The reaction A(g) leftharpoons B(g) + C(g) was initiated with the amount ' a ' of A(g) . At equilibrium it is found that the amount of A(g) remaining is (a - x) at a total pressure of p . The equilibrium constant K_p of

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Correct answer: B. (x^2)/(a^2 - x^2) × p

The reaction is A(g) leftharpoons B(g) + C(g) Initial moles: a for A , 0 for B , 0 for C Moles at equilibrium: (a - x) for A , x for B , x for C Total moles at equilibrium = (a - x) + x + x = a + x Partial pressures at equilibrium: p_A = (a - x)/(a + x) p p_B = (x)/(a + x) p p_C = (x)/(a + x) p The equilibrium constant K_p is given by: K_p = (p_B · p_C)/(p_A) K_p = (((x)/(a + x) p) ((x)/(a + x) p))/((a - x)/(a + x) p) K_p = (x^2)/((a + x)(a - x)) p K_p = (x^2)/(a^2 - x^2) p Answer: (x^2)/(a^2 - x^2) × p

The reaction A(g) leftharpoons B(g) + C(g) was initiated with the...

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