Vant' Hoff equation shows the effect of temperature on equilibrium constants $K_{c}$ and $K_{p} .$ The $K_{p}$ varies with temperature according to the relation:

l o g \frac{K_{p_{2}}}{K_{p_{1}}} = \frac{△ H}{2.303 R} [\frac{T_{1} - T_{2}}{T_{1} T_{2}}]

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l o g \frac{K_{p_{2}}}{K_{p_{1}}} = \frac{△ H}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]

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l o g \frac{K_{p_{2}}}{K_{p_{1}}} = \frac{△ H}{2.303 R} [\frac{T_{1} T_{2}}{T_{1} - T_{2}}]

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l o g \frac{K_{p_{2}}}{K_{p_{1}}} = \frac{△ H}{2.303 R} [\frac{T_{1} T_{2}}{T_{2} - T_{1}}]

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Solution

The correct option is B $l o g \frac{K_{p_{2}}}{K_{p_{1}}} = \frac{△ H}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
$K_{p} = A e^{- △ H / R T}$
$l o g K_{e q .} = l o g A - \frac{△ H^{o}}{2.303 R T}$
On applying the limit of initial and final states, we will get the option (b).

$l o g \frac{K_{p_{2}}}{K_{p_{1}}} = \frac{△ H}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

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Similar questions

l o g K_{2} - l o g K_{1} = \frac{- Δ H}{2.303 R}

[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

K_{c}

K_{p} .

K_{p}

Find the relation between $t_{1} a n d t_{2}$ if normals at $(a t_{1}^{2}, 2 a t_{1}) a n d (a t_{2}^{2}, 2 a t_{2})$ meet on the parabola.

Column I	Column II
(A) Zero order	(i) $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
(B) First order	(ii) $Δ H > 0$
(C) Endothermic reaction	(iii) $k = \frac{2.303}{t} log \frac{[A]_{0}}{[A]}$
(D) Activation energy	(iv) $k = \frac{1}{t} ([A]_{0} - [A])$

(T_{2} > T_{1})

l o g \frac{K_{2}}{K_{1}} = \frac{- Δ H}{2.303} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]

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