For the reaction X_2O_4(l) → 2XO_2(g) , U = 2.1 k cal and S = 20 cal K^-1 at 300 K . Hence, G is: - Chemistry MCQ Chemistry | ExamTest.live

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For the reaction X_{2} O_{4} (l) \to 2 XO_{2} (g), Δ U = 2.1 k cal and Δ S = 20 cal K^{- 1} at 300 K . Hence, Δ G is:

A $2.7 k cal$
B $- 2.7 k cal$
C $9.3 k cal$
D $- 9.3 k cal$

Solution & Step-by-step Explanation

First, find Δ H : Δ H = Δ U + Δ n_{g} RT For the reaction, Δ n_{g} = 2 - 0 = 2 . R = 2 cal K^{- 1} mol^{- 1} = 2 \times 1 0^{- 3} k cal K^{- 1} mol^{- 1} . Δ H = 2.1 + (2 \times 2 \times 1 0^{- 3} \times 300) = 2.1 + 1.2 = 3.3 k cal .Now find Δ G : Δ G = Δ H - T Δ S Δ S = 20 cal K^{- 1} = 0.02 k cal K^{- 1} . Δ G = 3.3 - (300 \times 0.02) = 3.3 - 6 = - 2.7 k cal .

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For the reaction X_{2} O_{4} (l) \to 2 XO_{2} (g), Δ U = 2.1 k cal and Δ S = 20 cal K^{- 1} at 300 K . Hence, Δ G is:

2.7 k cal

- 2.7 k cal

9.3 k cal

- 9.3 k cal

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For the reaction $X_{2} O_{4} (l) \to 2 XO_{2} (g)$ , $Δ U = 2.1 k cal$ and $Δ S = 20 cal K^{- 1}$ at $300 K$ . Hence, $Δ G$ is:

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For the reaction X2​O4​(l)→2XO2​(g), ΔU=2.1k cal and ΔS=20cal K−1 at 300K. Hence, ΔG is:

Solution & Step-by-step Explanation

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Discussion

For the reaction $X_{2} O_{4} (l) \to 2 XO_{2} (g)$ , $Δ U = 2.1 k cal$ and $Δ S = 20 cal K^{- 1}$ at $300 K$ . Hence, $Δ G$ is: