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Chemical Equilibrium question

2014 · Q111
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Chemical Equilibrium question

2014 · Q111

NEETChemistryChemical EquilibriumMCQ+4 / −1
For a given exothermic reaction, Kp and K'p are the equilibrium constants at temperatures T1 and T2, respectively. Assuming that heat of reaction is constant in temperature range between T1 and T2, it is readily observed that
  1. A
    Kp > K'p
  2. B
    Kp < K'p
  3. C
    Kp = K'p
  4. D
    Kp = 1kp′{1 \over {k{'_p}}}kp′​1​
View written solutionFree

Correct answer: A

log

K'p
Kp

= $$ - {{\Delta H} \over {2.303R}}\left[ {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right]$$

For exothermic reaction, $\Delta $H = -ve means the temperature T2 is higher than T1.

$ \therefore $ $$\left[ {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right]$$ is negative.

So log K'p - log Kp = -ve

$ \Rightarrow $ log Kp > log K'p

$ \Rightarrow $ Kp > K'p
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