Question

The equilibrium constant of the reaction $A_{2} \, \left(g\right) \, + \, B_{2} \, \left(g\right)\rightleftarrows2AB \, \left(g\right)$ at $373 \, K$ is $50$ . If $\text{1} \, \text{L}$ of flask containing $1 \, mole$ of $\left(\text{A}\right)_{\text{2}} \, \text{(g)}$ is connected to $\text{2} \, \text{L}$ flask containing $2 \, moles \, B_{2} \, \left(g\right)$ at $\text{100}^{\text{o}} \text{C}$ , the amount of $AB$ produced at equilibrium at $\text{100}^{\text{o}} \text{C}$ would be

• A. $0.93 \, mol$
• B. $1.87 \, mol$
• C. $2.80 \, mol$
• D. $3.74 \, mol$

Answer 1

Answer: (B)

$A_{2} \, \left(g\right) \, \, \, + \, B_{2} \, \left(g\right) \, \, \rightleftarrows \, \, \, 2AB \, \left(g\right)$
Initially: $\frac{1}{3}\text{M}\text{M}$ 2 3 $\text{M}$
At equilibrium: $\left(\frac{1}{3} - \text{x}\right) \text{M} \, \left(\frac{2}{3} - \text{x}\right) \left(\text{M}\right) \, 2 \text{x } \left(\text{M}\right)$
$K_{e q}=50 \, =\frac{4 \, x^{2}}{\left(\frac{1}{3} - x\right) \left(\frac{2}{3} - x\right)}$
or $50 \, =\frac{36 \, x^{2}}{\left(1 - 3 x\right) \left(2 - 3 x\right)}$
or $50 \, \left(9 x^{2} \, - \, 9 x \, + \, 2\right)=36 \, x^{2}$
or $450 \, x^{2} \, - \, 450x \, + \, 100=36x^{2}$
or $414x^{2} \, - \, 450x \, + \, 100=0$
or $\text{x} = \frac{+ 450 - \sqrt{\left(450\right)^{2} - 4 \times 414 \times 100}}{2 \times 414}$
or $x=\frac{+ 450 - \sqrt{202500 - 165600}}{2 \times 414}$
or $x=\frac{450 - \sqrt{36900}}{2 \times 414}=\frac{450 - 192.1}{2 \times 414}$
= $0.31 \, \left(M\right)$
$\therefore $ moles of $AB$ produced = $0.31 \, \times \, 6=1.86$