3 A_{(g)} rightarrow 2 B_{(g)} + 2 D_{(g)} + | vedclass

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(A) The reaction is $3 A_{(g)} \rightarrow 2 B_{(g)} + 2 D_{(g)} + E_{(g)}$.The change in moles of gaseous products is $\Delta n_g = (2 + 2 + 1) - 3 =

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Reaction is spontaneous and $\Delta G = -1.2 \ kcal / mol$

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(A) The reaction is $3 A_{(g)} ightarrow 2 B_{(g)} + 2 D_{(g)} + E_{(g)}$.The change in moles of gaseous products is $\Delta n_g = (2 + 2 + 1) - 3 = 2$.Given $\Delta U = 5.1 \ kcal / mol$,$\Delta S = 25 \ cal / mol \cdot K = 0.025 \ kcal / mol \cdot K$,and $T = 300 \ K$.Calculate $\Delta H$ using $\Delta H = \Delta U + \Delta n_g RT$:$\Delta H = 5.1 + \frac{2 	imes 2 	imes 300}{1000} = 5.1 + 1.2 = 6.3 \ kcal / mol$.Calculate $\Delta G$ using $\Delta G = \Delta H - T \Delta S$:$\Delta G = 6.3 - (300 	imes 0.025) = 6.3 - 7.5 = -1.2 \ kcal / mol$.Since $\Delta G < 0$,the reaction is spontaneous.

$3 A_{(g)} \rightarrow 2 B_{(g)} + 2 D_{(g)} + E_{(g)}$
For the above reaction,$\Delta U = 5.1 \ kcal / mol$ and $\Delta S = 25 \ cal / mol \cdot K$. Which of the following is the correct statement at $300 \ K$?

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$3 A_{(g)} \rightarrow 2 B_{(g)} + 2 D_{(g)} + E_{(g)}$For the above reaction,$\Delta U = 5.1 \ kcal / mol$ and $\Delta S = 25 \ cal / mol \cdot K$. Which of the following is the correct statement at $300 \ K$?