Q. 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

Correct Answer: 0.557, 0.418

Solution

The given equilibrium reaction is:

A(g) ⇌ B(g)

At temperature T, the equilibrium concentrations are:

[A] = 0.5 M [B] = 0.375 M

First, we calculate the equilibrium constant K_c.

K_c = [B] / [A]

K_c = 0.375 / 0.5 = 0.75

Now, 0.1 moles of A are added to a 1 L flask. Since the volume is 1 L, the concentration of A increases by 0.1 M.

New initial concentrations after addition:

[A] = 0.5 + 0.1 = 0.6 M [B] = 0.375 M

Let x be the amount of A that reacts to re-establish equilibrium.

At equilibrium:

[A] = 0.6 − x [B] = 0.375 + x

Using the equilibrium constant expression:

K_c = (0.375 + x) / (0.6 − x)

0.75 = (0.375 + x) / (0.6 − x)

Cross multiplying:

0.75(0.6 − x) = 0.375 + x

0.45 − 0.75x = 0.375 + x

0.45 − 0.375 = x + 0.75x

0.075 = 1.75x

x = 0.0429 ≈ 0.043

Final equilibrium concentrations:

[A] = 0.6 − 0.043 = 0.557 M [B] = 0.375 + 0.043 = 0.418 M

Hence, the new equilibrium concentrations of A and B are 0.557 M and 0.418 M.

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