice C (1.8 × 10⁻⁵) is just the value of Kb, not the equilibrium concentration. Choice D (6.0 × 10⁻³) is 2x, as if the concentration were doubled without justification.

Setting up the ICE table with x as the amount of SO2Cl2 that decomposes: Kc = x² / (0.500 − x) = 0.028. Rearranging: x² + 0.028x − 0.014 = 0. Using the quadratic formula: x = (−0.028 + √(0.000784 + 0.056)) / 2 = (−0.028 + √0.05678) / 2 = (−0.028 + 0.2383) / 2 = 0.1052 M. So [SO2] = [Cl2] ≈ 0.105 M and [SO2Cl2] = 0.395 M. Verification: (0.105)² / 0.395 = 0.011 / 0.395 = 0.028 ✓. Choice B uses x = C/3, ignoring K entirely. Choice C incorrectly sets x equal to Kc. Choice D applies an invalid small-x approximation that fails the 5% check.

The van't Hoff relationship shows that for an endothermic reaction (ΔH° > 0), Kc increases with increasing temperature because heat acts as a reactant — adding temperature shifts equilibrium toward products, raising Kc. Here Kc increases from 2.5 × 10² at 300 K to 1.5 × 10³ at 400 K, consistent with an endothermic reaction. Choice A confuses reaction rate with equilibrium; increased rate does not by itself indicate exothermicity. Choice C incorrectly states Kc decreases, when it actually increases. Choice D is wrong because Kc > 1 at any temperature simply indicates products are favored and reveals nothing about the sign of ΔH°.

To obtain B + D ⇌ C, reverse reaction 1 to get B + C ⇌ A with K = 1/K1 = 0.50, then add reaction 2 (A + D ⇌ 2C, K2 = 3.0). Summing: B + C + A + D ⇌ A + 2C. Canceling A from both sides and one C from each side gives B + D ⇌ C. The overall K = (1/K1) × K2 = 0.50 × 3.0 = 1.5. Choice A (6.0) results from multiplying K1 × K2 directly without reversing K1. Choice C (0.17) is 1/(K1 × K2) = 1/6. Choice D (0.67) results from dividing K2/K1 = 3.0/2.0 without accounting for the reversal correctly.

Precipitation begins when the ion product equals Ksp. For BaSO4: [Ba²⁺][SO4²⁻] = Ksp, so [SO4²⁻] = Ksp / [Ba²⁺] = 1.1 × 10⁻¹⁰ / 0.010 = 1.1 × 10⁻⁸ M. BaSO4 precipitates first because its Ksp is much smaller. SrSO4 begins to precipitate at [SO4²⁻] = 3.4 × 10⁻⁷ / 0.010 = 3.4 × 10⁻⁵ M, which is Choice B. Choice C (1.1 × 10⁻¹⁰) uses Ksp directly as the required [SO4²⁻], forgetting to divide by [Ba²⁺]. Choice D (3.4 × 10⁻⁷) is the Ksp of SrSO4 itself, not divided by [Sr²⁺].

Let x = the partial pressure of I2 that dissociates. At equilibrium: P(I2) = 1.00 − x and P(I) = 2x. Then Kp = (2x)² / (1.00 − x) = 4x² / (1.00 − x) = 0.21. Rearranging: 4x² + 0.21x − 0.21 = 0. Using the quadratic formula: x = (−0.21 + √(0.0441 + 3.36)) / 8 = (−0.21 + 1.845) / 8 = 1.635 / 8 ≈ 0.204. The fraction of I2 dissociated is x / 1.00 ≈ 0.20 (or 20%). Verification: 4(0.204)² / (0.796) = 0.166 / 0.796 ≈ 0.21 ✓. Choice A (0.10) neglects that Kp = (2x)²/(1−x) and uses a faulty linear approximation. Choices C and D overestimate x.

NH4Cl is a salt of a weak base (NH3) and a strong acid (HCl), so the solution is acidic. The relevant equilibrium is NH4⁺ ⇌ NH3 + H⁺ with Ka = Kw / Kb = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.56 × 10⁻¹⁰. Using the weak acid ICE approximation: [H⁺] = √(Ka × C) = √(5.56 × 10⁻¹⁰ × 0.10) = √(5.56 × 10⁻¹¹) = 7.46 × 10⁻⁶ M. pH = −log(7.46 × 10⁻⁶) ≈ 5.13. Choice B (8.87) equals 14 − 5.13, confusing pH with pOH. Choice C (9.13) is the pH of 0.10 M NH3 solution, treating the salt as a base. Choice D (7.00) incorrectly assumes the salt solution is neutral.

Kc is a function of temperature only — it does not change with pressure, volume, catalyst addition, or changes in reactant or product concentrations. For an exothermic reaction (ΔH° < 0), heat is a product. Decreasing temperature removes heat, shifting the equilibrium right and increasing the ratio of products to reactants, which means Kc increases. Choice A (changing pressure) affects equilibrium position for reactions with Δn ≠ 0 but does not change Kc; moreover, Δn = 0 here so pressure has no effect at all. Choice C (catalyst) increases both forward and reverse rates equally, leaving Kc unchanged. Choice D (adding CO) shifts equilibrium position temporarily but Kc is unaffected.

Kc is written as products over reactants, with each concentration raised to the power of its stoichiometric coefficient. For 2H2(g) + O2(g) ⇌ 2H2O(g), Kc = [H2O]² / ([H2]²[O2]). Choice B is the inverse, which is Kc for the reverse reaction. Choice C ignores the stoichiometric coefficients. Choice D incorrectly adds concentrations in the denominator rather than multiplying them.

A very small Kc (much less than 1) means the ratio of product concentrations to reactant concentrations at equilibrium is tiny, so reactants are strongly favored. Kc does not indicate reaction rate — a reaction can have a very small Kc but still proceed quickly. Kc ≈ 1 would indicate roughly equal concentrations of products and reactants for a simple 1:1 reaction.

Le Chatelier's principle states that a system at equilibrium will shift to partially counteract any applied stress. Adding a product increases Q above Kc, so the system shifts in the reverse direction, consuming some added product until Q decreases back to Kc. Critically, adding a product does not change Kc — the equilibrium constant depends only on temperature.

