NEET Solved Paper 2014 Question 15
Question: 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 [AIPMT 2014]
Options:
A) $ K _{p}>K_p^{’} $
B) $ K _{p}<K_p^{’} $
C) $ K _{p}=K_p^{’} $
D) $ K _{p}=\frac{1}{K_p^{’}} $
Show Answer
Answer:
Correct Answer: A
Solution:
The equilibrium constant at two different temperatures for a thermodynamic process is given by
$ \log \frac{k _2}{k _1}=\frac{\Delta {H^{{}^\circ }}}{2.303R}[ \frac{1}{T _1}-\frac{1}{T _2} ] $ Here, $ K _1 $ and $ K _2 $ are replaced by $ K _{p} $ and $ K{’ _{p}}. $
Therefore, $ \log \frac{K{’ _{p}}}{K _{p}}=\frac{\Delta H{}^\circ }{2.303R}[ \frac{1}{T _1}-\frac{1}{T _2} ] $
For exothermic reaction, $ T _2>T _1 $ and $ \Delta H=-ve $
$ \therefore $ $ K _{p}>K{’ _{p}} $
BETA
