See image | AITS & Test Series — Chem-Mantra

AITS & Test SerieshardNUMERICAL

Question

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Answer: 00000.91

💡 Solution & Explanation

Sol. C H3 CH3 C H3 CH3 CH3 1 mole 1 - x 0 x        0 0 ΔG ΔG iso-butane ΔG n butane f f = –21.39 + 15.69 = – 5.7 = 2.303 RT logK

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