Chemical and Ionic Equilibrium - Result Question 52
52. The equilibrium constant of the reaction $A _2(g)+B _2(g) \rightleftharpoons 2 A B(g)$ at $100^{\circ} C$ is $50$ . If a one litre flask containing one mole of $A _2$ is connected to a two litre flask containing two moles of $B _2$, how many moles of $A B$ will be formed at $373 K$ ?
$(1985,4 M)$
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Answer:
Correct Answer: 52. $(1.86)$
Solution:
- $A_2(g)+B_2(g) \rightleftharpoons 2 A B(g)\quad\quad\quad$ $\Delta n=0$
$ \begin{aligned} & K=\dfrac{[A B]^2}{\left[A_2\right]\left[B_2\right]}=\dfrac{\left(n_{A B}\right)^2}{n_{A_2} \cdot n_{B_2}}=\dfrac{(2 x)^2}{(1-x)(2-x)} \\ & \Rightarrow \quad 50=\dfrac{4 x^2}{x^2-3 x+2} \Rightarrow 23 x^2-75 x+50=0 \\ & \Rightarrow \quad x=\dfrac{75 \pm \sqrt{75^2-4 \times 23 \times 50}}{46}=0.93,2.32 \\ \end{aligned} $
$2.32$ is not acceptable because $x$ cannot be greater than $1$ .
Mole of $A B=2 x=2 \times 0.93=1.86$
BETA
