The gas phase reaction, $2A_{\left(g\right)}⟹A_{2\left(g\right)}$ at 400K has  $\Delta G^0=+25.2KJmo{l}^{-1}$ The equilibrium constant $K_c$ for this reaction is _____ $\times {10}^{-2}$

[Use: $R=8.3Jmo{l}^{-1}{K}^{-1},\ln 10=2.3,{\log }_{10}2=0.30,1atm=1bar]$

[antilog $\left(-0.3\right)=0.501$]

As the reaction is in gaseous phase,

$\Delta G^0=+25.2kJ/mo{l}^{-1}$

So, $K_p$will be calculated from equation,

$\Delta G^0=-2.303RT\log K_p$

$25.2\times 1000\frac{J}{mol}=-2.303\times 8.3\frac{J}{molK}\times 400K\times \log K_p$

$\log K_p=\frac{-25.2\times 1000}{2.303\times 8.3\times 400}=-3.30$

$K_p={10}^{-3.30};K_p={10}^{-0.30}\times {10}^{-3};K_p=0.501\times {10}^{-3}$

$K_p=5.01\times {10}^{-4};K_p=K_c{\left(RT\right)}^{\Delta n_g}$

$5.01\times {10}^{-4}=K_c{\left(8.3\times 400\right)}^{-1}$

$K_c=5.01\times {10}^{-4}\times 8.3\times 400=166.3\times {10}^{-2}\approx 166\times {10}^{-2}$