$$ \begin{aligned} & \text { Following data is given for the reaction } \\ & \mathrm{CaCO}_3(\mathrm{~s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \\ & \Delta_f H^{\ominus}[\mathrm{CaO}(s)]=-635.1 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_f H^{\ominus}\left[\mathrm{CO}_2(g)\right]=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_f H^{\ominus}\left[\mathrm{CaCO}_3(s)\right]=-1206.9 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$$
Predict the effect of temperature on the equilibrium constant of the above reaction.
Given that,
$\begin{aligned} \Delta_f H^{\ominus}[\mathrm{CaO}(s)] & =-635.1 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta_f H^{\ominus}\left[\mathrm{CO}_2(g)\right] & =-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta_f H^{\ominus}\left[\mathrm{CaCO}_3(s)\right] & =-1206.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$
In the reaction,
$\begin{gathered}\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(g) \\ \Delta_f H^{\ominus}=\Delta_f H^{\ominus}[\mathrm{CaO}(\mathrm{s})]+\Delta_f H^{\ominus}\left[\mathrm{CO}_2(g)\right]-\Delta_f H^{\ominus}\left[\mathrm{CaCO}_3(s)\right] \\ \therefore\quad\Delta_f H^{\ominus}=-635.1+(-393.5)-(-1206.9)=178.3 \mathrm{kJmol}^{-1}\end{gathered}$
Because $\Delta H$ value is positive, so the reaction is endothermic. Hence, according to Le-Chatelier's principle, reaction will proceed in forward direction on increasing temperature. Thus, the value of equilibrium constant for the reaction increases.
Match the following equilibria with the corresponding condition.
| A. | Liquid $$\rightleftharpoons$$ Vapour | 1. | Saturated solution |
|---|---|---|---|
| B. | Solid $$\rightleftharpoons$$ Liquid | 2. | Boiling point |
| C. | Solid $$\rightleftharpoons$$ Vapour | 3. | Sublimation point |
| D. | Solute (s) $$\rightleftharpoons$$ Solute (solution) | 4. | Melting point |
| 5. | Unsaturated solution |
A. $\rightarrow(2)$
B. $\rightarrow(4)$
C. $\rightarrow(3)$
D. $\rightarrow$ (1)
A. Liquid $\rightleftharpoons$ Vapour equilibrium exists at the boiling point.
B. Solid $\rightleftharpoons$ Liquid equilibrium exists at the melting point.
C. Solid $\rightleftharpoons$ Vapour equilibrium exists at the sublimation point.
D. Solute $(s) \rightleftharpoons$ Solute (solution) equilibrium exists at saturated solution.
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. $\mathrm{HSO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{SO}_4$