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Standard XII

Chemistry

Reaction Quotient

The K c for...

Question

The $K_{c}$ for, $A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$ at $100$ is $50$ . If one litre flask containing one mole of $A_{2}$ is connected with a two litre flask containing 2 mole of $B_{2}$ , how many mole of $A B$ will be formed at $100$ , rounded to the nearest integer?

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Solution

The equilibrium expression, the initial number of moles and the equilibrium number of moles are as given below. $A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$
Initial mole $120$
Equilibrium mole $(1 - x) (2 - x) 2 x$
The volume becomes 3 litre on joining two containers.
At equilibrium
$∴ [A_{2}] = \frac{1 - x}{3}; [B_{2}] = \frac{2 - x}{3}; [A B] = \frac{2 x}{3}$
The expression for the equilibrium constant becomes
$K_{c} = \frac{{[A B]}^{2}}{[A_{2}] [B_{2}]}$
$= \frac{{(2 x / 3)}^{2}}{\frac{(1 - x)}{3} \frac{(2 - x)}{3}} = \frac{4 x^{2}}{(2 - 3 x + x^{2})}$
or $50 = \frac{4 x^{2}}{(2 - 3 x + x^{2})}$ .(i)
Solving equation (i) $x = 0.935$ and $2.326$ ; the later value is not valid since $x ≯ 2$
$∴ [A B] = 2 x = 2 \times 0.935 = 1.87$
Hence, the number of moles of AB formed at equilibrium is $1.87$ .

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The Kc for, A2(g)+B2(g)⇌2AB(g) at 100 is 50. If one litre flask containing one mole of A2 is connected with a two litre flask containing 2 mole of B2, how many mole of AB will be formed at 100, rounded to the nearest integer?

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