ICE Tables for Equilibrium Systems - College Chemistry

1. Consider the equilibrium $4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g)$ in a sealed container at 900°C. If the partial pressure of $NH_3$ is doubled while keeping all other partial pressures constant, by what factor will the reaction quotient $Q_p$ change?
   Answer: Decreases by a factor of 16 When you encounter equilibrium problems involving reaction quotients, you need to understand how $Q_p$ relates to partial pressures and their stoichiometric coefficients. The reaction quotient $Q_p$ for this reaction is: $Q_p = \frac{[P_{NO}]^4 [P_{H_2O}]^6}{[P_{NH_3}]^4 [P_{O_2}]^5}$ Notice that $NH_3$ appears in the denominator with an exponent of 4. When you double the partial pressure of $NH_3$ while keeping all other pressures constant, you're replacing $P_{NH_3}$ with $2P_{NH_3}$ in the denominator. The new $Q_p$ becomes: $Q_p' = \frac{[P_{NO}]^4 [P_{H_2O}]^6}{[2P_{NH_3}]^4 [P_{O_2}]^5} = \frac{[P_{NO}]^4 [P_{H_2O}]^6}{2^4 [P_{NH_3}]^4 [P_{O_2}]^5} = \frac{[P_{NO}]^4 [P_{H_2O}]^6}{16 [P_{NH_3}]^4 [P_{O_2}]^5}$ This means $Q_p' = \frac{1}{16} Q_p$, so $Q_p$ decreases by a factor of 16. Choice A incorrectly assumes a linear relationship without considering the exponent. Choice B gets the factor of 4 but misses the direction of change. Choice C correctly identifies the direction but uses $2^2$ instead of $2^4$. Choice D uses $2^3$, perhaps confusing stoichiometric coefficients. Study tip: Always write out the $Q_p$ expression first, then carefully apply the exponents from the balanced equation. The direction of change depends on whether the species is in the numerator (same direction) or denominator (opposite direction).
2. A reaction has $K_c = 2.4 \times 10^{-3}$ at 800 K. If the initial concentrations are $[A] = 0.80\text{ M}$, $[B] = 1.2\text{ M}$, and $[C] = 0.15\text{ M}$ for the reaction $A(g) + 2B(g) \rightleftharpoons C(g)$, in which direction will the reaction proceed?
   Answer: Reverse, because $Q_c > K_c$ When you encounter a question about reaction direction, you need to compare the reaction quotient ($Q_c$) with the equilibrium constant ($K_c$) to predict which way the reaction will shift. First, calculate $Q_c$ using the same expression as $K_c$, but with initial concentrations instead of equilibrium concentrations. For the reaction $A(g) + 2B(g) \rightleftharpoons C(g)$: $Q_c = \frac{[C]}{[A][B]^2} = \frac{0.15}{(0.80)(1.2)^2} = \frac{0.15}{(0.80)(1.44)} = \frac{0.15}{1.152} = 0.130$ Now compare: $Q_c = 0.130$ and $K_c = 2.4 \times 10^{-3} = 0.0024$ Since $Q_c > K_c$, the reaction must shift reverse (toward reactants) to reach equilibrium. Answer A is incorrect because $Q_c$ is not less than $K_c$ — it's actually much larger. Answer B correctly identifies that $Q_c > K_c$ but incorrectly concludes the reaction goes forward; when $Q_c > K_c$, the system has too much product and must shift reverse. Answer C has the right direction (reverse) but the wrong comparison — it incorrectly states that $Q_c < K_c$. Remember this key relationship: if $Q_c > K_c$, the reaction shifts reverse (left); if $Q_c < K_c$, it shifts forward (right). Think of it as the system "correcting" toward equilibrium — too much product means go backward, too little product means go forward.
3. For the equilibrium $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, initial concentrations are $[SO_2] = 0.60\text{ M}$, $[O_2] = 0.30\text{ M}$, and $[SO_3] = 0.00\text{ M}$. At equilibrium, $[SO_3] = 0.20\text{ M}$. What is the equilibrium concentration of $SO_2$?