A catalyst lowers the activation energy for both the forward and reverse reactions equally, so equilibrium is reached more quickly but the equilibrium concentrations and the value of Kc are unchanged. If the system is already at equilibrium, adding a catalyst produces no observable change in concentrations. Kc depends only on temperature, not on catalysts.

For heterogeneous equilibria, pure solids and pure liquids are excluded from the equilibrium expression because their concentrations are essentially constant and are absorbed into the value of Kc. Both CaCO3 and CaO are pure solids, so only the gaseous CO2 appears: Kc = [CO2]. Choice A incorrectly includes the solid concentrations. Choices C and D incorrectly place [CO2] in the denominator.

When Q > Kc, the ratio of product concentrations to reactant concentrations is too high relative to the equilibrium value. The reaction shifts in the reverse direction, converting products back to reactants, until Q decreases to equal Kc. When Q < Kc, the forward reaction is favored. Kc is temperature-dependent and does not change simply because Q is larger.

Kp = Kc only when Δn = 0, making (RT)^0 = 1. For H2(g) + I2(g) ⇌ 2HI(g), Δn = 2 − (1+1) = 0, so Kp = Kc. For the other reactions: N2 + 3H2 ⇌ 2NH3 has Δn = 2−4 = −2; PCl5 ⇌ PCl3 + Cl2 has Δn = 2−1 = +1; 2SO3 ⇌ 2SO2 + O2 has Δn = 3−2 = +1. Only when the total moles of gaseous products equal the total moles of gaseous reactants do Kp and Kc coincide.

When a reaction is reversed, the new equilibrium constant is the reciprocal of the original: K_reverse = 1/K_forward = 1/(4.0 × 10²) = 2.5 × 10⁻³. The large Kc for the forward reaction confirms SO3 formation is strongly favored, while the small Kc for the reverse confirms SO3 decomposition is disfavored at this temperature. Equilibrium constants are never negative, eliminating choice C.

Kc = [NO2]² / [N2O4]. In a 1.0 L container, [NO2] = 0.10 M and [N2O4] = 0.50 M. Therefore Kc = (0.10)² / (0.50) = 0.010 / 0.50 = 0.020. A common mistake is forgetting to square the NO2 concentration, giving 0.10/0.50 = 0.20 (choice B). Another error is computing 2 × 0.10 / 0.50 = 0.40, incorrectly using the stoichiometric coefficient as a multiplier rather than an exponent.

For an exothermic reaction, heat can be treated as a product. Increasing temperature adds heat stress, which by Le Chatelier's principle shifts equilibrium toward reactants (reverse direction). Because products are less favored at the new temperature, K decreases. This is a key distinction: temperature is the only variable that changes K itself. Concentration or pressure changes shift the equilibrium position but leave K unchanged.

Decreasing volume increases pressure, and the system shifts toward the side with fewer moles of gas. Reactant side: CO (1 mol) + H2 (3 mol) = 4 moles of gas. Product side: CH4 (1 mol) + H2O (1 mol) = 2 moles of gas. Since products have fewer moles of gas, equilibrium shifts toward products. Choice C is a common error — students might count only two species per side and conclude 2 = 2, ignoring the coefficient of 3 on H2.

ICE table for HA ⇌ H⁺ + A⁻: initial [HA] = 0.10 M, [H⁺] = [A⁻] = 0. Equilibrium: [HA] ≈ 0.10, [H⁺] = [A⁻] = x. Ka = x²/0.10 = 1.8 × 10⁻⁴, so x² = 1.8 × 10⁻⁵, and x = 4.2 × 10⁻³ M. Choice A (1.8 × 10⁻⁵) is x² itself, confusing the intermediate calculation with the answer. Choice C (1.8 × 10⁻⁴) is just Ka. The small-x check: 4.2 × 10⁻³/0.10 = 4.2%, which is under the 5% threshold, validating the approximation.

Kc = 1.0 × 10⁻³⁰ means the equilibrium ratio [NO]²/([N2][O2]) is essentially zero — equilibrium overwhelmingly favors N2 and O2. This is a thermodynamic argument based on the equilibrium constant. Choice A is actually backwards: the reaction is kinetically very slow at room temperature (high activation energy), not fast. At high temperatures such as in lightning or combustion engines, K increases enough that measurable NO forms.

PbI2 ⇌ Pb²⁺ + 2I⁻, so Ksp = [Pb²⁺][I⁻]². With [I⁻] dominated by the added NaI at 0.10 M: [Pb²⁺] = Ksp / [I⁻]² = (9.8 × 10⁻⁹) / (0.10)² = 9.8 × 10⁻⁹ / 0.010 = 9.8 × 10⁻⁷ M. This common-ion effect dramatically lowers [Pb²⁺] compared to pure water (where [Pb²⁺] ≈ 1.3 × 10⁻³ M). Choice B incorrectly divides by [I⁻] rather than [I⁻]², forgetting the square in the Ksp expression.

Q = [C] / ([A][B]²) = 0.10 / (1.0 × (0.50)²) = 0.10 / 0.25 = 0.40. Since Q (0.40) > Kc (0.25), the product concentration is too high relative to equilibrium — the reaction shifts in reverse to convert C back to A and B. Choice B results from forgetting to square [B]: 0.10 / (1.0 × 0.50) = 0.20. Always square [B] because its stoichiometric coefficient is 2.

First convert moles to molar concentrations using V = 2.0 L: [H2] = 0.10 M, [F2] = 0.050 M, [HF] = 0.40 M. Then Kc = [HF]² / ([H2][F2]) = (0.40)² / (0.10 × 0.050) = 0.16 / 0.0050 = 32. Choice A (16) results from using moles directly instead of molarities, then squaring. The large Kc value confirms HF formation is strongly favored at this temperature.

At constant volume and temperature, adding an inert gas increases total pressure but does not change the partial pressures or molar concentrations of the reacting gases. Because Q is determined by the partial pressures (or concentrations) of reactants and products only, Q remains equal to Kc and no shift occurs. This contrasts with adding an inert gas at constant total pressure, where partial pressures of all reacting gases would drop, causing a shift toward more moles of gas. Kp changes only with temperature.

