A2 + B2 2AB. f H = —200 kJ mol–1 AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2 — JEE Mains Chemistry Past Papers Chemistry Question
Question
A2 + B2 2AB. f H = —200 kJ mol–1 AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2 , B2 and AB are in the ratio 1 : 0.5 : 1, then the bond enthalpy of A2 is ______________kJ mol–1 (Nearest integer)
Answer: .
💡 Solution & Explanation
A2 + B2 2AB x –2x 2A 2B x/2 H = x + x/2 – 2x = – 200 = – x/2 = – 200 x = 400 J | JEE(Main) 2023 | DATE : 13-04-2023 (SHIFT-1) | PAPER-1 | OFFICIAL | CHEMISTRY PAGE # 12
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