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 [20

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 [2026]

Smart Achievers · Accepted Answer

(2)

$\begin{matrix} A & ⇌ & B \\ 0.5 M & 0.375 M & (At equilibrium) \end{matrix}$

$K_{e q} = \frac{{[B]}_{e q}}{{[A]}_{e q}} = \frac{0.375}{0.5} = 0.75$

$Now 0.1 mole of A is added so reaction will move in forward direction.$

$\begin{matrix} A & ⇌ & B \\ 0.6 - x & 0.375 + x \end{matrix}$

$K_{e q} = 0.75 = \frac{0.375 + x}{0.6 - x}$

$0.45 - 0.75 x = 0.375 + x$

$1.75 x = 0.075$

$x = \frac{0.075}{1.75} = \frac{3}{70} = 0.043$

$Moles of A = 0.6 - 0.043 = 0.557$

$Moles of B = 0.375 + 0.043 = 0.418$

$Ans. (2) is correct.$

Solution

1 0.742, 0.557

2 0.557, 0.418

3 0.367, 0.275

4 0.53, 0.4

Want Us to Email you About Special Offers & Updates?