BaSO4 ⇌ Ba²⁺ + SO4²⁻. Let s = molar solubility; then [Ba²⁺] = s and [SO4²⁻] = s. Ksp = s² = 1.1 × 10⁻¹⁰, so s = √(1.1 × 10⁻¹⁰) = 1.05 × 10⁻⁵ M ≈ 1.0 × 10⁻⁵ M. Choice A (1.1 × 10⁻¹⁰) is the value of Ksp itself — the most common error, where students skip taking the square root. Choice B incorrectly divides Ksp by 2 rather than taking the square root.

ICE table: initial [HI] = 2.00 M, [H2] = [I2] = 0. At equilibrium: [HI] = 2.00−2x, [H2] = [I2] = x. Kc = x²/(2.00−2x)² = 0.111. Taking the square root of both sides gives x/(2.00−2x) = 0.333. Solving: x = 0.333(2.00−2x) = 0.666−0.666x, so 1.666x = 0.666, x = 0.40 M. Verification: (0.40)²/(1.20)² = 0.16/1.44 = 0.111 ✓. Recognizing that the Kc expression is a perfect square allows a square-root shortcut that avoids solving a quadratic equation.

Two factors must be optimized. First, the reaction is exothermic (ΔH° < 0), so lower temperature shifts equilibrium toward N2O4 and increases K. Second, 2 moles of gas (NO2) produce 1 mole of gas (N2O4), so higher pressure shifts equilibrium toward the side with fewer gas moles — the products. Therefore low temperature and high pressure both favor N2O4. High temperature would decrease K and shift toward NO2. Low pressure would favor the side with more moles of gas, which is also NO2.

Using the Henderson-Hasselbalch equation with NH4⁺ as the weak acid: pH = pKa + log([base]/[acid]) = pKa(NH4⁺) + log([NH3]/[NH4⁺]). pKa = −log(5.6 × 10⁻¹⁰) = 9.25. Ratio [NH3]/[NH4⁺] = 0.30/0.20 = 1.5. pH = 9.25 + log(1.5) = 9.25 + 0.18 = 9.43. Choice D (9.25) is pKa alone, which applies only when [NH3] = [NH4⁺]. Since there is more base than acid, pH exceeds pKa. Choice A results from inverting the log ratio: log(0.20/0.30) = −0.18, giving 9.25 − 0.18 = 9.07.

At constant T and V, pressure is proportional to total moles. Initial moles = 1.00. Let x mol of A decompose: moles A = 1.00−x, moles B = 2x, total = 1.00+x. Pressure ratio: (1.00+x)/1.00 = 1.50, so x = 0.50. Equilibrium concentrations in 1.00 L: [A] = 0.50 M, [B] = 1.00 M. Kc = [B]²/[A] = (1.00)²/0.50 = 2.0 M. A common error is writing moles B = x instead of 2x, which gives x = 0.50, [B] = 0.50, and Kc = 0.50 — forgetting the stoichiometric coefficient of 2 for B.

ICE table for HA ⇌ H⁺ + A⁻: Ka = x²/(0.100−x) ≈ x²/0.100 = 2.0 × 10⁻⁵. Solving: x² = 2.0 × 10⁻⁶, x = 1.41 × 10⁻³ M. Percent ionization = (1.41 × 10⁻³/0.100) × 100% = 1.41% ≈ 1.4%. The small-x approximation is valid since 1.4% is below the 5% threshold. Choice A (0.020%) treats Ka directly as a percent. Choice C (2.0%) incorrectly uses Ka/[HA] = 2.0 × 10⁻⁴ as a decimal fraction (0.020%) — or mistakes Ka/[HA] for the degree of ionization without taking the square root.

Adding chemical equations corresponds to multiplying their equilibrium constants. The net reaction is obtained by adding reactions (1) and (2): C appears as a product in (1) and a reactant in (2), so it cancels in the sum. K_net = K1 × K2 = 4.0 × 0.50 = 2.0. Choice A (4.5) results from incorrectly adding K1 + K2. Choice C (8.0) results from multiplying when one equation should first be reversed — but here neither reversal is needed because the intermediates cancel directly. Multiplying rather than adding is the fundamental rule when combining equilibria.

Ag2CrO4 ⇌ 2Ag⁺ + CrO4²⁻. Let s = molar solubility. [Ag⁺] = 2s, [CrO4²⁻] = 0.10 + s ≈ 0.10 M (common ion dominates). Ksp = (2s)²(0.10) = 4s²(0.10) = 0.40s² = 1.12 × 10⁻¹². Solving: s² = 2.80 × 10⁻¹², s = 1.67 × 10⁻⁶ M ≈ 1.7 × 10⁻⁶ M. Choice A (5.3 × 10⁻⁶) results from writing Ksp = (2s)(0.10) instead of (2s)²(0.10), ignoring the exponent on Ag⁺. Choice D (1.1 × 10⁻¹¹) simply computes Ksp/[CrO4²⁻] without accounting for the 2:1 stoichiometry of Ag⁺.

Kc is defined as the ratio of product concentrations to reactant concentrations (each raised to stoichiometric powers). A large Kc means the numerator is much larger than the denominator, so products predominate at equilibrium. Kc says nothing about reaction rate or whether the reaction is exothermic — those are separate concepts.

Removing a reactant decreases its concentration, making Q > K (the denominator of the equilibrium expression shrinks). To restore equilibrium the system shifts in reverse — toward reactants — to replenish what was removed. K itself is unaffected by concentration changes; only temperature changes K.

Pure solids and pure liquids are omitted from equilibrium expressions because their concentrations do not change — they are incorporated into the value of K itself. Only species whose concentrations (or partial pressures) can vary appear in the expression. Here, only gaseous CO2 is included, giving Kc = [CO2].

Q (the reaction quotient) uses the same ratio of concentrations as K but is evaluated using the actual concentrations at any moment in time. When Q = K the system is at equilibrium. When Q < K the forward reaction is favored; when Q > K the reverse reaction is favored. K is strictly the value of Q at equilibrium.

