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At constant temperature 60% AB dissociat...

At constant temperature 60% AB dissociates into `A_2` and `B_2`, then the equilibrium constant for `2AB(g)⇌A_2(g) +B_2(g)` is ?

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To solve the problem step-by-step, let's break down the process of finding the equilibrium constant \( K_c \) for the reaction: \[ 2AB(g) \rightleftharpoons A_2(g) + B_2(g) \] ### Step 1: Understand the dissociation We are given that 60% of \( AB \) dissociates. This means that if we start with 2 moles of \( AB \), then at equilibrium, 60% of these will have dissociated into \( A_2 \) and \( B_2 \). ### Step 2: Define the initial and equilibrium concentrations ...
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