At 25°C, for the reaction $2 \mathrm { H } _ { 2 } ( \mathrm {~g} ) + \mathrm { O } _ { 2 } ( \mathrm {~g} ) f \quad 2 \mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { l } )$ , the standard enthalpy change (?H°) is ?285.8 kJ/mol, and the absolute entropies of the products and reactants are S°(H₂O) = 70.00 J/(mol·K), S°(O₂) = 205.2 J/(mol·K), and S°(H₂) = 130.7 J/(mol·K). What is the standard free-energy change for the reaction?

Question

Quizplus · Accepted Answer

The standard free-energy change ($\Delta G^\circ$) for a reaction can be calculated using the equation $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, where $\Delta H^\circ$ is the standard enthalpy change, $T$ is the temperature in Kelvin, and $\Delta S^\circ$ is the standard entropy change. Given $\Delta H^\circ = -285.8$ kJ/mol, $T = 25°C = 298K$, and the absolute entropies, we first calculate $\Delta S^\circ$ for the reaction: $\Delta S^\circ = [2S°(H_2O) - (2S°(H_2) + S°(O_2))]$. Substituting the given values, $\Delta S^\circ = [2(70.00) - (2(130.7) + 205.2)] J/(mol·K)$ = $-326.6 J/(mol·K) = -0.3266 kJ/(mol·K)$. Then, $\Delta G^\circ = -285.8 kJ/mol - 298K(-0.3266 kJ/(mol·K)) = -285.8 kJ/mol + 97.4 kJ/mol = -188.4 kJ/mol$.

At 25°C, for the Reaction 2H2( g)+O2( g)f2H2O(l)2 \mathrm { H } _ { 2 } ( \mathrm {~g} ) + \mathrm { O } _ { 2 } ( \mathrm {~g} ) f \quad 2 \mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { l } )2H2​( g)+O2​( g)f2H2​O(l)

At 25°C, for the Reaction $2 \mathrm { H } _ { 2 } ( \mathrm {~g} ) + \mathrm { O } _ { 2 } ( \mathrm {~g} ) f \quad 2 \mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { l } )$