# Question 271f2

Jan 22, 2015

The answer is $\Delta {G}^{\circ} = + 24.5$ $\text{kJ/mol}$.

The standard-state free energy of reaction, or $\Delta {G}^{\circ}$, refers to the free energy of reaction at standard state conditions, and can be calculated using

$\Delta {G}^{\circ} = \Delta {H}^{\circ} - T \cdot \Delta {S}^{\circ}$

Again, this is as straighforward as it gets, just plug in your values and solve for $\Delta {G}^{\circ}$.

DeltaG^@ = "-50.0 kJ/mol" -"298.15 K" * ("-250" "J"/("mol" * "K"))#

$\Delta {G}^{\circ} = \text{-50.0 kJ/mol" + "74.5 kJ/mol" = "+24.5 kJ/mol}$