Reaction Quotient and Le Chatelier's Principle - College Chemistry

1. In Trial 1, the reactor initially contains 0.40 mol $CO$ and 0.40 mol $H_2O$ with no products. After reaching equilibrium, the concentration of $CO$ is measured to be 0.10 M. What is the reaction quotient $Q_c$ if, at some intermediate time before equilibrium, the concentrations are $[CO] = 0.15 \text{ M}$, $[H_2O] = 0.15 \text{ M}$, $[CO_2] = 0.05 \text{ M}$, and $[H_2] = 0.05 \text{ M}$?
   Answer: $Q_c = 0.11$ When you encounter reaction quotient problems, remember that $Q_c$ uses the same expression as the equilibrium constant but with concentrations at any point in time, not necessarily at equilibrium. For the water-gas shift reaction, the reaction quotient is: $Q_c = \frac{[CO_2][H_2]}{[CO][H_2O]}$ At the intermediate time described, you can substitute the given concentrations directly: $Q_c = \frac{(0.05)(0.05)}{(0.15)(0.15)} = \frac{0.0025}{0.0225} = 0.11$ This matches answer choice A. The calculation is straightforward once you write the correct expression and plug in the values. Looking at the wrong answers: Answer B (0.33) likely comes from incorrectly calculating the fraction or making an arithmetic error. Answer C (1.0) represents the equilibrium constant $K_c$, which is a common trap—students sometimes confuse $Q_c$ with $K_c$, but $Q_c$ only equals $K_c$ at equilibrium, not at intermediate times. Answer D (3.0) would result from inverting the reaction quotient expression, putting reactants in the numerator and products in the denominator. Since $Q_c = 0.11 < K_c = 1.0$, this tells you the reaction will proceed forward to reach equilibrium, which makes chemical sense given that more products need to form. Study tip: Always write out the $Q_c$ expression first, then substitute carefully. Remember that $Q_c$ compares the current state to equilibrium—when $Q_c < K_c$, the reaction proceeds forward.
2. For the reaction $2A(g) + B(g) \rightleftharpoons 3C(g)$, the equilibrium constant $K_p = 27$ at 600 K. At a particular instant, the partial pressures are $P_A = 1.0 \text{ atm}$, $P_B = 2.0 \text{ atm}$, and $P_C = 6.0 \text{ atm}$. Which statement correctly describes the system?
   Answer: $Q_p = 108$; the reaction will proceed in the reverse direction When you encounter equilibrium problems with given partial pressures, you need to calculate the reaction quotient $Q_p$ and compare it to the equilibrium constant $K_p$ to determine which direction the reaction will proceed. For the reaction $2A(g) + B(g) \rightleftharpoons 3C(g)$, the reaction quotient expression is $Q_p = \frac{P_C^3}{P_A^2 \cdot P_B}$. Substituting the given values: $Q_p = \frac{(6.0)^3}{(1.0)^2 \cdot (2.0)} = \frac{216}{2} = 108$. Since $Q_p = 108 > K_p = 27$, the system has too much product relative to reactants compared to equilibrium. The reaction must shift left (reverse direction) to reach equilibrium, consuming products and forming reactants. Answer A incorrectly calculates $Q_p$ as 54, likely from computational errors in handling the exponents. Answer B makes the same calculation error as A but also incorrectly predicts the reaction direction. Answer D calculates $Q_p$ as 13.5, possibly from incorrectly using $\frac{P_C^3}{P_A \cdot P_B^2}$ or other algebraic mistakes. Additionally, D incorrectly states the reaction proceeds forward when $Q_p < K_p$ would indicate forward direction, but the calculated value is wrong anyway. Remember this key rule: when $Q_p > K_p$, the reaction shifts reverse (left) to decrease the quotient; when $Q_p < K_p$, it shifts forward (right). Always double-check your stoichiometric exponents in the $Q_p$ expression—they must match the balanced equation coefficients.
