Daily Physicist

Frederick Reines image courtesy of Los Alamos National Laboratory.

Today is the birthday of

• Georg Ohm (1789-1854).
• Frederick Reines (1918-1998).

Today’s Problem

Calculate the change in the Gibbs free energy of one mole of glucose in the following reaction

\[C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O\]

using the fact that the Gibbs free energy of formation for \(CO_2\) gas is \(\Delta_f G=-394.36\) kJ, for \(H_2O\) liquid is \(\Delta_f G=-237.13\) kJ, and for glucose is \(\Delta_f G=-910\) kJ.

Answer

\[\begin{eqnarray} \Delta G &=& \Delta_f G(\text{products}) - \Delta_f G(\text{reactants})\nonumber\\ &=& 6(-394.36\,\text{kJ}) + 6(-237.13\,\text{kJ}) -(-910\,\text{kJ})\nonumber\\ &=& -2878.94\,\text{kJ}\nonumber\\ \end{eqnarray}\]