The standard enthalpy and standard entropy of | vedclass

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(A) The reaction is $N_2O_4(g) \rightarrow 2NO_2(g)$.Given: $\Delta H^{\circ} = 55.0 \ kJ \ mol^{-1} = 55000 \ J \ mol^{-1}$.Given: $\Delta S^{\circ}

Vedclass · Accepted Answer

$2850$

Vedclass · Answer

(A) The reaction is $N_2O_4(g) ightarrow 2NO_2(g)$.Given: $\Delta H^{\circ} = 55.0 \ kJ \ mol^{-1} = 55000 \ J \ mol^{-1}$.Given: $\Delta S^{\circ} = 175.0 \ J \ K^{-1} \ mol^{-1}$.Temperature $T = 25 + 273 = 298 \ K$.The Gibbs free energy change is given by the formula: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.Substituting the values: $\Delta G^{\circ} = 55000 \ J \ mol^{-1} - (298 \ K 	imes 175.0 \ J \ K^{-1} \ mol^{-1})$.$\Delta G^{\circ} = 55000 - 52150 = 2850 \ J \ mol^{-1}$.

The standard enthalpy and standard entropy of decomposition of $N_2O_4$ to $NO_2$ are $55.0 \ kJ \ mol^{-1}$ and $175.0 \ J \ K^{-1} \ mol^{-1}$ respectively. The standard free energy change for this reaction at $25 ^{\circ}C$ in $J \ mol^{-1}$ is $............$ $(Nearest \ integer)$.

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