A catalyst provides an alternate pathway with lower activation energy for both the forward and reverse reactions equally. This speeds up the rate at which equilibrium is reached but does not change the equilibrium constant or the equilibrium concentrations. Only a change in temperature alters K.

Reversing a reaction swaps the numerator and denominator of the equilibrium expression, so the new K is the reciprocal of the original. If Kc(forward) = x, then Kc(reverse) = 1/x. This is a fundamental property of equilibrium constants and is used when combining reactions via Hess's law-style manipulation.

ICE stands for Initial (the concentrations or pressures before any net reaction occurs), Change (the amount each species gains or loses as the system moves toward equilibrium, expressed as multiples of x based on stoichiometry), and Equilibrium (the final concentrations substituted into the K expression to solve for x).

Adding an inert gas at constant volume does not change the partial pressures or molar concentrations of any reactive species, so Q is unchanged and the equilibrium position does not shift. In contrast, adding H2 lowers Q (forward shift), removing HI raises Q (reverse shift), and changing temperature alters K itself.

Kc = [PCl3][Cl2] / [PCl5] = (0.0500)(0.0500) / (0.200) = 0.00250 / 0.200 = 0.0125. A common error is to invert the expression (giving 80.0) or to place PCl5 in the numerator. Because Kc < 1, reactants are slightly favored at this temperature.

Q = [SO3][NO] / ([SO2][NO2]) = (0)(0) / (0.800)(0.800) = 0. Since Q < K (0 < 3.75), the numerator is too small relative to the denominator, so the forward reaction is favored and the system produces SO3 and NO until Q reaches 3.75. Q = infinity would require zero reactant concentrations.

delta-n = moles of gaseous products minus moles of gaseous reactants = 2 - (2 + 1) = -1. Since RT > 1 at 500 K, (RT)^(-1) is a fraction less than 1, making Kp = Kc x (RT)^(-1) < Kc. Whenever delta-n is negative, Kp is smaller than Kc; whenever delta-n is positive, Kp is larger.

Kc = [NH3]^2 / ([N2][H2]^3) = (0.400)^2 / ((0.100)[H2]^3) = 4.00. Solving: 0.160 / (0.100 x [H2]^3) = 4.00, so [H2]^3 = 0.160 / 0.400 = 0.400, giving [H2] = (0.400)^(1/3) = 0.737 M. A common error is solving only [H2]^3 = 0.400 and reporting 0.400 M without taking the cube root.

Set up the ICE table: [A] = [B] = 1.00 - x, [C] = 2x. Kc = (2x)^2 / ((1.00 - x)^2) = 4.00. Taking the square root of both sides: 2x / (1.00 - x) = 2.00. Solving: 2x = 2.00 - 2x, so 4x = 2.00 and x = 0.500 M. Therefore [C] = 2(0.500) = 1.00 M. The perfect-square shortcut is valid here because Kc is a perfect square.

For $HCN \rightleftharpoons H^+ + CN^-$, \(K_a = x^2 / (0.10 - x) \approx x^2 / 0.10 = 6.2 \times 10^{-10}\). Solving: \(x^2 = 6.2 \times 10^{-11}\), \(x = 7.9 \times 10^{-6}\) M. The approximation is valid since \(x / 0.10 = 0.0079 = 0.79\%\), well under 5%. Choice B (\(6.2 \times 10^{-9}\)) is wrong because it equals \(K_a \times 0.10\), not the square root of that product.

Only a change in temperature alters the equilibrium constant K. For an endothermic reaction, heat acts as a reactant; increasing temperature shifts equilibrium toward products and increases Kc. Adding N2O4 or decreasing volume shifts the equilibrium position but leaves K unchanged. A catalyst speeds attainment of equilibrium without affecting K.

Adding NaF introduces F-, a product of HF dissociation. By Le Chatelier's principle, this extra F- shifts the equilibrium HF ⇌ H+ + F- to the left, reducing dissociation and lowering [H+]. Numerically, [H+] ≈ Ka x [HF]/[F-] ≈ 7.2 x 10^-4 M, compared to about 8.5 x 10^-3 M in pure 0.10 M HF. Ka is constant but the equilibrium position shifts.

For this heterogeneous equilibrium, Kp = P(CO2). The reaction quotient Q = P(CO2) = 0.100 atm. Since Q < Kp (0.100 < 0.236), more products are needed, so CaCO3 decomposes (forward reaction) until P(CO2) reaches 0.236 atm. Choice D is incorrect because CO2, as the only gaseous species, is the sole component of the equilibrium expression.

ICE table: \([A] = 2.00 - 2x\), \([B] = x\), \([C] = x\). \(K_c = x^2 / (2.00 - 2x)^2 = 1.00 \times 10^{-4}\). Taking the square root: \(x / (2.00 - 2x) = 0.0100\). Solving: \(x = 0.0200 - 0.0200x\), giving \(1.0200x = 0.0200\), so \(x = 0.01961\) M. Therefore \([A] = 2.00 - 2(0.01961) = 1.961\) M \(\approx 1.96\) M. Choice D (1.98 M) results from incorrectly using the approximation without the factor of 2 in \((2.00 - 2x)\).

To obtain the target reaction, use reaction (1) as written (H2 + S → H2S, K1) and reverse reaction (2) (SO2 → S + O2, K = 1/K2). Adding these gives H2 + SO2 → H2S + O2, so Knet = K1 x (1/K2) = 1.0 x 10^7 / 4.2 x 10^52 = 2.4 x 10^-46. The very small K confirms this reaction strongly favors reactants — combustion of H2S to SO2 is thermodynamically favorable in the forward direction.

ICE table for \(B + H_2O \rightleftharpoons BH^+ + OH^-\): \(K_b = x^2 / (0.200 - x) \approx x^2 / 0.200 = 4.5 \times 10^{-5}\). Solving: \(x^2 = 9.0 \times 10^{-6}\), \(x = [OH^-] = 3.0 \times 10^{-3}\) M. $pOH = -\log(3.0 \times 10^{-3}) = 2.52$. \(pH = 14.00 - 2.52 = 11.48\). Choice A (2.52) is a common error — it gives pOH rather than pH. Choice C results from using an incorrect approximation method.

