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.
Question
For which of the equilibrium systems represented below will the amount of product(s) at equilibrium increase if the volume of the reaction vessel is increased at a constant temperature?
(A) \(PCl_{5}(g) \rightleftharpoons PCl_{ 3}(g) + Cl_{2} \rightleftharpoons (g)\)
(B)\( 2 NO(g) + O_{2}(g) \rightleftharpoons 2 NO_{2} (g)\)
(C) \(N_{2}(g) + O_{2}(g) \rightleftharpoons 2 NO(g)\)
(D) \(2 CO(g) \rightleftharpoons C(s) + CO_{2}(g)\)
▶️Answer/Explanation
Ans:A
The equilibrium of a reaction can be shifted by changing the volume of the reaction vessel according to Le Chatelier’s principle. If the volume is increased, the system will shift to the side with more moles of gas to counteract the change and decrease the pressure.
Looking at the given reactions:
(A) \(PCl_{5}(g) \rightleftharpoons PCl_{ 3}(g) + Cl_{2}(g)\) – The right side has more moles of gas.
(B) \(2 NO(g) + O_{2}(g) \rightleftharpoons 2 NO_{2}(g)\) – Both sides have the same number of moles of gas.
(C) \(N_{2}(g) + O_{2}(g) \rightleftharpoons 2 NO(g)\) – Both sides have the same number of moles of gas.
(D) \(2 CO(g) \rightleftharpoons C(s) + CO_{2}(g)\) – The left side has more moles of gas.
So, if the volume of the reaction vessel is increased, the equilibrium will shift to the side with more moles of gas. Therefore, the amount of product(s) at equilibrium will increase for reaction
(A) \(PCl_{5}(g) \rightleftharpoons PCl_{ 3}(g) + Cl_{2}(g)\). For reaction (D), the equilibrium will shift to the left, favoring the reactants. For reactions (B) and (C), the change in volume will not affect the position of the equilibrium as the number of moles of gas is the same on both sides.
Questions
The reactions represented above are carried out in sealed, rigid containers and allowed to reach equilibrium. If the volume of each container is reduced from 1.0 L to 0.5 L at constant temperature, for which of the reactions will the amount of product(s) be increased?
(A) Reaction A
(B) Reaction B
(C) Reaction C
(D) Reaction D
▶️Answer/Explanation
Ans: A
To determine how the change in volume affects the equilibrium position for each reaction, we can use Le Chatelier’s Principle. This principle states that if a system at equilibrium is subjected to a change in temperature, pressure, or concentration of reactants or products, the equilibrium will shift to counteract the effect of that change.
In this case, the volume of each container is reduced from \(1.0 \, \text{L}\) to \(0.5 \, \text{L}\). According to Le Chatelier’s Principle:
- If the volume of the container decreases, the equilibrium will shift in the direction that reduces the total number of gas molecules (or moles) to counteract the decrease in volume.
- Conversely, if the volume of the container increases, the equilibrium will shift in the direction that increases the total number of gas molecules.
A. \(4 \text{ HCl}(g) + \text{O}_2(g) \rightleftarrows 2 \text{Cl}_2(g) + 2 \text{H}_2\text{O}(g)\)
- In this reaction, the total number of gas molecules on the left side of the equation (5) is greater than the total number of gas molecules on the right side (4). Therefore, decreasing the volume will favor the side with fewer gas molecules, which is the right side. As a result, the amount of products will increase. So, the answer is (A) Reaction A.
B. \(\text{N}_2\text{O}_4(g) \rightleftarrows 2 \text{NO}_2(g)\)
- This reaction involves the same number of gas molecules on both sides of the equation (2). Therefore, the change in volume will not favor either the reactants or the products. The amount of products will remain unchanged.
C. \(\text{H}_2(g) + \text{I}_2(g) \rightleftarrows 2 \text{HI}(g)\)
- Similar to reaction B, this reaction also involves the same number of gas molecules on both sides of the equation (2). Therefore, the change in volume will not favor either the reactants or the products. The amount of products will remain unchanged.
D. \(2 \text{NH}_3(g) \rightleftarrows \text{N}_2(g) + 3 \text{H}_2(g)\)
- In this reaction, the total number of gas molecules on the left side of the equation (2) is less than the total number of gas molecules on the right side (4). Therefore, decreasing the volume will favor the side with fewer gas molecules, which is the left side. As a result, the amount of reactants will increase.
So, the correct answer is (A) Reaction A.