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Standard XII

Chemistry

Reaction Quotient

The equilibri...

Question

The equilibrium constant of the reaction, $A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$ at $100^{o} C$ is $50$ . If a one-liter flask containing $1 m o l e$ of $A_{2}$ is connected to a two-liter flask containing $2 m o l e s$ of $B_{2}$ , how many moles of $A B$ will be formed at $373 K$ ?

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Solution

$A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$
Initial no. of moles $1$ $2$ $0$
No. of moles
at equilibrium $(1 - x)$ $(2 - x)$ $2 x$
(Total vol $= 3$ litre)
Active conc. $\frac{(1 - x)}{3}$ $\frac{(2 - x)}{3}$ $\frac{2 x}{3}$
Applying law of mass action
$K_{c} = \frac{{[A B]}^{2}}{[A_{2}] {[B_{2}]}} = \frac{{(\frac{2 x}{3})}^{2}}{(\frac{(1 - x)}{3}) (\frac{(2 - x)}{3})} = \frac{4 x^{2}}{(1 - x) (2 - x)}$
But,
$\frac{4 x^{2}}{(1 - x) (2 - x)} = 50$
or $23 x^{2} - 75 x + 50 = 0$
$x = \frac{75 \pm \sqrt{75^{2} - 4 \times 23 \times 50}}{2 \times 23}$
$x = 2.317$ or $0.943$
The value of $x$ cannot be more than $1$ i.e., greater than the number of moles of $A_{2}$ and hence $x = 0.943$
No. of moles of $A B = 2 x = (2 \times 0.934) = 1.868$

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The equilibrium constant of the reaction, A2(g)+B2(g)⇌2AB(g) at 100oC is 50. If a one-liter flask containing 1 mole of A2 is connected to a two-liter flask containing 2 moles of B2, how many moles of AB will be formed at 373 K?

The equilibrium constant of the reaction, $A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$ at $100^{o} C$ is $50$ . If a one-liter flask containing $1 m o l e$ of $A_{2}$ is connected to a two-liter flask containing $2 m o l e s$ of $B_{2}$ , how many moles of $A B$ will be formed at $373 K$ ?