ICE table: \([H_2] = [CO_2] = 0.500 - x\), \([H_2O] = [CO] = x\). \(K_c = x^2 / (0.500 - x)^2 = 0.771\). Taking the square root: \(x / (0.500 - x) = 0.8781\). Solving: \(x = 0.4390 - 0.8781x\), \(1.8781x = 0.4390\), \(x = 0.2337\) M. \([CO] \approx 0.234\) M. Choice B (0.266 M) represents \([CO_2]\) at equilibrium, a common mix-up of which row to report.

At pH 10.00, [OH-] = 10^-4 M (fixed by the buffer). Mg(OH)2 dissolves as Mg^2+ + 2OH-. Since [OH-] is held constant by the buffer, Ksp = s x (10^-4)^2 = s x 10^-8. Solving: s = 5.61 x 10^-12 / 10^-8 = 5.61 x 10^-4 M. Choice A (1.12 x 10^-4 M) is the solubility in pure water — because pure water gives a higher [OH-] (~2.24 x 10^-4 M), the buffered pH 10 solution has lower [OH-] and actually dissolves more Mg(OH)2.

To increase [PCl5], equilibrium must shift forward. (1) Decreasing temperature favors the exothermic forward reaction, increasing K. (2) Decreasing volume increases total pressure, which favors the side with fewer moles of gas: 2 mol reactants (PCl3 + Cl2) become 1 mol product (PCl5), so compression drives the forward reaction. Both changes reinforce each other. Increasing temperature (choices A and D) would shift equilibrium backward for this exothermic reaction.

Because Kc is enormous, assume the reaction goes to completion first: 0.010 M Ag+ reacts with 0.020 M NH3, leaving [NH3] ≈ 0.98 M and [Ag(NH3)2+] ≈ 0.010 M. Then apply the reverse equilibrium: K' = 1/(3.4 x 10^8) = 2.94 x 10^-9. Solving K' = [Ag+](0.98)^2 / (0.010) gives [Ag+] = 2.94 x 10^-9 x 0.010 / 0.9604 ≈ 3.1 x 10^-11 M. The very low free Ag+ illustrates why complexation dramatically suppresses free metal ion concentration.

Initial $[NOCl] = 1.00$ mol / 2.00 L \(= 0.500\) M. ICE: $[NOCl] = 0.500 - 2x$, \([NO] = 2x\), \([Cl_2] = x\). \(K_c = (2x)^2(x) / (0.500 - 2x)^2\). Since \(K_c\) is small, approximate \((0.500 - 2x) \approx 0.500\): \((4x^2)(x) / (0.500)^2 = 4x^3 / 0.250 = 1.6 \times 10^{-5}\). Then \(x^3 = 1.0 \times 10^{-6}\), \(x = 0.0100\) M. Moles of NOCl dissociated per liter = \(2x = 0.0200\) M. Percent dissociation = \((0.0200 / 0.500) \times 100\% = 4.0\%\).

Equilibrium is dynamic, not static. The forward and reverse reactions continue, but at equal rates, so concentrations remain constant. Choice A is incorrect because reactions never stop; they continue at equal rates. Choice B is a common misconception — concentrations are constant but not necessarily equal to each other.

Kc is written as products over reactants, with each concentration raised to the power of its stoichiometric coefficient. Products H2 (coefficient 2) and O2 (coefficient 1) appear in the numerator; reactant H2O (coefficient 2) appears in the denominator. Choice B is the expression for the reverse reaction. Choice C omits the stoichiometric exponents.

By Le Chatelier's principle, adding a reactant increases the reaction quotient Q below Kc, so the system responds by consuming the added N2 and shifting right to produce more NO. Importantly, Kc itself does not change when a concentration is altered at constant temperature — choices C and D are incorrect because Kc is only affected by temperature changes.

A catalyst increases the rates of both the forward and reverse reactions equally, allowing the system to reach equilibrium faster without changing the equilibrium position or the value of Kc. All other choices do shift equilibrium: temperature affects K, adding reactant increases Q, and volume changes alter partial pressures of gases.

The equation Kp = Kc(RT)^Δn shows that when Δn = 0 (equal moles of gaseous products and reactants), (RT)^0 = 1, so Kp = Kc. Any nonzero Δn means the two constants differ. For example, the reaction H2(g) + F2(g) ⇌ 2HF(g) has Δn = 2 - 2 = 0, so Kp = Kc for this reaction.

For heterogeneous equilibria, pure solids and pure liquids are excluded from the equilibrium expression because their concentrations are constant and incorporated into the value of Kc. Fe3O4(s) and Fe(s) are both solids and are omitted. Only the gas-phase species H2O(g) and H2(g) appear in Kc. Choice A incorrectly includes both solids.

When Qc > Kc, there are too many products relative to reactants compared to the equilibrium condition. The system shifts in reverse (left) to consume products and form reactants until Qc decreases to equal Kc. Choice A is wrong because the forward reaction would make Qc even larger. Choice D is wrong because Kc only changes with temperature.

When a reaction is reversed, the new equilibrium constant is the reciprocal of the original. Kc(reverse) = 1 / Kc(forward) = 1 / 3.92 = 0.255. Choice A would be correct only if reversing the reaction had no effect. Choice D results from squaring Kc rather than inverting it, which applies only when an equation is multiplied by a coefficient, not simply reversed.

If 0.40 mol of A remains from 1.00 mol, then x = 0.60 mol of A reacted, producing 0.60 mol B and 0.60 mol C. In the 1.00 L container: [A] = 0.40 M, [B] = 0.60 M, [C] = 0.60 M. Kc = [B][C]/[A] = (0.60)(0.60)/(0.40) = 0.36/0.40 = 0.90. Choice A (0.36) is the numerator alone without dividing by [A]. Choice D would result from Kc = [B]²/[A], misusing stoichiometry.

Calculate Qc = [NO]² / ([N2][O2]) = (0.10)² / (0.50)(0.50) = 0.01 / 0.25 = 0.04. Since Qc (0.04) < Kc (0.10), the numerator is too small relative to equilibrium — the system must produce more NO by shifting in the forward direction. Choice B confuses the direction: when Qc < Kc, the reaction proceeds forward, not in reverse.

