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
- 0.742, 0.557
- 0.367, 0.275
- 0.53, 0.4
- 0.557, 0.418
Answer
(4) 0.557, 0.418
Solution
Kc = [B][A] = 0.3750.5 = 0.75
After adding 0.1 mol A (1 L): [A] = 0.6 M, [B] = 0.375 M
Qc = 0.3750.6 = 0.625 < Kc reaction moves forward by x M.
0.375 + x0.6 - x = 0.75
0.375 + x = 0.45 - 0.75x
1.75x = 0.075 x = 0.0429
[A] = 0.6 − 0.0429 = 0.557 M, [B] = 0.375 + 0.0429 = 0.418 M
Answer: (4)