For a given exothermic reaction - K - p - and - - K - - - - p - are the equilibrium constants temperature - T - 1 - and - T - 2 - respectively- Assuming that heat of reaction is constant in temperature range between 183- - T - 1 - and - T - 2 - it is readily observed that- K - p - - - K - - - - p - K - p - - - - K - - - - p - K - p - - - - K - - - - p - K - p - 1- - K - - - - p -

Question

Toppr Admin · Accepted Answer

The equilibrium constant at two different temperatures for thermodynamic process is given by

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

Here $K_{1}$ and $K_{2}$ are replaced by $K_{p}$ and $K_{p}^{'}$
Therefore,

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

For exothermic reaction, $T_{2} > T_{1}$ and $Δ H = - v e$

$K_{p} > K_{p}^{'}$

Option C is correct.

For a given exothermic reaction Kp and K′p are the equilibrium constants at temperature T1 and T2 respectively. Assuming that heat of reaction is constant in temperature range between 183. T1 and T2 it is readily observed that:Kp=K′pKp>K′pKp<K′pKp=1/K′p

The equilibrium constant at two different temperatures for thermodynamic process is given bylogK2K1=ΔH2.303R[1T1−1T2]Here K1 and K2 are replaced by Kp and K′pTherefore,logK′pKp=ΔH2.303R[1T1−1T2]For exothermic reaction, T2>T1 and ΔH=−veKp>K′pOption C is correct.