Consider the following reaction approaching equilibrium at 27ºC and 1 atm pressure KF = 103 Kb = 102 — JEE Mains Chemistry Past Papers Chemistry Question
Question
Consider the following reaction approaching equilibrium at 27ºC and 1 atm pressure KF = 103 Kb = 102 A+B C+D The standard Gibb's energy change (, Gº) at 27ºC is (–) ___________ kJ mol–1 (Nearest integer). (Given: R = 8.3 J K–1mol–1 and ln 10 = 2.3)
Answer: .
💡 Solution & Explanation
Keq = 3 b f 0 K K = 10 Gº = –RT n keq = – 8.3 × 300 n (10) = – 8.3 × 300 × 2.3 = – 5.72 × 103 J mol–1 = – 5.72 KJ mol–1 6 KJ mol–1 | JEE(Main) 2023 | DATE : 29-01-2023 (SHIFT-1) | PAPER-1 | OFFICIAL PAPER | CHEMISTRY PAGE # 11
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