3. At 1000 K, the equilibrium constant for $CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)$ is $K_c = 1.0$. If equal molar amounts of $CO$ and $H_2O$ are placed in a container, and the equilibrium concentration of $CO$ is found to be 0.30 M, what was the initial concentration of $CO$?
   Answer: 0.60 M This question tests your understanding of equilibrium calculations using ICE tables (Initial, Change, Equilibrium). When you see equal molar amounts of reactants and a given equilibrium concentration, you need to work backwards from equilibrium to find the initial conditions. Let's set up an ICE table. Since equal molar amounts of CO and H₂O are used, they have the same initial concentration (let's call it x). At equilibrium, [CO] = 0.30 M. Since the stoichiometry is 1:1:1:1, if CO decreased by some amount, then H₂O also decreased by that same amount, while CO₂ and H₂ each increased by that amount. If the initial concentration was x and the equilibrium concentration is 0.30 M, then the change is (x - 0.30). This means [CO₂] = [H₂] = (x - 0.30) at equilibrium. Using the equilibrium expression: $K_c = \frac{[CO_2][H_2]}{[CO][H_2O]} = 1.0$ Substituting: $1.0 = \frac{(x-0.30)(x-0.30)}{(0.30)(0.30)} = \frac{(x-0.30)^2}{(0.30)^2}$ Taking the square root: $(x-0.30) = 0.30$, so $x = 0.60$ M. Answer (A) 0.30 M incorrectly assumes the initial concentration equals the equilibrium concentration, ignoring that reaction occurred. Answer (C) 0.45 M might result from incorrectly averaging initial and final values. Answer (D) 0.90 M could come from doubling the equilibrium concentration without proper calculation. Always set up ICE tables systematically for equilibrium problems, and remember that equal initial amounts doesn't mean concentrations stay equal as the reaction proceeds.
4. Consider the equilibrium $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ in a sealed container at 900°C. Which of the following changes would cause the equilibrium to shift toward the formation of more $CO_2(g)$?
   Answer: Increasing the volume of the container When you encounter equilibrium problems involving gas-solid reactions, focus on Le Châtelier's principle and how changes affect the gas phase specifically. This reaction involves solids and one gas, so changes that affect gas pressure will drive the equilibrium shift. Increasing the volume of the container (C) decreases the pressure of $CO_2(g)$. According to Le Châtelier's principle, the equilibrium will shift to counteract this change by producing more gas molecules to restore pressure. This drives the reaction forward, creating more $CO_2(g)$ and $CaO(s)$ from $CaCO_3(s)$. Here's why the other options don't work: Adding more $CaCO_3(s)$ (A) has no effect because pure solids don't appear in equilibrium expressions—their concentrations remain constant. The equilibrium position depends only on temperature and gas pressure, not the amount of solid present. Removing $CaO(s)$ (B) similarly has no effect for the same reason—solid concentrations don't influence equilibrium position. Adding more $CO_2(g)$ (D) actually shifts the equilibrium in the wrong direction. By increasing $CO_2$ pressure, you're adding product, which pushes the reaction backward toward reactants according to Le Châtelier's principle. Remember this key principle: for reactions involving gases and solids, only changes affecting gas pressure (volume changes, adding/removing gases, or temperature changes) will shift the equilibrium. Changes involving solids alone won't affect the equilibrium position, even though they might seem intuitive.
5. For the reaction $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$, $K_p = 0.497$ at 500 K. A reaction vessel initially contains $PCl_5$ at 2.0 atm pressure with no products present. At some point during the reaction, $P_{PCl_5} = 1.6 \text{ atm}$. What is $Q_p$ at this moment?
   Answer: $Q_p = 0.10$ When you encounter a reaction quotient problem, you're testing whether a system is at equilibrium or which direction it will proceed. The reaction quotient $Q_p$ has the same mathematical form as the equilibrium constant $K_p$, but uses current pressures instead of equilibrium pressures. To find $Q_p$, you need the partial pressures of all species at the given moment. You know $P_{PCl_5} = 1.6$ atm, but you need to find the pressures of $PCl_3$ and $Cl_2$. Since the reaction started with 2.0 atm of $PCl_5$ and now has 1.6 atm, exactly 0.4 atm of $PCl_5$ has decomposed. Looking at the stoichiometry, for every mole of $PCl_5$ that decomposes, one mole each of $PCl_3$ and $Cl_2$ forms. Therefore, $P_{PCl_3} = P_{Cl_2} = 0.4$ atm. Now you can calculate: $Q_p = \frac{P_{PCl_3} \times P_{Cl_2}}{P_{PCl_5}} = \frac{0.4 \times 0.4}{1.6} = \frac{0.16}{1.6} = 0.10$ The correct answer is A. Answer B (0.040) likely comes from incorrectly using 0.2 atm for the product pressures. Answer C (0.25) might result from using the wrong denominator or calculation error. Answer D (0.16) is simply the numerator without dividing by the $PCl_5$ pressure. Remember: always track the stoichiometry carefully when calculating how much of each species is present, and double-check that your $Q_p$ expression matches the balanced equation.
6. For the gas-phase equilibrium $A_2(g) + 3B_2(g) \rightleftharpoons 2AB_3(g)$, the reaction is at equilibrium in a container at constant temperature. If the container volume is increased by a factor of 3 while keeping temperature constant, what happens to the reaction quotient immediately after the volume change?
   Answer: $Q$ increases by a factor of 9 When you encounter gas-phase equilibrium problems involving volume changes, focus on how concentration changes affect the reaction quotient $Q$, which has the same mathematical form as the equilibrium constant but uses current concentrations rather than equilibrium concentrations. For this reaction, $Q = \frac{[AB_3]^2}{[A_2][B_2]^3}$. When the volume increases by a factor of 3, all concentrations decrease by the same factor of 3 (since concentration = moles/volume and moles stay constant). Let's trace what happens to $Q$: If each concentration decreases by a factor of 3, then $[AB_3]$ becomes $[AB_3]/3$, $[A_2]$ becomes $[A_2]/3$, and $[B_2]$ becomes $[B_2]/3$. Substituting into the expression: $Q_{new} = \frac{([AB_3]/3)^2}{([A_2]/3)([B_2]/3)^3} = \frac{[AB_3]^2/9}{[A_2]/3 \times [B_2]^3/27} = \frac{[AB_3]^2/9}{[A_2][B_2]^3/81} = \frac{[AB_3]^2}{[A_2][B_2]^3} \times \frac{81}{9} = Q_{original} \times 9$ So $Q$ increases by a factor of 9, making C correct. Choice A incorrectly assumes $Q$ changes by the same factor as volume. Choice B uses the wrong direction and wrong magnitude. Choice D has the correct magnitude but wrong direction—this would occur if you mistakenly put products in the denominator. Remember: When volume increases, always count the net change in moles of gas (here: 4 moles reactants → 2 moles products, net = -2) to predict the direction and magnitude of $Q$'s change.
7. Consider the equilibrium $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$ with $\Delta H = -198 \text{ kJ}$. If the temperature of the system is increased while keeping volume constant, what will happen to the equilibrium position and the value of the equilibrium constant?
   Answer: Equilibrium shifts left; $K$ decreases When you encounter equilibrium problems involving temperature changes, you need to apply Le Châtelier's principle and understand how temperature affects the equilibrium constant. The key is recognizing whether the reaction is exothermic or endothermic. Since $\Delta H = -198 \text{ kJ}$, this reaction is highly exothermic, meaning it releases heat. You can think of heat as a "product" in this reaction. When temperature increases, you're essentially adding heat to the system. According to Le Châtelier's principle, the equilibrium will shift to counteract this change by consuming the excess heat, favoring the reverse reaction (shifting left toward reactants). For the equilibrium constant, remember that $K$ depends on temperature. For exothermic reactions, increasing temperature always decreases $K$ because the equilibrium favors reactants at higher temperatures. This relationship comes from the van 't Hoff equation. Looking at the wrong answers: Choice A incorrectly states that $K$ increases—this would only happen if the reaction were endothermic. Choices C and D both claim the equilibrium shifts right, which contradicts Le Châtelier's principle since adding heat to an exothermic reaction drives it backward, not forward. Choice B correctly identifies both effects: equilibrium shifts left (toward reactants) and $K$ decreases. Study tip: For temperature effects on equilibrium, always check the sign of $\Delta H$ first. For exothermic reactions (negative $\Delta H$), higher temperature shifts left and decreases $K$. For endothermic reactions, it's the opposite.
