Question

Observe the following equilibrium in a 1 L flask: \[ \ce{A(g) <=> B(g)} \] 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. 0.742, 0.557
• B. 0.367, 0.275
• C. 0.53, 0.4
• D. 0.557, 0.418

Hint:

Always recalculate equilibrium concentrations using ICE table method after perturbation.

Answer 1

The Correct Option is A

Step 1: Calculate equilibrium constant.

\[ K_c = \frac{[B]}{[A]} = \frac{0.375}{0.5} = 0.75 \]
Step 2: After adding 0.1 mol of A.

Initial concentrations:
\[ [A] = 0.5 + 0.1 = 0.6, \quad [B] = 0.375 \]
Step 3: Let shift in equilibrium be \( x \).

\[ [A] = 0.6 - x, \quad [B] = 0.375 + x \]
At equilibrium:
\[ K_c = \frac{0.375 + x}{0.6 - x} = 0.75 \]
Step 4: Solve for \( x \).

\[ 0.375 + x = 0.75(0.6 - x) \]
\[ 0.375 + x = 0.45 - 0.75x \]
\[ 1.75x = 0.075 \Rightarrow x = 0.043 \]
Step 5: New equilibrium concentrations.

\[ [A] = 0.6 - 0.043 = 0.557, \quad [B] = 0.375 + 0.043 = 0.418 \]

Thus, the correct answer is option (4).
(Note: The given key suggests option (1), but calculation shows option (4).)