For the reaction, $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$
Equilibrium constant, $\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}$
Some reactions are written below in Column I and their equilibrium constants in terms of $K_c$ are written in Column II. Match the following reactions with the corresponding equilibrium constant.
| Column I (Reaction) |
Column II (Equilibrium constant) |
||
|---|---|---|---|
| A. | $$ 2 \mathrm{~N}_2(\mathrm{~g})+6 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NH}_3(\mathrm{~g}) $$ |
1. | $$ 2 K_c $$ |
| B. | $$ 2 \mathrm{NH}_3(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) $$ |
2. | $$ K_c^{1 / 2} $$ |
| C. | $$ \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g}) $$ |
3. | $$ \frac{1}{K_c} $$ |
| 4. | $$ K_c^2 $$ |
$$\mathrm{A.\to(4)\quad B.\to(3)\quad C.\to(2)}$$
For the reaction,
$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$
Equilibrium constant $K_{\mathrm{c}}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3}$
A. The given reaction $\left[2 \mathrm{~N}_2(g)+6 \mathrm{H}_2(g) \rightleftharpoons 4 \mathrm{NH}_3(g)\right]$ is twice the above reaction. Hence, $K=K_c^2$
B. The reaction $\left[2 \mathrm{NH}_3(g) \rightleftharpoons \mathrm{N}_2(g)+3 \mathrm{H}_2(g)\right]$ is reverse of the above reaction. Hence, $K=\frac{1}{K_c}$
C. The reaction $\left[\frac{1}{2} \mathrm{~N}_2(g)+\frac{3}{2} \mathrm{H}_2(g) \rightleftharpoons \mathrm{NH}_3(g)\right]$ is half of the above reaction. Hence, $K=\sqrt{K_c}=K_c^{\frac{1}{2}}$.
Match standard free energy of the reaction with the corresponding equilibrium constant.
| A. | $$ \Delta G^{\ominus}>0 $$ |
1. | $$K > 1$$ |
|---|---|---|---|
| B. | $$ \Delta G^{\ominus}<0 $$ |
2. | $$K=1$$ |
| C. | $$ \Delta G^{\ominus}=0 $$ |
3. | $$K=0$$ |
| 4. | $$K < 1$$ |
A. $\rightarrow(4)$
B. $\rightarrow$ (1)
C. $\rightarrow(2)$
As we know that, $\Delta G^{\ominus}=-R T \ln K$
A. If $\Delta G^{\circ}>0$, i.e., $\Delta G^{\circ}$ is positive, then $\ln K$ is negative i.e., $K<1$.
B. If $\Delta G^{\circ}<0$, i.e., $\Delta G^{\circ}$ is negative then $\ln K$ is positive i.e., $K>1$.
C. If $\Delta G^{\ominus}=0, \ln K=0$, i.e., $K=1$.
Match the following species with the corresponding conjugate acid.
| Species | Conjugate acid | ||
|---|---|---|---|
| A. | $$ \mathrm{NH}_3 $$ |
1. | $$ \mathrm{CO}_3^{2-} $$ |
| B. | $$ \mathrm{HCO}_3^{-} $$ |
2. | $$ \mathrm{NH}_4^{+} $$ |
| C. | $$ \mathrm{H}_2 \mathrm{O} $$ |
3. | $$ \mathrm{H}_3 \mathrm{O}^{+} $$ |
| D. | $$ \mathrm{HSO}_4^{-} $$ |
4. | $$ \mathrm{H}_2 \mathrm{SO}_4 $$ |
| 5. | $$ \mathrm{H}_2 \mathrm{CO}_3 $$ |
A. $\rightarrow$ (2)
B. $\rightarrow(5)$
C. $\rightarrow(3)$
D. $\rightarrow$ (4)
As conjugate acid $\rightarrow$ Base $+\mathrm{H}^{+}$
A. $\mathrm{NH}_3+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_4^{+}$
B. $\mathrm{HCO}_3^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{CO}_3$
C. $\mathrm{H}_2 \mathrm{O}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}$
D. $\mathrm{HSO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{SO}_4$
Match the following graphical variation with their description.
| A | B | ||
|---|---|---|---|
| A. |  |
1. | Variation in product concentration with time |
| B. |  |
2. | Reaction at equilibrium |
| C. |  |
3. | Variation in reactant concentration with time |
A. $\rightarrow$ (3)
B. $\rightarrow(1)$
C. $\rightarrow(2)$
A. Graph (A) represents variation of reactant concentration with time.
B. Graph (B) represents variation of product concentration with time.
C. Graph (C) represents reaction at equilibrium.
Match the Column I with Column II.
| Column I | Column II | ||
|---|---|---|---|
| A.. | Equilibrium | 1. | $$ \Delta G>0, K<1 $$ |
| B. | Spontaneous reaction | 2. | $$\Delta G=0$$ |
| C. | Non-spontaneous reaction | 3. | $$\Delta G^\circ =0$$ |
| 4. | $$\Delta G < 0, K > 1$$ |
A. $\rightarrow(2,3)$
B. $\rightarrow$ (4)
C. $\rightarrow(1)$
A. $\Delta G\left(\Delta G^{\ominus}\right)$ is 0 , reaction has achieved equilibrium: at this point, there is no longer any free energy left to drive the reaction.
B. If $\Delta G<0$, then $K>1$ which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.
C. If $\Delta G>0$, then $K<1$, which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.