Observe the following equilibrium in a 1 L flask. A(g) B(g) At T(K), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to

Question

Observe the following equilibrium in a 1 L flask. A(g) B(g) At T(K), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to T(K) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively.

(A) 0.367, 0.275
(B) 0.53, 0.4
(C) 0.742, 0.557
(D) 0.557, 0.418

Prashant Kotian · Accepted Answer

Correct answer: D. A B 0.5M 0.375 M (At equilibrium) eq eq eq [B] 0.375 K 0.75 [A] 0.5    Now 0.1 mole of A is added so reaction will move in forward direction. A B 0.6–x 0.375+x Keq = 0.75 = 0.375 x 0.6 x   0.45 – 0.75x = 0.375 + x 1.75x = 0.075 X = 0.075 1.75  = 0.043 Moles of A = 0.043 = 0.557 Moles of B = 0.418 Ans. (4) is correct. ALLEN 

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