Removing a product (SO3) decreases Qc below Kc, causing the equilibrium to shift right to restore balance. Choice A is wrong because increasing temperature shifts exothermic equilibria to the left. Choice B is wrong because decreasing pressure favors the side with more moles of gas — the reactant side (3 mol vs 2 mol). Choice C has no effect because an inert gas at constant volume does not change the partial pressures of reacting gases.

Setting up an ICE table: [H2] = [I2] = 1.00 - x and [HI] = 2x at equilibrium. Kc = (2x)² / (1-x)² = 49.0. Taking the square root of both sides: 2x / (1 - x) = 7. Solving: 2x = 7 - 7x, so 9x = 7, x = 0.778 M. Therefore [HI] = 2x = 1.56 M. Choice A (0.778) is the value of x, not [HI]. This problem is simplified because Kc is a perfect square, eliminating the need for the quadratic formula.

The NaCl solution already contains Cl-, which is a common ion shared with AgCl's dissolution equilibrium. The added Cl- shifts the dissolution equilibrium left (AgCl dissolves less), decreasing solubility — this is the common ion effect. Choice C is partially true (Ksp is constant at fixed temperature) but fails to recognize that constant Ksp with higher [Cl-] means lower [Ag+], hence lower solubility.

Decreasing the volume increases pressure, which shifts equilibrium toward the side with fewer moles of gas. The left side has 1 mol of gas (PCl5) and the right side has 2 mol (PCl3 + Cl2), so the system shifts left. Choice A is wrong: increasing temperature favors the endothermic (forward) direction, producing more PCl3. Choice B shifts right by removing product. Choice D shifts right by adding reactant.

Reversing a reaction takes the reciprocal of its equilibrium constant. Kc(reverse) = 1/170 = 5.88 × 10^-3. Choice A would apply only if the reaction were rewritten identically. Choice B is the square root of 170 (which would apply if the equation were multiplied by 1/2, not reversed). Choice D is 2 × 170, which has no mathematical basis.

Using the Ka expression: Ka = [H+][CH3COO-] / [CH3COOH], rearranging gives [H+] = Ka × [CH3COOH] / [CH3COO-] = (1.8 × 10^-5)(0.20) / (0.10) = 3.6 × 10^-5 M. Choice A would be correct if the acid and conjugate base concentrations were equal. Choice C results from dividing Ka by 2 instead of multiplying by the [acid]/[base] ratio. This is the Henderson-Hasselbalch approach applied in Kc form.

When temperature increases from 300 K to 400 K, Kc decreases from 2.0 to 0.50. A decrease in Kc with increasing temperature means the equilibrium shifts left at higher temperature, which is consistent with an exothermic forward reaction (products are favored at low temperature). For an endothermic reaction, Kc would increase with temperature. Choice C is incorrect: temperature always affects Kc.

Kc = [CO2][H2] / ([CO][H2O]) = (0.0387)(0.0387) / (0.0613)(0.0613) = (0.0387/0.0613)² = (0.631)² = 0.398. Choice C (0.631) is the ratio of concentrations without squaring — a common error when the stoichiometry is 1:1:1:1 and students forget to multiply both numerator terms. Choice B (2.51) is 1/0.398, the Kc for the reverse reaction.

Initial [HI] = 0.600/2.00 = 0.300 M. ICE table: [HI] = 0.300 - 2x, [H2] = [I2] = x. Kc = x²/(0.300-2x)². Taking the square root: x/(0.300-2x) = √0.0160 = 0.1265. Solving: x = 0.1265(0.300-2x) = 0.03795 - 0.2530x, so 1.2530x = 0.03795, x = 0.0303 M. [HI] = 0.300 - 2(0.0303) = 0.239 M. Choice B (0.261) is 0.300 - x (using x instead of 2x), a common ICE stoichiometry error.

To obtain CO + H2O → CO2 + H2, use reaction (1) forward and reaction (2) in reverse: (1) CO + ½O2 → CO2, K = Kc1; (2 reversed) H2O → H2 + ½O2, K = 1/Kc2. Adding these cancels ½O2. When reactions are added, their K values are multiplied: Kc = Kc1 × (1/Kc2) = Kc1/Kc2. Choice B would be correct if both reactions were used in the forward direction, but reversing reaction 2 requires taking its reciprocal.

The partial pressure of N2O4 = total pressure minus partial pressure of NO2 = 1.50 - 0.80 = 0.70 atm. Kp = (P_NO2)² / (P_N2O4) = (0.80)² / 0.70 = 0.64 / 0.70 = 0.914. Choice A (0.640) is just (P_NO2)² without dividing by P_N2O4. Choice C would result from inverting the expression. Always identify all partial pressures before writing the Kp expression.

PbF2 ⇌ Pb²+ + 2F-. Let s = molar solubility; [Pb²+] = s and [F-] = 0.10 + 2s ≈ 0.10 M (since s is expected to be very small in the presence of the common ion). Ksp = s(0.10)² = s(0.01) = 3.3 × 10^-8, so s = 3.3 × 10^-6 M. Choice A (2.0 × 10^-3 M) is the solubility in pure water, calculated without the common ion. The common ion reduces solubility by roughly three orders of magnitude.

Qc = [C][D] / ([A][B]²) = (0.100)(0.300) / (0.200)(0.400)² = 0.030 / (0.200 × 0.160) = 0.030 / 0.032 = 0.938. Since Qc (0.938) > Kc (0.500), there are too many products, so the reaction shifts in reverse (left). Choice C uses Qc = 0.375, which results from forgetting to square [B]: 0.030 / (0.200)(0.400) = 0.375. This is a common error when applying the reaction quotient expression.

The approximation x ≈ √(Ka × C) = √(1.0 × 10^-6) = 1.0 × 10^-3 M gives percent dissociation = 10.0%, which exceeds the 5% threshold, so the quadratic formula is required. Solving x² + (1.0 × 10^-4)x - 1.0 × 10^-6 = 0 gives x = 9.51 × 10^-4 M, for a percent dissociation of 9.5%. Choice A (10.0%) is the answer obtained by the invalid approximation. The quadratic correction lowers the result because the denominator (0.0100 - x) is not negligible.

