Question

For the reaction, $X_2{O}_4\left(l\right)\to {2XO}_2\left(g\right)$, $\Delta U=2.1$ Kcal, $\Delta S=20cal{K}^{-1}$ at 300 K.
Hence, $\Delta G$ is:

Answer 1

The change in Gibbs Free energy is given by

$\Delta H=\Delta U+\Delta n_{g}RT$

where

$\Delta H$ is the enthalpy of the reaction

$\Delta S$ is the entropy of the reaction

and $\Delta U$ is the change in internal energy

$\Delta n_g$ is the (number of gaseous moles in product) - (number of gaseous moles in reactant)=2-0=2

R is the gas constant =2 cal

But,

$\begin{array}{c}\Delta H=\left(2.1\times {10}^3\right)+\left(2\times 2\times 300\right)=3300cal\\ Hence,\\ \Delta G=\Delta H=T\Delta S\\ \Delta G=3300-\left(300\times 20\right)\\ \Delta G=-2700cal=-2.7cal\end{array}$