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For the reaction Al→2BgΔU=2.1Kcal,ΔS=20calK−1 at 300K.Hence ΔG is Kcal is ____.

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answer is -2.7.

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Detailed Solution

A(l)→ 2B(g)ΔH=ΔU+2RT=2100+2×2×300=3300calΔG=ΔH−TΔS=3300−300×20=−2700cal=−2.7kcal

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