Question

If ${K_1}$ and ${K_2}$ are the equilibrium constants at temperatures ${T_1}$ and ${T_2}$ , where ${T_2} > {T_1}$ , then-
A.${K_1} = {K_2}$ when $\Delta H = 0$
B.${K_2} > {K_1}$ when $\Delta H$ is positive
C.${K_2} < {K_1}$ when $\Delta H$ is negative
D.${K_2} < {K_1}$ when $\Delta H$ is positive

Answer 1

We will use Van’t Hoff’s equation which gives the relation between the equilibrium constants,${K_1}$ and ${K_2}$, at temperatures ${T_1}$ and ${T_2}$ and enthalpy of reaction $\Delta H$ which is given as-
$\Rightarrow \log \dfrac{{{K_2}}}{{{K_1}}} = \dfrac{{\Delta H}}{{2.303R}}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]$ where the symbols have usual notations. Now we will check each option by putting the values of each option in the formula.

Complete step by step answer:
Given if ${K_1}$ and ${K_2}$ are the equilibrium constants at temperatures ${T_1}$ and ${T_2}$ , where ${T_2} > {T_1}$ then we have find the relation between ${K_1}$ and ${K_2}$.
We know Van’t Hoff equation which gives the relation between${K_1}$, ${K_2}$, ${T_1}$ ,${T_2}$ and enthalpy of reaction $\Delta H$ is given as-
$\Rightarrow \log \dfrac{{{K_2}}}{{{K_1}}} = \dfrac{{\Delta H}}{{2.303R}}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]$ where the symbols have usual notations
On simplifying the equation, we get-
$\Rightarrow \log \dfrac{{{K_2}}}{{{K_1}}} = \dfrac{{\Delta H}}{{2.303R}}\left[ {\dfrac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right]$-- (i)
Now, it is given that ${T_2} > {T_1}$
Now when we put $\Delta H = 0$in eq. (i), we see that the term of the right side will become zero so we can write-
$\Rightarrow \log \dfrac{{{K_2}}}{{{K_1}}} = 0$
Now on applying the logarithm rule-$\log \dfrac{m}{n} = \log m - \log n$ , we get-
$\Rightarrow \log {K_2} - \log {K_1} = 0$
On simplifying, we get-
$\Rightarrow \log {K_2} = \log {K_1}$
On further simplifying, we get-
$\Rightarrow {K_2} = {K_1}$
So option A is correct.
Now on putting $\Delta H > 0$ in eq. (i), we get-
$\Rightarrow \dfrac{{\Delta H}}{{2.303R}}\left[ {\dfrac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right] > 0$
Then since $\log \dfrac{{{K_2}}}{{{K_1}}} = \dfrac{{\Delta H}}{{2.303R}}\left[ {\dfrac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right]$ then we can write-
$\Rightarrow \log \dfrac{{{K_2}}}{{{K_1}}} > 0$
On simplifying, we get-
$\Rightarrow \log {K_2} - \log {K_1} > 0$
On simplifying further, we get-
$\Rightarrow \log {K_2} > \log {K_1}$
On removing log, we get-
$\Rightarrow {K_2} > {K_1}$
Hence option B is also correct.
Now, on putting $\Delta H < 0$ in eq. (i), we get-
$\Rightarrow \dfrac{{\Delta H}}{{2.303R}}\left[ {\dfrac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right] < 0$
Then since $\log \dfrac{{{K_2}}}{{{K_1}}} = \dfrac{{\Delta H}}{{2.303R}}\left[ {\dfrac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right]$ then we can write-
$\Rightarrow \log \dfrac{{{K_2}}}{{{K_1}}} < 0$
On simplifying, we get-
$\Rightarrow \log {K_2} - \log {K_1} < 0$
On simplifying further, we get-
$\Rightarrow \log {K_2} < \log {K_1}$
On removing log, we get-
$\Rightarrow {K_2} < {K_1}$
So option C is correct.

The correct options are A, B and C.

Note:
The Van’t Hoff equation tells about the following,-
-The equilibrium constants are dependent on the temperature for a given reaction.
- If the reaction is endothermic then the enthalpy of reaction is positive as energy is taken to facilitate the reaction.
- If the reaction is exothermic then the enthalpy of reaction is negative as energy is released during the reaction.