Question

The equilibrium constant of the reaction $A_2\left(g\right)+B_2\left(g\right)\rightleftarrows 2AB\left(g\right)$ at $373K$ is $50$ . If $\text{1}\text{L}$ of flask containing $1mole$ of ${\left(\text{A}\right)}_{\text{2}}\text{(g)}$ is connected to $\text{2}\text{L}$ flask containing $2moles{B}_2\left(g\right)$ at ${\text{100}}^{\text{o}}\text{C}$ , the amount of $AB$ produced at equilibrium at ${\text{100}}^{\text{o}}\text{C}$ would be

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

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: $\left(\frac{1}{3}-\text{x}\right)\text{M}\left(\frac{2}{3}-\text{x}\right)\left(\text{M}\right)2\text{x }\left(\text{M}\right)$
$K_{eq}=50=\frac{4x^2}{\left(\frac{1}{3}-x\right)\left(\frac{2}{3}-x\right)}$
or $50=\frac{36x^2}{\left(1-3x\right)\left(2-3x\right)}$
or $50\left(9x^2-9x+2\right)=36x^2$
or $450x^2-450x+100=36x^2$
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 = $0.31\times 6=1.86$