van't Hoffs equation shows the effect of temperature on equilibrium constants $K_{c} and K_{p}$ the K _pvaries with temperature according to the relation

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$\log \frac{K_{p_{2}}}{K_{p_{1}}} = \frac{{ΔH}^{\circ}}{2 .303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$

$\log \frac{K_{p_{2}}}{K_{p_{1}}} = \frac{{ΔH}^{\circ}}{2 .303 R} (\frac{T_{1} T_{2}}{T_{1} - T_{2}})$

$\log \frac{K_{p_{2}}}{K_{p_{1}}} = \frac{{ΔH}^{\circ}}{2 .303 R} (\frac{T_{1} - T_{2}}{T_{1} T_{2}})$

$\log \frac{K_{p_{2}}}{K_{p_{1}}} = \frac{∆ H^{°}}{R} (\frac{T_{1} - T_{2}}{T_{1} T_{2}})$

answer is B.

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Detailed Solution

Vant hoff's equation =

$\frac{{dlnK}_{p}}{dT} = \frac{∆ H^{°}}{{RT}^{2}}$

On integrating and simplifying it changes to

$\log_{10} \frac{K_{2}}{K_{1}} = \frac{∆ H^{°}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

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