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Calculate \mathrm{\Delta G^{0 }} (in kJ) for the following reaction at \mathrm{298 \: K}

\mathrm{CO(g)+ \frac{1}{2} O_{2}(g)\rightarrow CO_{2}(g);\Delta H^{0}=-280\; kJ/mol}

Given : \mathrm{S^{0}_{CO_{2}}=215JK^{-1}mol^{-1},S^{0}_{CO(g)}=200JK^{-1}mol^{-1},S^{0}_{O_{2}}=205JK^{-1}mol^{-1}}

Option: 1

-306


Option: 2

306


Option: 3

254


Option: 4

-254


Answers (1)

best_answer

As we have learned,

\mathrm{\Delta S^{0}=\sum S^{0}_{(products)}-\sum S^{0}_{(reactants)}}

            \mathrm{=\left [ S^{0}_{CO_{2}} \right ]-\left [ S^{0 }_{CO}+\frac{1}{2}S^{0 }_{O_{2}} \right ]}

\mathrm{\Delta S^{0}=215-\left ( 200+\frac{1}{2}\times 205 \right )}

\mathrm{\Delta S^{0}=-87.5\; J/mol.K}

Now, as we learnt

\mathrm{\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}}

\Delta G^{0 }=-280 -298 \times (-87.5) \times10^{-3}

             \mathrm{=-254\; kJ}

 

Posted by

Nehul

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