AP Chemistry 7.9 Introduction to Le Châtelier’s Principle - MCQs - Exam Style Questions
▶️ Answer/Explanation
1. Analyze the Stress:
Reducing the volume of the container (from \(1.0\) L to \(0.5\) L) at constant temperature causes an increase in pressure.
2. Apply Le Chatelier’s Principle:
According to Le Chatelier’s principle, when pressure is increased, the equilibrium will shift to the side of the reaction with the fewer total moles of gas to relieve the pressure.
3. Analyze the Goal:
We are looking for the reaction where the amount of product(s) is increased. This means we need the reaction that shifts to the right (product side) when pressure increases.
4. Count Moles of Gas for Each Reaction:
- (A) Reaction A:
- Reactant moles = \(4 + 1 = 5\) moles
- Product moles = \(2 + 2 = 4\) moles
- Result: Shifts right (to products) because \(4 < 5\).
- (B) Reaction B:
- Reactant moles = \(1\) mole
- Product moles = \(2\) moles
- Result: Shifts left (to reactants) because \(1 < 2\).
- (C) Reaction C:
- Reactant moles = \(1 + 1 = 2\) moles
- Product moles = \(2\) moles
- Result: Moles are equal. No shift.
- (D) Reaction D: (Assumed balanced: \(2~NH_3\))
- Reactant moles = \(2\) moles
- Product moles = \(1 + 3 = 4\) moles
- Result: Shifts left (to reactants) because \(2 < 4\).
Only Reaction A shifts to the product side.
✅ Answer: (A)
Questions
\(Ge(g)+2Cl_{2}(g)\rightleftharpoons GeCl_{4}(g)\)
The value of the equilibrium constant for the reaction represented above is 1 × 1010. What is the value of the equilibrium constant for the following reaction?
\(2Ge(g)Cl_{4}(g)\rightleftharpoons 2Ge(g)+4Cl_{2}(g)\)
(A) 1 × \(10^{-20}\)
(B) 1 × \(10^{-10}\)
(C) 1 × \(10^{10}\)
(D) 1 ×\( 10^{20}\)
▶️Answer/Explanation
Ans: A
To find the equilibrium constant (\(K_{\text{eq}}\)) for the second reaction from the equilibrium constant of the first reaction, we use the concept of equilibrium constant expressions and stoichiometry.
Given the first reaction:
\[\text{Ge(g)} + 2\text{Cl}_2(g) \rightleftharpoons |text{GeCl}_4(g) \]
with equilibrium constant \(K_1 =1 times 10^{10}\).
Now, for the second reaction:
\[2|text{Ge(g)} +4\text{Cl}_2(g) \rightleftharpoons 2|text{Ge(g)} +4\text{Cl}_2(g) \]
We can see that the second reaction is the reverse of the first reaction, multiplied by 2.
According to Le Chatelier’s principle, if we reverse a reaction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.
Therefore, for the second reaction:
\[ K_{\text{eq,2}} = \frac{1}{K_{1}^{2}} \]
Substituting the given value of \(K_1\):
\[K_{\text{eq,2}}=\frac{1}{(1\times 10^{10})^2} \]
\[K_{\text{eq,2}} = \frac{1}{1 \times 10^{20}} \]
Thus, the equilibrium constant for the second reaction is \(1 \times 10^{-20}\).
So, the answer is (A) \(1 \times 10^{-20}\)/
Question
\[
\mathrm{Li}_3 \mathrm{~N}(s)+2 \mathrm{H}_2(g) \rightleftarrows \mathrm{LiNH}_2(s)+2 \mathrm{LiH}(s) \quad \Delta H^{\circ}=-192 \mathrm{~kJ} / \mathrm{mol}_{r x n}
\]
Because pure \(\mathrm{H}_2\) is a hazardous substance, safer and more cost effective techniques to store it as a solid for shipping purposes have been developed. One such method is the reaction represented above, which occurs at \(200^{\circ} \mathrm{C}\).
The amount of\( H_2\)(g) present in a reaction mixture at equilibrium can be maximized by
(A) increasing the temperature and increasing the pressure by decreasing the volume
(B) increasing the temperature and decreasing the pressure by increasing the volume
(C) decreasing the temperature and increasing the pressure by decreasing the volume
(D) decreasing the temperature and decreasing the pressure by increasing the volume
▶️Answer/Explanation
Ans:C
\[ \mathrm{Li}_3 \mathrm{~N}(s) + 2 \mathrm{H}_2(g) \rightleftarrows \mathrm{LiNH}_2(s) + 2 \mathrm{LiH}(s) \]
This reaction involves the decomposition of lithium amide (\( \mathrm{Li}_3 \mathrm{N} \)) and the formation of lithium imide (\( \mathrm{LiNH}_2 \)) and lithium hydride (\( \mathrm{LiH} \)).
From the stoichiometry of the reaction, we can see that \( 2 \) moles of \( H_2(g) \) are consumed to form the products. According to Le Chatelier’s Principle, to maximize the amount of \( H_2(g) \) present in the equilibrium mixture, we need to shift the equilibrium to the left, favoring the formation of more reactants.
Now, let’s analyze the given options:
(A) Increasing the temperature and increasing the pressure by decreasing the volume: This would likely favor the forward reaction according to Le Chatelier’s Principle because the forward reaction is endothermic (heat is absorbed). Therefore, this option would not maximize the amount of \( H_2(g) \) present in the equilibrium mixture.
(B) Increasing the temperature and decreasing the pressure by increasing the volume: Increasing the temperature would favor the endothermic reaction, but decreasing the pressure would not favor the formation of more reactants because there are fewer moles of gas on the reactant side compared to the product side. So, this option would also not maximize the amount of \( H_2(g) \) present.
(C) Decreasing the temperature and increasing the pressure by decreasing the volume: Decreasing the temperature would favor the exothermic reaction (the reverse reaction), and increasing the pressure by decreasing the volume would favor the side with fewer moles of gas molecules, which is the reactant side in this case. This option would shift the equilibrium to the left, favoring the formation of more \( H_2(g) \). Therefore, this option seems promising.
(D) Decreasing the temperature and decreasing the pressure by increasing the volume: Similar to option (B), decreasing the pressure would not favor the formation of more reactants, so this option would not maximize the amount of \( H_2(g) \) present.
Therefore, the correct answer is:
(C) Decreasing the temperature and increasing the pressure by decreasing the volume.