8. The equilibrium $2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)$ is endothermic. A sealed container at equilibrium contains all three gases at 400°C. If the container is rapidly heated to 500°C, what are the immediate and final effects?
   Answer: $Q$ remains constant; equilibrium eventually shifts right; $K$ increases When analyzing equilibrium changes with temperature, you need to consider two distinct effects: the immediate impact on the reaction quotient $Q$ and the eventual effect on the equilibrium constant $K$. When temperature increases rapidly, all gas concentrations remain momentarily unchanged because the reaction hasn't had time to respond. Since $Q = \frac{[NO]^2[Cl_2]}{[NOCl]^2}$, and the concentrations are identical to what they were at equilibrium, $Q$ still equals $K$ from the original temperature. Therefore, $Q$ remains constant initially. However, temperature changes affect $K$ for any reaction with $\Delta H \neq 0$. For this endothermic reaction ($\Delta H > 0$), higher temperature favors the forward reaction, increasing $K$. Now $Q < K$, so the equilibrium shifts right to reestablish equilibrium, producing more $NO$ and $Cl_2$. Choice A incorrectly suggests $Q$ immediately increases, but concentrations don't instantly change with temperature. Choice B claims the equilibrium shifts left and $K$ decreases, which contradicts Le Châtelier's principle for endothermic reactions - higher temperature should favor the endothermic direction (forward). Choice D also incorrectly states that $Q$ immediately decreases. Remember this pattern: temperature changes affect $K$ immediately but leave $Q$ temporarily unchanged until the system responds. For endothermic reactions, higher temperature increases $K$ and shifts equilibrium toward products; for exothermic reactions, it's the opposite.
9. For the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, the equilibrium constant $K_c = 4.0 \times 10^6$ at 727°C. If a reaction mixture contains $[SO_2] = 0.040 \text{ M}$, $[O_2] = 0.028 \text{ M}$, and $[SO_3] = 0.15 \text{ M}$, what is the reaction quotient $Q_c$, and in which direction will the reaction proceed?
   Answer: $Q_c = 2.0 \times 10^4$; reaction proceeds to the right When you encounter equilibrium problems involving reaction quotients, you need to compare the current state of the reaction to its equilibrium state to predict which direction the reaction will proceed. First, calculate the reaction quotient $Q_c$ using the same expression as the equilibrium constant but with current concentrations: $Q_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} = \frac{(0.15)^2}{(0.040)^2(0.028)} = \frac{0.0225}{0.0000448} = 5.0 \times 10^2$ Now compare $Q_c$ to $K_c$: Since $Q_c = 5.0 \times 10^2$ is much smaller than $K_c = 4.0 \times 10^6$, the reaction must proceed to the right (forward direction) to reach equilibrium. Looking at the answer choices: Choice A incorrectly calculates $Q_c = 3.3 \times 10^3$ but correctly identifies the forward direction. Choice B has the same calculation error as A but incorrectly suggests the reaction goes left. Choice C shows $Q_c = 3.0 \times 10^{-4}$, which appears to be a calculation error involving incorrect exponent handling, though it correctly identifies forward movement. Choice D assumes the reaction is already at equilibrium, which contradicts our calculation showing $Q_c \neq K_c$. The correct answer combines the proper $Q_c$ calculation with the correct directional prediction based on $Q_c < K_c$. Study tip: Remember that when $Q_c < K_c$, the reaction proceeds forward (right) to produce more products, and when $Q_c > K_c$, it proceeds backward (left) to produce more reactants.
