The standard enthalpies of formation of (g), (s) and (l) are -46, -155 and -285 kJ/mol, respectively — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

The reaction of ammonia with $CuO$ is: 2$NH_3$(g) + 3$CuO$(s) -> $N_2$(g) + 3Cu(s) + 3$H_2O$(l). The enthalpy of this reaction is: δ H° = [3 * delta_f H°($H_2O$, l) + 0 + 0] - [2 * delta_f H°($NH_3$, g) + 3 * delta_f H°($CuO$, s)] = [3 * (-285)] - [2 * (-46) + 3 * (-155)] = -855 - [-92 - 465] = -855 + 557 = -298 kJ. This is for 2 moles (34 g) of $NH_3$. For 6.80 g of $NH_3$: δ H = (-298 kJ / 34 g) * 6.80 g = -59.6 kJ.