Question

The value of Kp, for water gas reaction is 1.06×105 at 298 K. Calculate the standard state Gibbs energy change of reaction at 298 K. Where value of R is 8.314 J K-1mol-1.

Answer 1

Given, $K_p=1.06\times 10^5$

$T=298K, R=8.314 JK^{-1}mol^{-1}$

The standard state Gibbs energy-

$\Delta G^{o}=-RTlnk_p$

$=-8.314\times 298\times ln (1.06\times 10^5)\\ =2477.572\times 5.0253 \\ =62.69KJ$