Variation of $K$ with temperature as given by Van't Hoff equation can be written as:

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

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

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

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Solution

The correct option is B $log \frac{K_{2}}{K_{1}} = - \frac{Δ H}{2.303} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$
Variation of $K$ with temperature as given by Van't Hoff equation can be written as
$log \frac{K_{2}}{K_{1}} = - \frac{Δ H}{2.303} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$
or
$log \frac{K_{2}}{K_{1}} = \frac{Δ H}{2.303} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
or
$log \frac{K_{2}}{K_{1}} = - \frac{Δ H}{2.303} [\frac{T_{1} - T_{2}}{T_{1} T_{2}}]$
Here, $K_{1}$ and $K_{2}$ are the values of the equilibrium constants at temperatures $T_{1}$ and $T_{2}$ . $Δ H$ is the enthalpy change for the reaction.

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Similar questions

(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}}]

a A + b B ⇌ c C + d D

K

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

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) $k = \frac{1}{t} ([A]_{0} - [A])$

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

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

a A + b B ⇌ c C + d D

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

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