For a given exothermic reaction, and are the equilibrium constants at temperatures and , respectively. Assuming that heat of reaction is constant in temperature range between and , it is readily observed that [2014]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
For a given exothermic reaction, $K_{p}$ and $K_{p}'$ are the equilibrium constants at temperatures $T_{1}$ and $T_{2}$ , respectively. Assuming that heat of reaction is constant in temperature range between $T_{1}$ and $T_{2}$ , it is readily observed that [2014]

1 $K_{p} > K_{p}'$

2 $K_{p} < K_{p}'$

3 $K_{p} = K_{p}'$

4 $K_{p} = \frac{1}{K_{p}'}$

Ans.
(1)

$\log \frac{K'_{p}}{K_{p}} = - \frac{Δ H}{2.303 R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

For exothermic reaction, $Δ H = - ve$ , i.e., heat is evolved. The temperature $T_{2}$ is higher than $T_{1}$ .

Thus, $(\frac{1}{T_{2}} - \frac{1}{T_{1}})$ is negative.

So, $\log K'_{p} - \log K_{p} = - ve$ or $\log K_{p} > \log K'_{p}$

or $K_{p} > K'_{p}$

Want Us to Email you About Special Offers & Updates?