At 298 K,Delta_r U^{ominus} and Delta_r S^{o | vedclass

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(B) The reaction is $2 \ X_{(g)} + Y_{(g)} \longrightarrow 2 \ Z_{(g)}$.First,calculate the change in the number of moles of gaseous products and reac

Vedclass · Accepted Answer

$-26.119$

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(B) The reaction is $2 \ X_{(g)} + Y_{(g)} \longrightarrow 2 \ Z_{(g)}$.First,calculate the change in the number of moles of gaseous products and reactants: $\Delta n_g = 2 - (2 + 1) = -1$.Next,calculate $\Delta_r H^{\ominus}$ using the relation $\Delta_r H^{\ominus} = \Delta_r U^{\ominus} + \Delta n_g RT$.$\Delta_r H^{\ominus} = -10.5 \ kJ + (-1 	imes 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1} 	imes 298 \ K) = -10.5 - 2.477 = -12.977 \ kJ$.Now,calculate $\Delta_r G^{\ominus}$ using the Gibbs-Helmholtz equation: $\Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus}$.Given $\Delta_r S^{\ominus} = +44.1 \ J \ K^{-1} \ mol^{-1} = 0.0441 \ kJ \ K^{-1} \ mol^{-1}$.$\Delta_r G^{\ominus} = -12.977 \ kJ - (298 \ K 	imes 0.0441 \ kJ \ K^{-1} \ mol^{-1}) = -12.977 - 13.1418 = -26.1188 \ kJ \approx -26.119 \ kJ$.

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