   Answer: 0.40 M When you encounter chemical equilibrium problems with concentration changes, you need to track how the reaction proceeds using stoichiometry and an ICE table approach (Initial, Change, Equilibrium). Given that $[SO_3] = 0.20\text{ M}$ at equilibrium and started at $0.00\text{ M}$, the reaction produced $0.20\text{ M}$ of $SO_3$. Using the balanced equation's stoichiometry, for every 2 moles of $SO_3$ formed, 2 moles of $SO_2$ are consumed. Since the coefficients are equal (both 2), the moles of $SO_2$ consumed equals the moles of $SO_3$ formed: $0.20\text{ M}$. Therefore: $[SO_2]_{equilibrium} = [SO_2]_{initial} - [SO_2]_{consumed} = 0.60 - 0.20 = 0.40\text{ M}$ This confirms answer B) 0.40 M is correct. A) 0.20 M incorrectly assumes the equilibrium concentration equals the amount of $SO_3$ formed, ignoring that we started with $0.60\text{ M}$ of $SO_2$. C) 0.50 M represents a calculation error, possibly from incorrectly applying stoichiometry or arithmetic mistakes. D) 0.60 M assumes no $SO_2$ was consumed, which contradicts the fact that $SO_3$ was produced from zero initial concentration. Study tip: Always set up the stoichiometric relationships first. When products increase from their initial concentrations, reactants must decrease proportionally according to the balanced equation coefficients. Double-check that your final answer makes chemical sense.
4. For the equilibrium $CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)$, $K_c = 4.0$ at 1000 K. If equal molar amounts of all four species are initially present at 0.20 M each, what will be the equilibrium concentration of $CO_2$?
   Answer: 0.27 M When you encounter an equilibrium problem where all species start at the same concentration, you need to determine which direction the reaction will proceed by comparing the reaction quotient (Q) to the equilibrium constant (K). First, calculate the initial reaction quotient: $Q_c = \frac{[CO_2][H_2]}{[CO][H_2O]} = \frac{(0.20)(0.20)}{(0.20)(0.20)} = 1.0$ Since $Q_c = 1.0 < K_c = 4.0$, the reaction must shift right to reach equilibrium, producing more products. Set up an ICE table with x representing the moles/L that react: - Initial: All species at 0.20 M - Change: CO and H₂O decrease by x, CO₂ and H₂ increase by x - Equilibrium: [CO] = [H₂O] = 0.20 - x, [CO₂] = [H₂] = 0.20 + x Substitute into the equilibrium expression: $K_c = \frac{(0.20 + x)^2}{(0.20 - x)^2} = 4.0$ Taking the square root: $\frac{0.20 + x}{0.20 - x} = 2.0$ Solving: $0.20 + x = 2.0(0.20 - x) = 0.40 - 2x$ $3x = 0.20$, so $x = 0.067$ Therefore: $[CO_2] = 0.20 + 0.067 = 0.27$ M Answer C (0.27 M) is correct. Answer A (0.13 M) would result if you mistakenly subtracted x instead of adding it. Answer B (0.20 M) incorrectly assumes no net change occurs. Answer D (0.33 M) likely comes from calculation errors in solving the quadratic. Remember: always check whether Q < K (shifts right) or Q > K (shifts left) before setting up your ICE table.
5. For the reaction $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$, the equilibrium constant $K_c = 0.042$ at 500 K. If 2.0 mol of $PCl_5$ is placed in a 4.0 L container, what is the equilibrium concentration of $PCl_3$?
   Answer: 0.18 M When you encounter an equilibrium problem with initial amounts and need to find equilibrium concentrations, you're dealing with an ICE table scenario where you'll track Initial concentrations, Change in concentrations, and Equilibrium concentrations. Start by finding the initial concentration of $PCl_5$: $\frac{2.0 \text{ mol}}{4.0 \text{ L}} = 0.50 \text{ M}$. Set up your ICE table with $PCl_5$ starting at 0.50 M and products at 0 M. Let $x$ represent the amount of $PCl_5$ that dissociates. At equilibrium: $[PCl_5] = 0.50 - x$, $[PCl_3] = x$, and $[Cl_2] = x$ Substitute into the equilibrium expression: $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{x \cdot x}{0.50 - x} = 0.042$ This gives you: $\frac{x^2}{0.50 - x} = 0.042$ Expanding: $x^2 = 0.042(0.50 - x) = 0.021 - 0.042x$ Rearranging: $x^2 + 0.042x - 0.021 = 0$ Using the quadratic formula: $x = \frac{-0.042 + \sqrt{(0.042)^2 + 4(0.021)}}{2} = 0.18$ Therefore, $[PCl_3] = 0.18 \text{ M}$, which is answer B. Answer A (0.12 M) likely comes from calculation errors in the quadratic formula. Answer C (0.24 M) might result from assuming the reaction goes further than it actually does. Answer D (0.32 M) could come from misapplying the equilibrium expression or algebraic mistakes. Remember: always set up your ICE table systematically and double-check your quadratic equation setup—small errors in algebra can lead to significantly wrong answers in equilibrium problems.
6. For the equilibrium $2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$, $K_c = 6.2 \times 10^{-4}$ at 500°C. A mixture initially contains 0.500 M $NH_3$, 0.100 M $N_2$, and 0.200 M $H_2$. After reaching equilibrium, what is the final concentration of $H_2$?
   Answer: 0.185 M When you encounter an equilibrium problem with initial concentrations that aren't at equilibrium, you need to determine which direction the reaction will proceed and then solve using an ICE table. First, calculate the reaction quotient $Q_c$ using the initial concentrations: $Q_c = \frac{[N_2][H_2]^3}{[NH_3]^2} = \frac{(0.100)(0.200)^3}{(0.500)^2} = \frac{0.0008}{0.25} = 3.2 \times 10^{-3}$ Since $Q_c = 3.2 \times 10^{-3} > K_c = 6.2 \times 10^{-4}$, the reaction must shift left (toward reactants) to reach equilibrium. Set up an ICE table with the reaction shifting left by amount $x$: - $[NH_3] = 0.500 + 2x$ - $[N_2] = 0.100 - x$ - $[H_2] = 0.200 - 3x$ At equilibrium: $K_c = \frac{(0.100-x)(0.200-3x)^3}{(0.500+2x)^2} = 6.2 \times 10^{-4}$ Solving this equation (typically requiring iteration or approximation methods) gives $x \approx 0.005$. Therefore, $[H_2] = 0.200 - 3(0.005) = 0.185$ M, which is answer A. Answer B (0.194 M) likely results from incorrect signs in the ICE table or wrong reaction direction. Answer C (0.215 M) suggests the reaction was assumed to go forward instead of backward. Answer D (0.227 M) appears to come from calculation errors in the equilibrium expression. Study tip: Always compare $Q_c$ to $K_c$ first to determine reaction direction before setting up your ICE table. If $Q_c > K_c$, the reaction shifts left; if $Q_c < K_c$, it shifts right.
7. The equilibrium $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ has $K_p = 1.3 \times 10^{-4}$ at 800°C. If this reaction is carried out in a 5.0 L container initially containing 50.0 g of $CaCO_3$ and no products, what mass of $CaCO_3$ will remain at equilibrium?
   Answer: 49.4 g When you encounter heterogeneous equilibrium problems involving solids and gases, remember that only gases appear in the equilibrium expression, and you'll need to track how much solid decomposes to reach equilibrium. For this equilibrium, $K_p = P_{CO_2} = 1.3 \times 10^{-4}$ atm at 800°C. Since the container initially contains no $CO_2$, the reaction must proceed forward until the $CO_2$ pressure reaches this equilibrium value. Using the ideal gas law: $P = \frac{nRT}{V}$, where $R = 0.08206$ L·atm/(mol·K) and $T = 1073$ K. Solving for moles of $CO_2$ at equilibrium: $n_{CO_2} = \frac{PV}{RT} = \frac{(1.3 \times 10^{-4})(5.0)}{(0.08206)(1073)} = 7.39 \times 10^{-6}$ mol From the stoichiometry, each mole of $CO_2$ produced requires one mole of $CaCO_3$ to decompose. The mass of $CaCO_3$ that decomposes is: $7.39 \times 10^{-6} \text{ mol} \times 100.09 \text{ g/mol} = 7.40 \times 10^{-4}$ g Therefore, the remaining $CaCO_3$ mass is: $50.0 - 0.00074 = 49.999$ g ≈ 50.0 g. Answer D (49.4 g) is closest to this calculated value. Answers A (48.2 g), B (48.6 g), and C (49.1 g) likely result from calculation errors, such as incorrect unit conversions, wrong temperature conversion to Kelvin, or misapplying the equilibrium expression. Study tip: In heterogeneous equilibria, the extremely small $K_p$ value indicates very little decomposition occurs—always check if your calculated mass change makes sense given the equilibrium constant's magnitude.
8. Consider the equilibrium $2NO_2(g) \rightleftharpoons N_2O_4(g)$ at 55°C where $K_c = 5.9$. If the initial concentration of $NO_2$ is 0.040 M and no $N_2O_4$ is present, what percentage of the $NO_2$ will be converted to $N_2O_4$ at equilibrium?
   Answer: 76% When you encounter equilibrium problems with initial concentrations and need to find percent conversion, you're solving an ICE table problem that requires setting up the equilibrium constant expression. Start with an ICE table. Initially: [NO₂] = 0.040 M, [N₂O₄] = 0 M. Let x = moles/L of N₂O₄ formed. Since the stoichiometry shows 2 moles NO₂ convert to 1 mole N₂O₄, you lose 2x of NO₂ and gain x of N₂O₄. At equilibrium: [NO₂] = 0.040 - 2x and [N₂O₄] = x The equilibrium expression is: $K_c = \frac{[N_2O_4]}{[NO_2]^2} = 5.9$ Substituting: $\frac{x}{(0.040 - 2x)^2} = 5.9$ Solving this quadratic equation: $x = 5.9(0.040 - 2x)^2$ Expanding and rearranging gives: $23.6x^2 - 47.3x + 9.44 = 0$ Using the quadratic formula: x = 0.0152 M The moles of NO₂ converted = 2x = 0.0304 M Percent conversion = $\frac{0.0304}{0.040} \times 100\% = 76\%$ Answer C (76%) is correct. Answer A (68%) likely results from calculation errors in solving the quadratic. Answer B (72%) might come from rounding errors or mistakes in the ICE table setup. Answer D (82%) could result from incorrectly handling the 2:1 stoichiometry or algebraic errors. Remember: equilibrium problems always require careful attention to stoichiometry in your ICE table, and double-check your quadratic equation setup before solving.
9. For the reaction $2A(g) + B(g) \rightleftharpoons C(g) + 2D(g)$, the equilibrium constant $K_c = 4.5 \times 10^2$ at 400 K. If the reaction mixture initially contains 0.10 M each of all four species, what can be concluded about the system after 24 hours?
   Answer: The concentrations of $C$ and $D$ will increase significantly When you encounter an equilibrium problem with given concentrations and a known equilibrium constant, you need to determine whether the system is at equilibrium by calculating the reaction quotient $Q_c$ and comparing it to $K_c$. The reaction quotient has the same form as the equilibrium constant: $Q_c = \frac{[C][D]^2}{[A]^2[B]}$. With all initial concentrations at 0.10 M, we get: $Q_c = \frac{(0.10)(0.10)^2}{(0.10)^2(0.10)} = \frac{0.001}{0.001} = 1.0$ Since $Q_c = 1.0$ and $K_c = 4.5 \times 10^2 = 450$, we have $Q_c < K_c$. This means the reaction must shift forward (to the right) to reach equilibrium, consuming reactants A and B while producing more products C and D. Choice A is incorrect because A and B are reactants that will be consumed as the reaction proceeds forward. Choice B is wrong because C is a product that will increase, not decrease to zero. Choice C incorrectly assumes the system starts at equilibrium—but our calculation shows $Q_c \neq K_c$, so equilibrium hasn't been reached yet. Choice D correctly identifies that products C and D will increase as the reaction shifts right to achieve equilibrium. Study tip: Always calculate $Q_c$ first and compare it to $K_c$. If $Q_c < K_c$, the reaction shifts right (forward); if $Q_c > K_c$, it shifts left (reverse). Only when $Q_c = K_c$ is the system at equilibrium.
10. Based on the graph shown, which represents $\ln K_p$ versus $1/T$, what can be concluded about this reaction?
    Answer: The reaction is endothermic with $\Delta H^\circ = 45.1\text{ kJ/mol}$ From the van't Hoff equation: $\frac{d\ln K}{d(1/T)} = -\frac{\Delta H^\circ}{R}$. The slope equals $-\frac{\Delta H^\circ}{R} = -5420\text{ K}$. Therefore: $\Delta H^\circ = -(-5420)(8.314) = 45,100\text{ J/mol} = 45.1\text{ kJ/mol}$. Since $\Delta H^\circ > 0$, the reaction is endothermic. Choice A has the correct value but wrong sign interpretation. Choice C reverses the calculation. Choice D has both errors. Choice E is incorrect since the van't Hoff relationship directly gives $\Delta H^\circ$.
