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

\mathrm{2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)}

Given the standard enthalpy change \mathrm{(\Delta H^{\circ})} as -483.6 kJ/mol and the standard entropy change \mathrm{(\Delta S^{\circ})} as 188.8 J/(mol·K) for the reaction.

Option: 1

\mathrm{-540 kJ/mol}


Option: 2

\mathrm{-300 kJ/mol}


Option: 3

\mathrm{4000 kJ/mol}


Option: 4

\mathrm{-900\: kJ\: mol}


Answers (1)

best_answer

The standard Gibbs free energy change(\Delta G\degree) for a reaction can be calculated using the equation:

\mathrm{(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ})}

where \mathrm{\Delta H^{\circ}}is the standard enthalpy change, T is the temperature in Kelvin, and \mathrm{\Delta S^{\circ}}
is the standard entropy change.
Given:

\mathrm{\begin{aligned} \Delta H^{\circ} & =-483.6 \mathrm{~kJ} / \mathrm{mol} \\ \Delta S^{\circ} & =188.8 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \end{aligned}}

Assuming the temperature (T) is 298 K, substitute the values:

\mathrm{\Delta G^{\circ}=-483.6 \mathrm{~kJ} / \mathrm{mol}-(298 \mathrm{~K} \times 0.1888 \mathrm{~kJ} / \mathrm{mol})}

Solving for \mathrm{\Delta G^{\circ}}

\mathrm{\Delta G^{\circ}=-483.6 \mathrm{~kJ} / \mathrm{mol}-56.2704 \mathrm{~kJ} / \mathrm{mol}=-539.8704 \mathrm{~kJ} / \mathrm{mol}}

Therefore, the standard Gibbs free energy change \mathrm{\Delta G^{\circ}} for the reaction is approximately -539.8704 kJ/mol.
So, otion A is correct

Posted by

vinayak

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