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For the reaction at $25 ° C, X_{2} O_{4} (l) ⟶ 2 {XO}_{2} (g), ΔU$ and $ΔS$ are $2.1k.Cal$ and $20 Cal / K$ respectively. What is $ΔG$ for the reaction at the same temperature?
$(R = 2 {CalK}^{- 1} {mol}^{- 1})$

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$+ 3.67 k.Cal$

$+ 2.67 k.Cal$

$- 2.67 k.Cal$

$- 1.67 k.Cal$

answer is A.

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Detailed Solution

$∆ H = ∆ U + ∆ n_{g} RT ∆ H = 2.1 + 1 \times 2 \times 10^{- 3} \times 300 = 3.3 k . Cal ∆ G = ∆ H - T ∆ S ∆ H = 3.3 - 300 \times 20 \times 10^{- 3} = - 2.7 k . Cal$

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