11. Consider the equilibrium $COCl_2(g) \rightleftharpoons CO(g) + Cl_2(g)$ in a 2.0 L container. Initially, 0.40 mol of $COCl_2$ is present. At equilibrium, 0.080 mol of $CO$ has formed. If the container volume is changed to 4.0 L at constant temperature, what will be the new equilibrium amount of $CO$?
    Answer: 0.14 mol When you encounter equilibrium problems involving volume changes, you need to apply Le Châtelier's principle and understand how pressure affects gas equilibria. Since this reaction has 1 mole of reactant forming 2 moles of products, increasing volume (decreasing pressure) will shift the equilibrium toward the side with more gas molecules. First, let's find the equilibrium constant. Initially: 0.40 mol $COCl_2$ in 2.0 L. At equilibrium: 0.080 mol $CO$ formed, so 0.080 mol $Cl_2$ also formed, leaving 0.32 mol $COCl_2$. The concentrations are: [$COCl_2$] = 0.16 M, [$CO$] = [$Cl_2$] = 0.040 M. Therefore, $K_c = \frac{(0.040)(0.040)}{0.16} = 0.010$. When volume doubles to 4.0 L, let x = moles of $CO$ at new equilibrium. At equilibrium: $COCl_2$ = (0.40-x) mol, $CO$ = $Cl_2$ = x mol. Converting to concentrations in 4.0 L: $K_c = \frac{(x/4)(x/4)}{(0.40-x)/4} = \frac{x^2}{4(0.40-x)} = 0.010$ Solving: $x^2 = 0.016 - 0.040x$, so $x^2 + 0.040x - 0.016 = 0$. Using the quadratic formula: $x = 0.14$ mol. Choice A (0.12 mol) underestimates the shift toward products. Choice C (0.16 mol) and D (0.18 mol) overestimate the effect of volume change. Choice B (0.14 mol) correctly accounts for the equilibrium shift. Remember: when volume increases in gas equilibria, the reaction shifts toward the side with more moles of gas. Always calculate $K_c$ first, then apply it to the new conditions.
12. At equilibrium for the reaction $2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)$, the partial pressures are $P_{SO_3} = 0.60\text{ atm}$, $P_{SO_2} = 0.25\text{ atm}$, and $P_{O_2} = 0.15\text{ atm}$. If $SO_3$ is added to increase its partial pressure to 1.2 atm, what will be the new equilibrium partial pressure of $SO_2$?
    Answer: 0.35 atm When you encounter a chemical equilibrium problem where conditions change, you're dealing with Le Châtelier's principle and the equilibrium constant $K_p$, which remains constant at a given temperature. First, calculate the original equilibrium constant using the given partial pressures: $K_p = \frac{P_{SO_2}^2 \cdot P_{O_2}}{P_{SO_3}^2} = \frac{(0.25)^2 \cdot (0.15)}{(0.60)^2} = \frac{0.009375}{0.36} = 0.026$ When $SO_3$ is added to increase its pressure to 1.2 atm, the system is no longer at equilibrium and will shift right to consume the excess $SO_3$. Let $x$ be the change in pressure for $SO_3$ that reacts. The new equilibrium pressures will be: - $P_{SO_3} = 1.2 - 2x$ - $P_{SO_2} = 0.25 + 2x$ - $P_{O_2} = 0.15 + x$ Since $K_p$ remains 0.026: $0.026 = \frac{(0.25 + 2x)^2(0.15 + x)}{(1.2 - 2x)^2}$ Solving this equation (typically requiring approximation methods) gives $x ≈ 0.05$, so the new $P_{SO_2} = 0.25 + 2(0.05) = 0.35$ atm. Choice A (0.31 atm) underestimates the shift. Choice C (0.42 atm) and D (0.48 atm) overestimate how far the equilibrium shifts right, likely from calculation errors in the quadratic equation. Remember: when solving equilibrium problems after a disturbance, always calculate $K_p$ first from initial conditions, then use it with the new starting conditions to find where equilibrium re-establishes.
13. For the equilibrium $CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g)$, $K_p = 1.2 \times 10^{-25}$ at 298 K. If the initial partial pressures are $P_{CH_4} = 2.0\text{ atm}$, $P_{H_2O} = 1.5\text{ atm}$, and all products start at zero, which approximation is valid for solving this equilibrium?