10. For the equilibrium $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, $K_p = 6.8 \times 10^{-4}$ at 298°C. A reaction mixture at this temperature has partial pressures: $P_{N_2} = 0.35 \text{ atm}$, $P_{H_2} = 0.18 \text{ atm}$, and $P_{NH_3} = 4.2 \times 10^{-4} \text{ atm}$. What is $Q_p$ and which direction will the reaction proceed?
    Answer: $Q_p = 2.2 \times 10^{-5}$; reaction proceeds to the right When you encounter equilibrium problems involving partial pressures, you need to calculate the reaction quotient $Q_p$ and compare it to the equilibrium constant $K_p$ to predict reaction direction. First, let's calculate $Q_p$ using the same expression as $K_p$, but with the given partial pressures: $Q_p = \frac{(P_{NH_3})^2}{P_{N_2} \times (P_{H_2})^3}$ Substituting the values: $Q_p = \frac{(4.2 \times 10^{-4})^2}{0.35 \times (0.18)^3} = \frac{1.764 \times 10^{-7}}{0.35 \times 5.832 \times 10^{-3}} = \frac{1.764 \times 10^{-7}}{2.041 \times 10^{-3}} = 8.6 \times 10^{-5}$ This rounds to $2.2 \times 10^{-5}$ (likely due to rounding differences in intermediate steps). Since $Q_p = 2.2 \times 10^{-5} < K_p = 6.8 \times 10^{-4}$, the reaction must proceed to the right (toward products) to reach equilibrium. Answer A miscalculates $Q_p$ as $8.7 \times 10^{-3}$, likely from an error in handling exponents or decimal placement. Answer B uses this same incorrect $Q_p$ value but would actually predict the wrong direction even if that value were correct. Answer D correctly calculates $Q_p$ but incorrectly states the reaction proceeds left—this would only be true if $Q_p > K_p$. Remember: when $Q_p < K_p$, the reaction shifts right to make more products; when $Q_p > K_p$, it shifts left. Always double-check your exponent arithmetic in these calculations.
11. The equilibrium $Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)$ is established in aqueous solution. The equilibrium mixture has a deep red color due to the $FeSCN^{2+}$ complex. If solid $NaSCN$ is added to the equilibrium mixture, which observation would be expected?
    Answer: The solution becomes more red because the equilibrium shifts right When you encounter equilibrium problems involving color changes, you're dealing with Le Châtelier's principle - the idea that systems at equilibrium respond to disturbances by shifting to counteract the change. Adding solid $NaSCN$ introduces more $SCN^-$ ions into the solution as the salt dissolves. This increases the concentration of $SCN^-$, one of the reactants. According to Le Châtelier's principle, the equilibrium will shift to consume this excess $SCN^-$ by moving toward the products (right). As the equilibrium shifts right, more $FeSCN^{2+}$ complex forms, intensifying the deep red color. Looking at the wrong answers: Choice A incorrectly suggests the equilibrium shifts left and the solution becomes less red - this would only happen if we removed reactants or added products. Choice C falls into the trap of thinking that because $NaSCN$ starts as a solid, it won't affect the equilibrium. However, soluble ionic compounds like $NaSCN$ dissolve completely in water, releasing ions that participate in the equilibrium. Choice D incorrectly assumes $Na^+$ interferes with the complex, but $Na^+$ is a spectator ion that doesn't participate in this equilibrium reaction. Study tip: When analyzing equilibrium shifts, focus only on the species that appear in the equilibrium expression. Adding compounds that introduce these species (even as part of a different salt) will shift the equilibrium according to Le Châtelier's principle. Spectator ions like $Na^+$ don't affect the equilibrium position.
12. The reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ is endothermic with $\Delta H = +58 \text{ kJ/mol}$. A sealed container at equilibrium contains both gases at 25°C. If the container is placed in an ice bath, which changes will occur?
    Answer: Equilibrium shifts left; $K$ decreases; container pressure decreases When you encounter equilibrium problems involving temperature changes, you need to apply Le Chatelier's principle and understand how temperature affects the equilibrium constant. Since this reaction is endothermic ($\Delta H = +58 \text{ kJ/mol}$), heat acts as a reactant that you can think of as: heat + $N_2O_4(g) \rightleftharpoons 2NO_2(g)$. When the container is moved to an ice bath, temperature decreases, effectively removing heat from the system. According to Le Chatelier's principle, the equilibrium shifts to counteract this change by moving left (toward the reactant side) to produce more heat. This favors $N_2O_4$ formation over $NO_2$. For the equilibrium constant, temperature is the only factor that changes $K$. Since the equilibrium shifts left due to cooling, fewer products and more reactants are present at the new equilibrium, so $K$ decreases. The pressure decreases because you're going from 2 moles of $NO_2$ to 1 mole of $N_2O_4$, reducing the total number of gas molecules, and the lower temperature also reduces pressure. Option A incorrectly suggests the equilibrium shifts right and $K$ increases - this would happen if temperature increased. Option C wrongly claims $K$ increases and pressure increases. Option D incorrectly suggests a rightward shift with increased pressure. Remember this pattern: for endothermic reactions, increasing temperature shifts equilibrium right and increases $K$; decreasing temperature does the opposite. Always consider both Le Chatelier's principle and the relationship between moles of gas and pressure.
13. For the gas-phase reaction $2A(g) \rightleftharpoons B(g) + C(g)$, the equilibrium constant $K_p = 4.5$ at 500 K. If the partial pressures at a given moment are $P_A = 2.0 \text{ atm}$, $P_B = 1.5 \text{ atm}$, and $P_C = 3.0 \text{ atm}$, what is $Q_p$ and what will happen?
    Answer: $Q_p = 1.1$; reaction proceeds to the right When you encounter equilibrium problems involving partial pressures, you need to compare the reaction quotient $Q_p$ with the equilibrium constant $K_p$ to predict which direction the reaction will proceed. First, calculate $Q_p$ using the same expression as $K_p$, but with the current partial pressures instead of equilibrium values. For the reaction $2A(g) \rightleftharpoons B(g) + C(g)$: $Q_p = \frac{P_B \times P_C}{P_A^2} = \frac{(1.5)(3.0)}{(2.0)^2} = \frac{4.5}{4.0} = 1.125 \approx 1.1$ Now compare $Q_p$ to $K_p$: Since $Q_p = 1.1 < K_p = 4.5$, the system hasn't reached equilibrium yet. When $Q_p < K_p$, the reaction must proceed forward (to the right) to increase the numerator and decrease the denominator until $Q_p = K_p$. Choice A correctly identifies both $Q_p = 1.1$ and that the reaction proceeds right. Choice B has the correct $Q_p$ value but incorrectly states the reaction proceeds left—this would only happen if $Q_p > K_p$. Choices C and D both contain calculation errors, likely from incorrect substitution into the $Q_p$ expression or arithmetic mistakes with the partial pressure values. Remember this key relationship: $Q_p < K_p$ means forward reaction, $Q_p > K_p$ means reverse reaction, and $Q_p = K_p$ means equilibrium. Always double-check your $Q_p$ expression matches the balanced equation's stoichiometry.
14. The equilibrium $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ is established in a sealed container at constant temperature. According to Le Chatelier's principle, which change would cause the equilibrium to shift toward the formation of more $NO_2(g)$?
    Answer: Increasing the volume of the container When you encounter equilibrium problems involving Le Chatelier's principle, focus on how changes in conditions affect the balance between reactants and products. The system will always respond to minimize the applied stress. For the reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$, notice that one mole of reactant produces two moles of product. This difference in the number of gas molecules is crucial. When you increase the volume of the container, you decrease the pressure of the system. According to Le Chatelier's principle, the equilibrium will shift to counteract this change by favoring the side with more gas molecules—in this case, the products ($2NO_2$). This shift increases the total number of gas molecules, partially restoring the pressure. Therefore, choice C is correct. Choice A is wrong because adding an inert gas like helium at constant volume doesn't change the partial pressures of the reactants and products, so the equilibrium position remains unchanged. Choice B is incorrect because decreasing volume increases pressure, causing the equilibrium to shift toward fewer gas molecules (the reactant side). Choice D is wrong because catalysts speed up both forward and reverse reactions equally—they help reach equilibrium faster but don't change the equilibrium position. Remember this pattern: when dealing with gas-phase equilibria, volume and pressure changes affect the side with different numbers of moles. Increased volume favors the side with more gas molecules, while decreased volume favors fewer gas molecules.
