The effect of temperature on equilibrium constant is expressed  as $\log \left[\frac{K_2}{K_1}\right]=\frac{-\mathrm{\Delta H}}{2.303}\left[\frac{1}{T_2}-\frac{1}{T_1}\right];\left(T_2>T_1\right)$

For endothermic reaction false statement is :

Given That ,$\log \frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=\frac{-\mathrm{\Delta H}}{2.303}\left[\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right]$

For endothermic reaction $\mathrm{\Delta H}>0$,

Now for,  ${\mathrm{T}}_2>{\mathrm{T}}_1$

$\Rightarrow \log \left(\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}\right)=\frac{\mathrm{\Delta H}}{2.303}\left[\frac{{\mathrm{T}}_2-{\mathrm{T}}_1}{{\mathrm{T}}_1{\mathrm{T}}_2}\right]$

$\Rightarrow \log \left(\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}\right)>0\Rightarrow \log {\mathrm{K}}_2-\log {\mathrm{K}}_1>0$

$\begin{array}{r}\Rightarrow \log {\mathrm{K}}_2>\log {\mathrm{K}}_1\\ \Rightarrow {\mathrm{K}}_2>{\mathrm{K}}_1\end{array}$

$\left[\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right]<0\text{ since }{\mathrm{T}}_2>{\mathrm{T}}_1$