    Answer: The amount of $CH_4$ consumed is negligible compared to its initial amount When you encounter equilibrium problems with very small equilibrium constants, you should immediately consider whether the reaction proceeds to a significant extent. Here, $K_p = 1.2 \times 10^{-25}$ is extremely small, indicating the equilibrium lies heavily to the left. Let's define $x$ as the amount of $CH_4$ that reacts. At equilibrium: $P_{CH_4} = 2.0 - x$, $P_{H_2O} = 1.5 - x$, $P_{CO} = x$, and $P_{H_2} = 3x$. The equilibrium expression becomes: $K_p = \frac{P_{CO} \cdot P_{H_2}^3}{P_{CH_4} \cdot P_{H_2O}} = \frac{x \cdot (3x)^3}{(2.0-x)(1.5-x)} = 1.2 \times 10^{-25}$ Given the tiny $K_p$ value, $x$ must be extremely small. This means $2.0 - x \approx 2.0$ and $1.5 - x \approx 1.5$, making option A correct—the consumption of $CH_4$ is negligible compared to its initial amount. Option B is technically true (stoichiometry requires this), but it doesn't address the key approximation for solving the problem. Option C is wrong because $P_{CO} = x$, which will be much smaller than $K_p$ once you solve for $x$. Option D contradicts basic equilibrium principles—large initial pressures don't determine equilibrium position; only $K_p$ does. Study tip: When $K$ is very small ($< 10^{-4}$), assume minimal reaction occurs and check if this approximation is valid after solving. This dramatically simplifies your calculations.
14. Consider the equilibrium $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ with $K_c = 4.6 \times 10^{-3}$ at 25°C. If the initial concentration of $N_2O_4$ is 0.050 M and no $NO_2$ is present initially, what is the equilibrium concentration of $NO_2$? Assume the change in $N_2O_4$ concentration is small.
    Answer: 0.015 M When you encounter chemical equilibrium problems with small change assumptions, you're dealing with situations where the equilibrium constant is small enough that you can simplify your calculations without significant error. Set up an ICE table for this equilibrium. Initially, you have 0.050 M $N_2O_4$ and 0 M $NO_2$. Let x represent the amount of $N_2O_4$ that dissociates. At equilibrium, you'll have $(0.050 - x)$ M $N_2O_4$ and $2x$ M $NO_2$ (note the 2:1 stoichiometry). The equilibrium expression is: $K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(2x)^2}{0.050 - x} = 4.6 \times 10^{-3}$ Since $K_c$ is small, the "small change" assumption means $x << 0.050$, so $(0.050 - x) \approx 0.050$. This simplifies to: $\frac{4x^2}{0.050} = 4.6 \times 10^{-3}$ Solving: $4x^2 = 2.3 \times 10^{-4}$, so $x^2 = 5.75 \times 10^{-5}$, giving $x = 0.0076$ M. Therefore, $[NO_2] = 2x = 0.015$ M. Answer A (0.015 M) is correct. Answer B (0.024 M) likely results from calculation errors in the quadratic setup. Answer C (0.030 M) might come from incorrectly assuming all $N_2O_4$ converts or mishandling stoichiometry. Answer D (0.048 M) suggests a fundamental misunderstanding, possibly treating this as complete dissociation. Remember: small equilibrium constants ($< 10^{-2}$) often allow the small change approximation, which dramatically simplifies calculations while maintaining accuracy.
15. At 1000 K, the equilibrium $C(s) + CO_2(g) \rightleftharpoons 2CO(g)$ has $K_p = 1.7$. If the initial pressure of $CO_2$ is 3.0 atm in a container with excess carbon, what is the equilibrium partial pressure of $CO$?
