The effect of temperature on equilibrium constant is expressed as $\log [\frac{K_{2}}{K_{1}}] = \frac{- ΔH}{2.303} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]; (T_{2} > T_{1})$
For endothermic reaction false statement is :

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$[\frac{1}{T_{2}} - \frac{1}{T_{1}}] = positive$

$K_{2} > K_{1}$

$\log K_{2} > \log K_{1}$

$ΔH = positive$

answer is A.

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Detailed Solution

Given That , $\log \frac{K_{2}}{K_{1}} = \frac{- ΔH}{2.303} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$

For endothermic reaction $ΔH > 0$ ,

Now for, $T_{2} > T_{1}$

$\Rightarrow \log (\frac{K_{2}}{K_{1}}) = \frac{ΔH}{2.303} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

$\Rightarrow \log (\frac{K_{2}}{K_{1}}) > 0 \Rightarrow \log K_{2} - \log K_{1} > 0$

$\begin{aligned} \Rightarrow \log K_{2} > \log K_{1} \\ \Rightarrow K_{2} > K_{1} \end{aligned}$

$[\frac{1}{T_{2}} - \frac{1}{T_{1}}] < 0 since T_{2} > T_{1}$

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