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  • Calculate the standard Gibbs free energy change <img alt="\mathrm{\left(\Delta G^{\circ}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%28%5CDelta%20G%5E%7B%5Ccir

Calculate the standard Gibbs free energy change \mathrm{\left(\Delta G^{\circ}\right)} for the following eaction at \mathrm{298 \mathrm{~K} :}

                         \mathrm{ 2 A(g)+3 B(g) \rightarrow 4 C(g)+2 D(g) }

 Given the standard Gibbs free energy of formation values:
     

                         \mathrm{ \begin{aligned} & \Delta G_f^{\circ}(\mathrm{A})=100 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta G_f^{\circ}(\mathrm{B})=50 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta G_f^{\circ}(\mathrm{C})=-75 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta G_f^{\circ}(\mathrm{D})=-50 \mathrm{~kJ} / \mathrm{mol} \end{aligned} }

Option: 1

-1250 \mathrm{~kJ} / \mathrm{mol}


Option: 2

-750 \mathrm{~kJ} / \mathrm{mol}


Option: 3

-500 \mathrm{~kJ} / \mathrm{mol}


Option: 4

-250 \mathrm{~kJ} / \mathrm{mol}


Answers (1)

best_answer

Solution: The standard Gibbs free energy change \mathrm{\left(\Delta G^{\circ}\right)} can be calculated using the formula:

       \mathrm{ \left.\Delta G^{\circ}=\sum \nu_i \Delta G_f^{\circ}(\text { products })-\sum \nu_i \Delta G_f^{\circ} \text { (reactants }\right) }
For the given reaction:

     \mathrm{ \Delta G^{\circ}=\left(4 \times \Delta G_f^{\circ}(\mathrm{C})+2 \times \Delta G_f^{\circ}(\mathrm{D})\right)-\left(2 \times \Delta G_f^{\circ}(\mathrm{A})+3 \times \Delta G_f^{\circ}(\mathrm{B})\right) }
Substitute the given values:

\mathrm{ \begin{gathered} \Delta G^{\circ}=(4 \times-75 \mathrm{~kJ} / \mathrm{mol}+2 \times-50 \mathrm{~kJ} / \mathrm{mol})-(2 \times 100 \mathrm{~kJ} / \mathrm{mol}+3 \times 50 \mathrm{~kJ} / \mathrm{mol}) \\\\ \Delta G^{\circ}=-750 \mathrm{~kJ} / \mathrm{mol} \end{gathered} }
So, the correct answer is:

B) \mathrm{-750 \mathrm{~kJ} / \mathrm{mol}}

Posted by

Divya Prakash Singh

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