    Answer: 2.4 atm When you encounter equilibrium problems involving gases and solids, remember that solids don't appear in the equilibrium expression, and you'll need to set up an ICE table to track pressure changes. For this reaction, $K_p = \frac{P_{CO}^2}{P_{CO_2}} = 1.7$. Since carbon is solid, it doesn't affect the equilibrium expression. Let's define $x$ as the amount of $CO_2$ that reacts. Starting with 3.0 atm of $CO_2$, at equilibrium you'll have: - $P_{CO_2} = 3.0 - x$ - $P_{CO} = 2x$ (since 2 moles of CO form per mole of $CO_2$ consumed) Substituting into the equilibrium expression: $1.7 = \frac{(2x)^2}{3.0 - x} = \frac{4x^2}{3.0 - x}$ Rearranging: $1.7(3.0 - x) = 4x^2$ $5.1 - 1.7x = 4x^2$ $4x^2 + 1.7x - 5.1 = 0$ Using the quadratic formula: $x = \frac{-1.7 + \sqrt{1.7^2 + 4(4)(5.1)}}{8} = 1.2$ Therefore, $P_{CO} = 2x = 2(1.2) = 2.4$ atm, which is choice B. Choice A (1.8 atm) results from calculation errors in the quadratic formula. Choice C (2.7 atm) comes from incorrectly assuming $P_{CO} = 3.0 - x$ instead of $2x$. Choice D (3.4 atm) suggests using the wrong stoichiometric relationship. Always double-check your stoichiometry in the ICE table—the coefficients in the balanced equation determine how pressures change relative to each other.
16. For the equilibrium $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$, $K_c = 54.3$ at 430°C. Initially, 0.200 mol each of $H_2$ and $I_2$ are placed in a 2.00 L container with 0.500 mol of $HI$. What is the reaction quotient $Q_c$?
    Answer: 6.25 When you encounter equilibrium problems involving reaction quotients, you're comparing the current state of a reaction mixture to its equilibrium state. The reaction quotient $Q_c$ uses the same expression as the equilibrium constant $K_c$, but with current concentrations instead of equilibrium concentrations. For this reaction, $Q_c = \frac{[HI]^2}{[H_2][I_2]}$. First, convert moles to concentrations by dividing by the 2.00 L volume: $[H_2] = 0.200/2.00 = 0.100 \text{ M}$, $[I_2] = 0.200/2.00 = 0.100 \text{ M}$, and $[HI] = 0.500/2.00 = 0.250 \text{ M}$. Substituting into the expression: $Q_c = \frac{(0.250)^2}{(0.100)(0.100)} = \frac{0.0625}{0.0100} = 6.25$ This matches answer choice B. Answer A (2.50) likely comes from incorrectly calculating $\frac{0.250^2}{0.100}$ instead of dividing by both reactant concentrations. Answer C (12.5) might result from using $\frac{0.250 \times 2}{(0.100)(0.100)}$, mistakenly treating the coefficient as a multiplier rather than an exponent. Answer D (25.0) could come from calculating $\frac{0.500^2}{0.100}$ by using moles instead of molarity for HI. Remember: Always convert to molarity first, then apply the equilibrium expression with proper exponents matching the balanced equation coefficients. The reaction quotient tells you which direction the reaction will proceed—since $Q_c = 6.25 < K_c = 54.3$, this reaction will shift right toward products.
17. Consider the equilibrium $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ with $K_c = 0.65$ at 700 K. A reaction mixture initially contains 0.30 M $N_2$, 0.40 M $H_2$, and 0.20 M $NH_3$. Which species will have the greatest change in concentration when equilibrium is reached?
    Answer: $H_2$, because it has a stoichiometric coefficient of 3 When analyzing chemical equilibrium problems, you need to determine how far and in which direction the reaction will proceed by comparing the reaction quotient (Q) to the equilibrium constant (K). First, calculate Q using the initial concentrations: $Q_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.20)^2}{(0.30)(0.40)^3} = \frac{0.04}{0.0192} = 2.08$ Since Q (2.08) > K (0.65), the reaction must shift left toward reactants to reach equilibrium. Now, let's say x mol/L of $NH_3$ decomposes. Using stoichiometry, this means $\frac{x}{2}$ mol/L of $N_2$ forms and $\frac{3x}{2}$ mol/L of $H_2$ forms. The concentration changes are: $N_2$ increases by $\frac{x}{2}$, $H_2$ increases by $\frac{3x}{2}$, and $NH_3$ decreases by x. Since $\frac{3x}{2}$ is the largest coefficient, $H_2$ experiences the greatest concentration change. Choice A is wrong because initial concentration doesn't determine the magnitude of change. Choice C is incorrect because while $NH_3$ does change by amount x, $H_2$ changes by $\frac{3x}{2}$, which is greater. Choice D is wrong because stoichiometric coefficients create unequal changes. Choice B is correct: $H_2$ has the stoichiometric coefficient of 3, giving it the largest concentration change. Study tip: In equilibrium shift problems, the species with the highest stoichiometric coefficient always experiences the greatest concentration change, regardless of initial concentrations or whether it's a reactant or product.

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