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NEET · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
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
- A \(K_p > K'_p\)
- B \(K_p < K'_p\)
- C \(K_p = K'_p\)
- D \(K_p = \frac{1}{K'_p}\)
Answer & Solution
Correct Answer
(A) \(K_p > K'_p\)
Step-by-step Solution
Detailed explanation
\(\log \frac{K_{p}^{\prime}}{K_{p}}=-\frac{\Delta H}{2.303 R}\left[\frac{1}{T_{2}}-\frac{1}{T_{1}}\right]\) For exothermic reaction, \(\Delta \mathrm{H}= -ve\) i.e. heat is evolved. The temperature \(\mathrm{T}_{2}\) is higher than \(\mathrm{T}_{1}\)
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