For the reaction X_2Y_{4(l)} rightarrow 2XY_{ | vedclass

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Question

(C) For the reaction $X_2Y_{4(l)} ightarrow 2XY_{2(g)}$$\Delta n_g = 	ext{number of gaseous products} - 	ext{number of gaseous reactants} = 2 - 0

Vedclass · Accepted Answer

$-2800$

Vedclass · Answer

(C) For the reaction $X_2Y_{4(l)} ightarrow 2XY_{2(g)}$$\Delta n_g = 	ext{number of gaseous products} - 	ext{number of gaseous reactants} = 2 - 0 = 2$Given, $\Delta U = 2 \ kcal = 2000 \ cal$, $\Delta S = 20 \ cal \ K^{-1}$, $T = 300 \ K$, and $R = 2 \ cal \ K^{-1} \ mol^{-1}$Using the relation $\Delta H = \Delta U + \Delta n_g RT$$\Delta H = 2000 + (2 	imes 2 	imes 300) = 2000 + 1200 = 3200 \ cal$Now, using the Gibbs free energy equation $\Delta G = \Delta H - T \Delta S$$\Delta G = 3200 - (300 	imes 20) = 3200 - 6000 = -2800 \ cal$

For the reaction $X_2Y_{4(l)} \rightarrow 2XY_{2(g)}$ at $300 \ K$, the values of $\Delta U$ and $\Delta S$ are $2 \ kcal$ and $20 \ cal \ K^{-1}$ respectively. The value of $\Delta G$ for the reaction is (in $cal$)

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