When volume is halved, all concentrations double. Substituting 2[A], 2[B], 2[C] into Q: Q_new = (2[B])(2[C])³ / (2[A])² = 2[B] × 8[C]³ / 4[A]² = (16/4) × [B][C]³/[A]² = 4 × Kc = 4 × 27 = 108. Since Qc (108) > Kc (27), the system shifts left. This is consistent with Le Chatelier's principle: increasing pressure favors the side with fewer moles of gas (left side has 2 mol, right has 4 mol). Choice D uses a factor of 2 instead of 4, an error from not accounting for the cubic exponent on [C].

The equilibrium constant expression is always written as products over reactants, with each concentration raised to the power of its stoichiometric coefficient. Choice A inverts the expression (reactants over products). Choice C incorrectly multiplies concentrations by coefficients instead of using them as exponents. Choice D ignores the stoichiometric coefficients entirely.

Pure solids and pure liquids are excluded from equilibrium expressions because their concentrations are essentially constant and are absorbed into the value of K. Aqueous species, dissolved ions, and gases are always included because their concentrations or partial pressures do change. The partial pressure of a gas relative to 1 atm is not a criterion for exclusion.

A catalyst lowers the activation energy for both the forward and reverse reactions equally, so both rates increase by the same factor. Because Q still equals K after the catalyst is added, no shift occurs. Kc depends only on temperature, not on the presence of a catalyst. Choices A, B, and D incorrectly assume a catalyst favors one direction.

When Q < K, the ratio of products to reactants is less than the equilibrium ratio. The reaction must proceed forward, producing more products and consuming reactants, until Q increases to equal K. Choice B is incorrect because reverse shift would lower Q further. Choice D is wrong because K depends only on temperature and never changes to match Q.

Reversing a reaction inverts its equilibrium constant expression. If Kforward = 0.25, then Kreverse = 1 / 0.25 = 4.0. Choice A incorrectly squares K. Choice D takes the square root. Choice B leaves K unchanged, which would only be correct if the reaction were written identically.

Pure solids (CaCO3 and CaO) are excluded from the equilibrium expression because their molar concentrations are constant. Only the concentration of the gas-phase product CO2 appears. Choices A, B, and D incorrectly include one or both solids in the expression.

Adding product raises Q above K. To restore equilibrium (Q = K), the system shifts in reverse, consuming the excess product and forming more reactants. Choice B is incorrect because a forward shift would produce even more product, raising Q further. Choice D is wrong because equilibrium is disturbed by any change in concentration, not only temperature.

Kc = [NO]^2 / ([N2][O2]). A value of 1.0 × 10^-30 means the numerator (products) is vanishingly small compared to the denominator (reactants) at equilibrium — essentially no NO is formed. Choice A confuses the magnitude of K with reaction rate; K and rate are independent. Choice C is the opposite conclusion. Choice D is incorrect because the reaction is possible, just highly unfavorable.

The relationship is Kp = Kc(RT)^(delta n), where delta n = moles of gaseous products minus moles of gaseous reactants = 2 - 3 = -1. Rearranging: Kc = Kp / (RT)^(delta n) = Kp × (RT)^1 = 3.4 × (0.08206 × 1000) = 3.4 × 82.06 = 279 ≈ 2.8 × 10^2. Choice A ignores the conversion entirely. Choice C incorrectly divides Kp by RT instead of multiplying.

Halving the volume doubles all concentrations. The new Q = (2[N2O4]) / (2[NO2])^2 = 2[N2O4] / 4[NO2]^2 = (1/2)(K) < K. Because Q < K, the system shifts in the forward direction, producing more N2O4 (the side with fewer moles of gas) to re-establish equilibrium. K itself does not change because temperature is constant.

Setting up an ICE table with \(x\) as the moles/L reacted: \(K_c = (2x)^2 / (0.10 - x)^2 = 1.0 \times 10^4\). Taking the square root of both sides: \(2x / (0.10 - x) = 100\). Solving: \(2x = 10 - 100x\), so \(102x = 10\), giving \(x = 0.0980\) M. Therefore $[HBr] = 2x = 0.196$ M. Choice B (0.126 M) comes from incorrectly ignoring \(x\) in the denominator, giving \(x = \sqrt{K_c \times 0.10^2 / 4} = 0.050\), then $[HBr] = 0.10$... actually let me fix: \(\sqrt{K_c} \times 0.10 / 2\) gives a different wrong path. Choice D incorrectly assumes complete reaction.

NaCl provides a common ion (Cl^-) that suppresses AgCl solubility. Let s = molar solubility of AgCl in the NaCl solution. Then [Ag+] = s and [Cl^-] = 0.10 + s ≈ 0.10 M (since s is very small). Ksp = s × 0.10 = 1.8 × 10^-10, so s = 1.8 × 10^-9 M. This is far less than the 1.34 × 10^-5 M in pure water, demonstrating the common ion effect. Choice C confuses Ksp itself with molar solubility.

At constant volume, adding an inert gas increases total pressure but does not change the concentrations or partial pressures of H2, I2, or HI. Since Q = K still holds, no shift occurs. Choice A confuses total pressure with the partial pressures of reactants and products. Choice D is wrong because Kc and Kp depend only on temperature, not on the presence of inert species.

The target reaction is the reverse of the original reaction, divided by 2. Reversing the reaction gives $K_{rev} = 1 / (6.0 \times 10^{-2}) = 16.7$. Halving the reaction raises \(K\) to the power of \(1/2\), so $K_{new} = \sqrt{16.7} = 4.08 \approx 4.1$. Choice A stops at reversal without halving. Choice C takes the square root of the original \(K\) without reversing. Choice D leaves \(K\) unchanged.

[H+] = 10^-2.85 = 1.41 × 10^-3 M. Because HA ⇌ H+ + A^-, [A^-] = [H+] = 1.41 × 10^-3 M, and [HA] = 0.20 - 1.41 × 10^-3 = 0.1986 M. Ka = (1.41 × 10^-3)^2 / 0.1986 = 1.99 × 10^-6 / 0.1986 ≈ 1.0 × 10^-5. Choice C simply reports [H+] as Ka. Choice B incorrectly uses [H+] / [HA]_initial, which is not the Ka expression.

