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  • The reaction <img alt="\mathrm{ A(g)+B(g) \rightleftharpoons C(g)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20A%28g%29&plus;B%28g%29%20%5Crightleftharpoons%20C%28

The reaction \mathrm{ A(g)+B(g) \rightleftharpoons C(g)} is at equilibrium at \mathrm{ 500 \mathrm{~K}} with a value of \mathrm{ K p=0.020 } atm. If the reaction is

exothermic with \mathrm{ \Delta H^{\circ}=-100 \mathrm{~kJ} / \mathrm{mol}}, calculate the standard entropy change \mathrm{ \left(\Delta S^{\circ}\right) } for the reaction at \mathrm{ 500 \mathrm{~K}.}

Option: 1

\mathrm{-168 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}


Option: 2

\mathrm{-200 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}


Option: 3

\mathrm{168 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}


Option: 4

\mathrm{200 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}


Answers (1)

best_answer

Step 1: Use the relation between \mathrm{K p} and \mathrm{\Delta G^{\circ}} to calculate \mathrm{\Delta G^{\circ}} at \mathrm{500 \mathrm{~K}.}

 

                                                         \mathrm{ \Delta G^{\circ}=-R T \ln (K p) }

Step 2: Convert the temperature to Kelvin: \mathrm{T=500 \mathrm{~K}}

Step 3: Convert \mathrm{K p} to a dimensionless quantity: \mathrm{K p=0.020}

Step 4: Calculate \mathrm{ \Delta G^{\circ}} at  \mathrm{ 500 \mathrm{~K}:}

                              \mathrm{ \begin{gathered} \Delta G^{\circ}=-(8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \times 500 \mathrm{~K}) \ln (0.020) \\\\ \Delta G^{\circ}=16262.28 \mathrm{~J} / \mathrm{mol} \end{gathered} }

Step 5: Use the standard Gibbs free energy equation to calculate $\Delta H^{\circ}$ at $500 \mathrm{~K}$.

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

Step 6: Rearrange and solve for \mathrm{\Delta S^{\circ} :}

                      \mathrm{ \begin{gathered} \Delta S^{\circ}=\frac{\Delta H^{\circ}-\Delta G^{\circ}}{T} \\ \Delta S^{\circ}=\frac{(-100 \mathrm{~kJ} / \mathrm{mol} \times 1000 \mathrm{~J} / \mathrm{kJ}-228.3 \mathrm{~J} / \mathrm{mol})}{500 \mathrm{~K}} \\ \Delta S^{\circ}=-168 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \end{gathered} }

So, the coorect option is (1)

Posted by

manish

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