Answer:
\[\Delta {{H}^{{}^\circ
}}=\Delta {{U}^{{}^\circ }}+\Delta {{n}_{g}}RT\]
\[=-10.5+(-1)\times
8.314\times {{10}^{-3}}\times 298\]
\[=-12.978kJ\]
According to
Gibbs-Helmholtz equation:
\[\Delta
{{G}^{{}^\circ }}=\Delta {{H}^{{}^\circ }}-T\Delta {{S}^{{}^\circ }}\]
\[=-12.978-298(-44.1\times
{{10}^{-3}})\]
\[=+0.164kJ\]
Since, \[\Delta
G{}^\circ \text{ }=+ve\], hence the reaction is non-spontaneous.
You need to login to perform this action.
You will be redirected in
3 sec