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

                               \mathrm{ 2 A(g)+3 B(g) \rightleftharpoons 4 C(g) }
given the following thermodynamic data at \mathrm{298 \mathrm{~K} :}

                                  \mathrm{ \begin{aligned} \Delta H_f^{\circ}(A) & =-100 \mathrm{~kJ} / \mathrm{mol} \\ \Delta H_f^{\circ}(B) & =-50 \mathrm{~kJ} / \mathrm{mol} \\ \Delta H_f^{\circ}(C) & =75 \mathrm{~kJ} / \mathrm{mol} \\ \Delta S_f^{\circ}(A) & =150 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ \Delta S_f^{\circ}(B) & =100 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ \Delta S_f^{\circ}(C) & =50 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ R & =8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \end{aligned} }

Option: 1

\mathrm{769.2 \, k J}


Option: 2

\mathrm{200.4 \, k J}


Option: 3

\mathrm{500.6 \, k J}


Option: 4

\mathrm{450.8 \, k J}


Answers (1)

best_answer

Step 1: Calculate \mathrm{\Delta H_{\text {rxn }}^{\circ}}

\mathrm{ \begin{gathered} \Delta H_{\mathrm{rxn}}^{\circ}=\sum n \Delta H_f^{\circ}(\text { products })-\sum n \Delta H_f^{\circ}(\text { reactants }) \\\\ \Delta H_{\mathrm{rxn}}^{\circ}=(4 \mathrm{~mol})\left(\Delta H_f^{\circ}(C)\right)-(2 \mathrm{~mol})\left(\Delta H_f^{\circ}(A)\right)-(3 \mathrm{~mol})\left(\Delta H_f^{\circ}(B)\right) \\\\ \Delta H_{\mathrm{rxn}}^{\circ}=(4 \mathrm{~mol})(75 \mathrm{~kJ} / \mathrm{mol})-(2 \mathrm{~mol})(-100 \mathrm{~kJ} / \mathrm{mol})-(3 \mathrm{~mol})(-50 \mathrm{~kJ} / \mathrm{mol}) \\\\ \Delta H_{\mathrm{rxn}}^{\circ}=300 \mathrm{~kJ}+200 \mathrm{~kJ}+150 \mathrm{~kJ}=650 \mathrm{~kJ} \end{gathered} }

Step 2: Calculate \mathrm{\Delta S_{\text {rxn }}^{\circ}}

\mathrm{ \begin{gathered} \Delta S_{\mathrm{rxn}}^{\circ}=\sum n \Delta S_f^{\circ}(\text { products })-\sum n \Delta S_f^{\circ}(\text { reactants }) \\\\ \Delta S_{\mathrm{rxn}}^{\circ}=(4 \mathrm{~mol})\left(\Delta S_f^{\circ}(C)\right)-(2 \mathrm{~mol})\left(\Delta S_f^{\circ}(A)\right)-(3 \mathrm{~mol})\left(\Delta S_f^{\circ}(B)\right) \\\\ \Delta S_{\mathrm{r} \times n}^{\circ}=(4 \mathrm{~mol})(50 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})-(2 \mathrm{~mol})(150 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})-(3 \mathrm{~mol})(100 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}) \\\\ \Delta S_{\mathrm{r} x \mathrm{n}}^{\circ}=200 \mathrm{~J} / \mathrm{K}-300 \mathrm{~J} / \mathrm{K}-300 \mathrm{~J} / \mathrm{K}=-400 \mathrm{~J} / \mathrm{K} \end{gathered} }

Step 3: Calculate \mathrm{\Delta G_{\mathrm{rxn}}^{\circ}} at \mathrm{298 \mathrm{~K}} using \mathrm{\Delta G_{\mathrm{rxn}}^{\circ}=\Delta H_{\mathrm{rxn}}^{\circ}-T \Delta S_{\mathrm{rxn}}^{\circ}}

\mathrm{ \begin{gathered} \Delta G_{\mathrm{rxn}}^{\circ}=650 \mathrm{~kJ}-(298 \mathrm{~K})(-0.4 \mathrm{~kJ} / \mathrm{K}) \\\\ \Delta G_{\mathrm{rxn}}^{\circ}=650 \mathrm{~kJ}+119.2 \mathrm{~kJ}=769.2 \mathrm{~kJ} \end{gathered} }

Thus, the change in Gibbs free energy for the reaction \mathrm{2 A(g)+3 B(g) \rightleftharpoons4 C(g)} at \mathrm{298 \mathrm{~K}} is \mathrm{769.2 \mathrm{~kJ}.}
Therefore, the correct option is 1 .

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mansi

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