Question 6.7: Calculate ΔG°R for the reaction in Example Problem 6.6 at 3...

Thermodynamics, Statistical Thermodynamics, & Kinetics

Calculate $\Delta G°_{R}$ for the reaction in Example Problem 6.6 at 325 K. Is the reaction spontaneous at this temperature?

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To calculate $\Delta G°_{R}$ at the elevated temperature, we use Equation (6.36) assuming that $\Delta H°_{R}$ is independent of T:

$\frac{\Delta G\left( T_{2} \right)}{T_{2}}=\frac{\Delta G\left( T_{1} \right)}{T_{1}} +\Delta H\left( T_{1} \right)\left( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right)$ (6.36)

\Delta G°_{R}\left( T_{2} \right)=T_{2}\left[ \frac{\Delta G°_{R}\left( 298.15 K \right)}{298.15 K} +\Delta H°_{R} \left( 298.15 K \right)\left( \frac{1}{T_{2}} -\frac{1}{298.15 K}\right)\right]

\Delta H°_{R}\left( 298.15 K \right) =\Delta H°_{f} \left( N_{2}O_{4},g \right) -2\Delta H°_{f}\left( NO_{2},g \right)

=11.1 KJ mol^{-1} -2 × |33.2 KJ mol^{-1} =-55.3 KJ mol^{-1}

\Delta G°_{R}\left( 325 K \right) =325 K × \left[ ^{\frac{-2.80 × 10^{3} J mol^{-1}}{298.15 K}}_{-55.3 × 10^{3} J mol^{-1} \left( \frac{1}{325 K}-\frac{1}{298.15 K} \right)} \right]

\Delta G°_{R}\left( 325 K \right) =1.92 KJ mol^{-1}

Because $\Delta G°_{R}\gt 0$ , the reaction is not spontaneous at 325 K.

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