Daily Physicist

Today is the birthday of

• Joseph Fourier (1768-1830).
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Today’s Problem

Calculate the change in the Gibbs free energy of two moles of ammonia in the following reaction

\[2NH_3+\frac{3}{2}O_2 \rightarrow N_2 + 3H_2O\]

using the fact that the Gibbs free energy of formation for \(NH_3\) gas is \(\Delta_f G=-16.45\) kJ, for \(H_2O\) gas is \(\Delta_f G=-228.57\) kJ, and is zero for both \(O_2\) and \(N_2\).

Answer

\[\begin{eqnarray} \Delta G &=& \Delta_f G(\text{products}) - \Delta_f G(\text{reactants})\nonumber\\ &=& 3(-228.57\,\text{kJ})-2(-16.45\,\text{kJ})\nonumber\\ &=& -652.81\,\text{kJ}\nonumber\\ \end{eqnarray}\]