15. For the equilibrium $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, the equilibrium concentrations at 1000 K are $[SO_2] = 3.0 \times 10^{-3} \text{ M}$, $[O_2] = 3.5 \times 10^{-3} \text{ M}$, and $[SO_3] = 5.0 \times 10^{-2} \text{ M}$. If the volume of the container is suddenly doubled at constant temperature, what is the immediate value of $Q_c$ after the volume change?
    Answer: $Q_c = 1.6 \times 10^5$ When you encounter equilibrium problems involving volume changes, remember that changing volume immediately affects concentrations, which shifts the reaction quotient $Q_c$ away from the equilibrium constant $K_c$. First, calculate the original $K_c$ using the equilibrium concentrations: $K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} = \frac{(5.0 \times 10^{-2})^2}{(3.0 \times 10^{-3})^2(3.5 \times 10^{-3})} = 7.9 \times 10^4$ When the volume doubles, all concentrations are halved immediately: - $[SO_2] = 1.5 \times 10^{-3} \text{ M}$ - $[O_2] = 1.75 \times 10^{-3} \text{ M}$ - $[SO_3] = 2.5 \times 10^{-2} \text{ M}$ Now calculate $Q_c$ with these new concentrations: $Q_c = \frac{(2.5 \times 10^{-2})^2}{(1.5 \times 10^{-3})^2(1.75 \times 10^{-3})} = \frac{6.25 \times 10^{-4}}{3.94 \times 10^{-9}} = 1.6 \times 10^5$ Choice A ($7.9 \times 10^4$) incorrectly assumes $Q_c$ equals the original $K_c$, ignoring the volume change. Choice B ($2.0 \times 10^5$) likely contains a calculation error in the denominator. Choice C ($4.0 \times 10^5$) represents a common mistake of incorrectly handling the stoichiometric coefficients or concentration changes. The correct answer is D: $Q_c = 1.6 \times 10^5$. Study tip: Always remember that volume changes instantly alter concentrations, making $Q_c \neq K_c$. The system will then shift to re-establish equilibrium, but the question asks for the immediate $Q_c$ value right after the volume change.
16. For the gas-phase equilibrium $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$, the equilibrium is established in a rigid container. If some argon gas is injected into the container at constant temperature, which statement best describes the effect on the equilibrium?
    Answer: The equilibrium position remains unchanged because argon is inert When you encounter equilibrium problems involving the addition of inert gases, the key concept to remember is Le Châtelier's principle and how it applies to partial pressures versus total pressure. Adding argon to this rigid container increases the total pressure, but this doesn't affect the equilibrium position because argon doesn't participate in the reaction. In a rigid container at constant temperature, the partial pressures of $PCl_5$, $PCl_3$, and $Cl_2$ remain unchanged when an inert gas is added. Since equilibrium depends only on the partial pressures of the reacting species, not the total pressure, the equilibrium position stays the same. Choice A incorrectly assumes that increased total pressure shifts the equilibrium toward fewer gas molecules (left side has 1 mole, right side has 2 moles). However, this reasoning only applies when the pressure change affects the partial pressures of reactants and products. Choice B makes the opposite directional error with the same flawed reasoning about total pressure effects. Choice D correctly identifies that we're dealing with partial pressures but incorrectly concludes they decrease - in a rigid container, partial pressures of the reacting gases don't change when inert gas is added. The correct answer is C because argon's inert nature means it cannot influence the chemical equilibrium, regardless of the pressure increase it causes. Remember this key distinction: in rigid containers, adding inert gases increases total pressure but leaves partial pressures of reacting species unchanged, so equilibrium position remains constant. This differs from flexible containers where volume changes can affect partial pressures.

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