The standard free energy of formation of at , representing the synthesis reaction , is exactly . Cal — Chemical Equilibrium Chemistry Question
Question
The standard free energy of formation of $NH_3(g)$ at $298\text{ K}$, representing the synthesis reaction $\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightleftharpoons NH_3(g)$, is exactly $-16.5\text{ kJ mol}^{-1}$. Calculate the numerical absolute value of $\Delta G^\circ$ in $\text{kJ mol}^{-1}$ for the Haber's process reaction written conventionally as $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$.
💡 Solution & Explanation
Standard free energy is an extensive property. The target reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$ is obtained by multiplying the given formation reaction by 2. Therefore, its standard free energy change $\Delta G^\circ = 2 \times (-16.5\text{ kJ mol}^{-1}) = -33.0\text{ kJ mol}^{-1}$. The question asks for the absolute value, which is 33.