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  • Consider the following reaction: <img alt="\mathrm{ A(g)+B(g) \rightleftharpoons C(g)+D(g) }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20A%28g%29&plus;B%28g%2

Consider the following reaction:

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

The standard enthalpy change \mathrm{\left(\Delta H^{\circ}\right)} for the reaction is \mathrm{-150 \mathrm{~kJ} / \mathrm{mol}} and the standard entropy change \mathrm{\left(\Delta S^{\circ}\right)} is \mathrm{-200 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}. Calculate the standard Gibbs free energy change \mathrm{\left(\Delta G^{\circ}\right)} for the reaction at \mathrm{298 \mathrm{~K}}.

Option: 1

\mathrm{-90.4 \mathrm{~kJ} / \mathrm{mol}}


Option: 2

\mathrm{-100 \mathrm{~kJ} / \mathrm{mol}}


Option: 3

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


Option: 4

\mathrm{-350 \mathrm{~kJ} / \mathrm{mol}}


Answers (1)

best_answer

Step 1: Use the standard Gibbs free energy equation to calculate \mathrm{\Delta G^{\circ}} at \mathrm{298 \mathrm{~K}.}

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

Step 2: Convert the given entropy change from \mathrm{\mathrm{J} / \mathrm{mol} \cdot \mathrm{K}} to \mathrm{\mathrm{kJ} / \mathrm{mol} \cdot \mathrm{K}} :

                  \mathrm{ \Delta S^{\circ}=-200 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} / 1000=-0.2 \mathrm{~kJ} / \mathrm{mol} \cdot \mathrm{K} }

Step 3: Substitute the given values and solve for \mathrm{\Delta G^{\circ}:}

\mathrm{ \Delta G^{\circ}=-150 \mathrm{~kJ} / \mathrm{mol}-(298 \mathrm{~K} \times(-0.2 \mathrm{~kJ} / \mathrm{mol} \cdot \mathrm{K}))=-150 \mathrm{~kJ} / \mathrm{mol}+59.6 \mathrm{~kJ} / \mathrm{mol} \approx-90.4 \mathrm{~kJ} / \mathrm{mol} }

Therefore, the correct option is (1)

Posted by

HARSH KANKARIA

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