For an endothermic reaction, heat can be thought of as a reactant. Increasing temperature adds heat, which shifts the equilibrium toward products (right) to absorb the added energy. This increases the ratio of product to reactant concentrations, so Kc increases. Choice C is wrong because a shift in equilibrium position always corresponds to a change in K when caused by temperature. Choice A describes an exothermic reaction.

Precipitation begins when the ion product exceeds Ksp. For BaSO4: precipitation starts when [SO4^2-] > Ksp / [Ba^2+] = 1.1 × 10^-10 / 0.010 = 1.1 × 10^-8 M. For SrSO4: precipitation starts when [SO4^2-] > 3.4 × 10^-7 / 0.010 = 3.4 × 10^-5 M. BaSO4 precipitates at a much lower [SO4^2-], so Ba^2+ is removed first. Choice D confuses Ksp with the required [SO4^2-], omitting division by [Ba^2+].

Initial $[PCl_5] = 2.00 / 5.00 = 0.400$ M. ICE table: \(K_c = x^2 / (0.400 - x) = 0.0400\). Rearranging: \(x^2 + 0.0400x - 0.0160 = 0\). Using the quadratic formula: \(x = (-0.0400 + \sqrt{0.00160 + 0.0640}) / 2 = (-0.0400 + 0.2561) / 2 = 0.108\) M. Verification: \((0.108)^2 / (0.400 - 0.108) = 0.01166 / 0.292 = 0.0399 \approx 0.0400\). Choice B (0.126 M) results from using the approximation \(x = \sqrt{K_c \times 0.400}\) without the quadratic, which is invalid here because \(x\) is not negligible compared to 0.400.

\(C(s)\) is excluded. ICE (pressures): \(CO_2\) starts at 2.00 atm, CO at 0. Let \(x\) = decrease in \(P(CO_2)\). At equilibrium \(P(CO_2) = 2.00 - x\) and \(P(CO) = 2x\). \(K_p = (2x)^2 / (2.00 - x) = 167.5\). Expanding: \(4x^2 + 167.5x - 335 = 0\). Quadratic formula: \(x = (-167.5 + \sqrt{167.5^2 + 4 \times 4 \times 335}) / 8 = (-167.5 + 182.8) / 8 = 1.91\) atm. Therefore \(P(CO) = 2x = 3.82\) atm. Choice B forgets to multiply \(x\) by 2 (the stoichiometric coefficient). Verification: \((3.82)^2 / (2.00 - 1.91) = 14.59 / 0.09 \approx 162\) — close enough within rounding.

The target reaction equals reaction (2) minus reaction (1): subtracting reaction (1) from reaction (2) gives the desired equation. When reactions are combined by subtraction, their K values are divided. K_target = K2 / K1 = (1.0 × 10^3) / (1.1 × 10^3) = 0.91. Choice B incorrectly multiplies K1 and K2 (used for adding reactions). Choice C inverts the ratio (K1/K2). Choice D reports only K1.

Mg(OH)2 ⇌ Mg^2+ + 2OH^-. Let s = molar solubility. Ksp = (s)(2s)^2 = 4s^3 = 5.6 × 10^-12. So s^3 = 1.4 × 10^-12 and s = 1.119 × 10^-4 M. Then [OH^-] = 2s = 2.237 × 10^-4 M. pOH = -log(2.237 × 10^-4) = 3.65. pH = 14 - 3.65 = 10.35. Choice B uses s instead of 2s for [OH^-]. Choice C reports pOH instead of pH. The dissociation stoichiometry (1 Mg^2+, 2 OH^-) is the key step that must be applied correctly.

HCl reacts completely with NH3: NH3 + H+ → NH4+. After reaction: mol NH3 = 0.25 - 0.050 = 0.20 mol; mol NH4+ = 0.15 + 0.050 = 0.20 mol. Using Henderson-Hasselbalch (mole ratio = concentration ratio since volume is the same): pH = pKa + log([NH3]/[NH4+]) = 9.25 + log(0.20/0.20) = 9.25 + 0 = 9.25. Choice B (9.47) is the original buffer pH before adding HCl, calculated as 9.25 + log(0.25/0.15). The added HCl equalized the moles of acid and base, bringing pH exactly to pKa.

ICE table: \(K_c = (2x)^2 / (0.500 - x) \approx 4x^2 / 0.500\) (valid because \(K_c\) is very small, so \(x \ll 0.500\)). Solving: \(x^2 = (1.0 \times 10^{-6} \times 0.500) / 4 = 1.25 \times 10^{-7}\), so \(x = 3.54 \times 10^{-4}\) M. \([B] = 2x = 7.07 \times 10^{-4}\) M \(\approx 7.1 \times 10^{-4}\) M. The approximation is valid: \(x / 0.500 = 0.071\% \ll 5\%\). Choice B forgets to multiply \(x\) by 2 to get \([B]\). Choice C results from using \(K_c\) directly as a concentration.

For an exothermic reaction, increasing temperature shifts equilibrium toward reactants, decreasing K. Using the van't Hoff equation: ln(K2/K1) = -(delta H°/R)(1/T2 - 1/T1) = -(-50000 / 8.314)(1/500 - 1/400) = 6015 × (-0.0005) = -3.007. K2/K1 = e^(-3.007) = 0.0495. K2 = 2.0 × 10^3 × 0.0495 ≈ 99. Choice A incorrectly applies a positive sign (as if endothermic). Choice C confuses the effect of pressure (delta n) with temperature on K.

Concentrations: [NOCl] = 0.800/2.00 = 0.400 M, [NO] = 0.300/2.00 = 0.150 M, [Cl2] = 0.100/2.00 = 0.050 M. Q = [NO]^2[Cl2] / [NOCl]^2 = (0.150)^2(0.050) / (0.400)^2 = (0.02250)(0.050) / 0.160 = 1.125 × 10^-3 / 0.160 = 7.03 × 10^-3. Since Q (7.0 × 10^-3) > K (4.0 × 10^-4), the reaction must shift in reverse to decrease the product concentrations and re-establish equilibrium. Choice B reaches the same Q but draws the wrong directional conclusion.