For the oxidation of ammonia at 298 K,the st | vedclass

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(D) The standard Gibbs energy change $(\Delta G^{\circ})$ is calculated using the relation: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.

Vedclass · Accepted Answer

$-339.3$

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(D) The standard Gibbs energy change $(\Delta G^{\circ})$ is calculated using the relation: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$.Given:$\Delta H^{\circ} = -382.64 \ kJ \ mol^{-1}$$\Delta S^{\circ} = -145.6 \ J \ K^{-1} \ mol^{-1} = -0.1456 \ kJ \ K^{-1} \ mol^{-1}$$T = 298 \ K$Substituting the values:$\Delta G^{\circ} = -382.64 \ kJ \ mol^{-1} - (298 \ K 	imes -0.1456 \ kJ \ K^{-1} \ mol^{-1})$$\Delta G^{\circ} = -382.64 + 43.3888$$\Delta G^{\circ} = -339.2512 \ kJ \ mol^{-1} \approx -339.3 \ kJ \ mol^{-1}$.

For the oxidation of ammonia at $298 \ K$,the standard enthalpy and standard entropy changes are $-382.64 \ kJ \ mol^{-1}$ and $-145.6 \ J \ K^{-1} \ mol^{-1}$ respectively. The standard Gibbs energy change for the same reaction at $298 \ K$ is $..... \ kJ \ mol^{-1}$.

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