Question

Effect of increasing temperature on equilibrium constant is given by
$\log {K_2} - \log {K_1} = \dfrac{{ - \Delta H}}{{2.303R}}\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]$
Then for an endothermic reaction the false statement is:
A. $\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right] = $ positive
B. $\log {K_2} > \log {K_1}$
C. $\Delta H = $ positive
D. ${K_2} > {K_1}$

Answer 1

Hint:This question gives the knowledge about the temperature dependent activation energy equation. Activation energy is the least amount of energy required to energize or activate atoms or molecules in order to proceed a chemical reaction.

Formula used:
The temperature dependent activation energy equation is as follows:
$\log {K_2} - \log {K_1} = \dfrac{{ - \Delta H}}{{2.303R}}\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]$
Where ${K_2}$ is second rate, ${K_1}$ is the first rate, $\Delta H$ is the change in enthalpy, $R$ is the gas constant, ${T_2}$ is the final temperature and ${T_1}$ is the initial temperature.

Complete step by step answer:
Activation energy is the least amount of energy required to energize or activate atoms or molecules in order to proceed a chemical reaction. With the increase in temperature, randomness increases which means the atoms or molecules start moving faster. Therefore, with this increase in temperature the probability of molecules to move with appropriate activation energy for collision also increases.

The temperature dependent activation energy equation is as follows:
$\log {K_2} - \log {K_1} = \dfrac{{ - \Delta H}}{{2.303R}}\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]$
In endothermic reactions generally absorption of heat energy takes place which means the enthalpy will be positive.
$ \Rightarrow \Delta H = $ positive
Also for endothermic reactions final temperature is higher than initial temperature which means
$ \Rightarrow {T_2} > {T_1}$

Consider the temperature change from the equation of activation energy as follows:
$ \Rightarrow \left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]$
On rearrangement, we get
$ \Rightarrow \left[ {\dfrac{{{T_1} - {T_2}}}{{{T_2}{T_1}}}} \right] < 0$
This concludes that $\left[ {\dfrac{1}{{{T_2}}} - \dfrac{1}{{{T_1}}}} \right]$ is negative.
Consider the rate from the equation of activation energy as follows:
$ \Rightarrow \log {K_2} - \log {K_1} > 0$
On rearranging, we get
$ \Rightarrow \log {K_2} > \log {K_1}$

Hence, option $A$ is the correct option.

Note:
Always remember the concept that with the increase in temperature, randomness increases which means the atoms or molecules start moving faster. Therefore, with this increase in temperature the probability of molecules to move with appropriate activation energy for collision also increases.