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Calculate the standard Gibbs free energy change \mathrm{\left(\Delta G^{\circ}\right)}  for the reaction:

\mathrm{2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)}
Given the standard Gibbs free energy of formation \left(\Delta G_{f}^{\circ}\right)$ values: $\Delta G_{f}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right)=-237.2 \mathrm{~kJ} / \mathrm{mol}

\Delta G_{f}^{\mathrm{o}}\left(\mathrm{H}_{2}\right)=0 \mathrm{~kJ} / \mathrm{mol}
\Delta G_{f}^{\mathrm{o}}\left(\mathrm{O}_{2}\right)=0 \mathrm{~kJ} / \mathrm{mol}
 

Option: 1

474.4 \mathrm{~kJ} / \mathrm{mol}


Option: 2

11.70 \mathrm{~kJ} / \mathrm{mol}


Option: 3

20.40 \mathrm{~kJ} / \mathrm{mol}


Option: 4

85.20 \mathrm{~kJ} / \mathrm{mol}


Answers (1)

best_answer

The standard Gibbs free energy change \left(\Delta G^{\circ}\right) for the reaction can be calculated using the formula:
\Delta G^{\circ}=\sum \Delta G_{f}^{\circ}(\text { products })-\sum \Delta G_{f}^{\circ}(\text { reactants })

Given the reaction:
2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)

The change in standard Gibbs free energy is:
\Delta G^{\circ}=\left(2 \cdot \Delta G_{f}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right)\right)-\left(2 \cdot \Delta G_{f}^{\circ}\left(\mathrm{H}_{2}\right)+\Delta G_{f}^{\circ}\left(\mathrm{O}_{2}\right)\right)
Substitute the given values:
\Delta G^{\circ}=(2 \cdot(-237.2 \mathrm{~kJ} / \mathrm{mol}))-(2 \cdot 0 \mathrm{~kJ} / \mathrm{mol}+0 \mathrm{~kJ} / \mathrm{mol}) \Delta G^{\circ}=-474.4 \mathrm{~kJ} / \mathrm{mol}

Therefore, the standard Gibbs free energy change for the reaction is \Delta G^{\circ}=$ $-474.4 \mathrm{~kJ} / \mathrm{mol}

Therefore, the correct option is (1).

Posted by

vishal kumar

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