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  • Consider a chemical reaction that takes place at 298 K and standard pressure: <img alt="\mathrm{A(g)+2 B(g) \rightarrow C(g)+D(g)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5

Consider a chemical reaction that takes place at 298 K and standard pressure:

\mathrm{A(g)+2 B(g) \rightarrow C(g)+D(g)}

The standard molar enthalpy change \mathrm{(\Delta H^{\circ})} for the reaction is -400 kJ/mol. The standard molar entropy change \mathrm{(\Delta S^{\circ})} for the reaction is 600 J/(mol·K).
Calculate the standard Gibbs free energy change \mathrm{(\Delta G^{\circ})} for the reaction.

 

Option: 1

\mathrm{\Delta G^{\circ}=-400 \mathrm{~kJ} / \mathrm{mol}}


Option: 2

\mathrm{\Delta G^{\circ}} =-580 \mathrm{~kJ} / \mathrm{mol} \\


Option: 3

\mathrm{\Delta G^{\circ}} =-578.8 \mathrm{~kJ} / \mathrm{mol} \\


Option: 4

\mathrm{\Delta G^{\circ}} =400 \mathrm{~kJ} / \mathrm{mol}


Answers (1)

best_answer

The standard Gibbs free energy change \mathrm{(\Delta G^{\circ})} for a reaction is related to the standard enthalpy change \mathrm{(\Delta H^{\circ})} and the standard entropy change \mathrm{(\Delta S^{\circ})} by the equation:

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

Where: - \mathrm{\Delta G^{\circ}} is the standard Gibbs free energy change. - \mathrm{\Delta H^{\circ}} is the standard enthalpy change. - \mathrm{\Delta S^{\circ}} is the standard entropy change. - T is the temperature in Kelvin (298 K in this case). Given: \mathrm{\Delta H^{\circ}=-400 \mathrm{~kJ} / \mathrm{mol}-\Delta S^{\circ}=600 \mathrm{~J} /(\mathrm{mol\: K})}

Now let's calculate \mathrm{\Delta G^{\circ}} :

                                       \mathrm{\Delta G^{\circ}=-400 \times 1000 \mathrm{~J} / \mathrm{mol}-(298 \times 600 \mathrm{~J} / \mathrm{molK})}

\begin{aligned} & \mathrm{\Delta G^{\circ}} \approx-400000-178800 \approx-578800 \mathrm{~J} / \mathrm{mol} \\ & \mathrm{\Delta G^{\circ}} \approx-578.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}

So, the standard Gibbs free energy change \mathrm{(\Delta G^{\circ})} for the reaction is approximately −578.8 kJ/mol.
So, correct option (3)

Posted by

shivangi.shekhar

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