Chemical and Ionic Equilibrium 1 Question 50
52. The equilibrium constant of the reaction $A_{2}(g)+B_{2}(g) \rightleftharpoons 2 A B(g)$ at $100^{\circ} \mathrm{C}$ 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 $373 \mathrm{~K}$ ?
$(1985,4 \mathrm{M})$
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Solution:
- $A_{2}(g)+B_{2}(g) \rightleftharpoons 2 A B(g)$
$$ \begin{array}{rc} & K=\frac{[A B]^{2}}{\left[A_{2}\right]\left[B_{2}\right]}=\frac{\left(n_{A B}\right)^{2}}{n_{A_{2}} \cdot n_{B_{2}}}=\frac{(2 x)^{2}}{(1-x)(2-x)} \ \Rightarrow \quad & 50=\frac{4 x^{2}}{x^{2}-3 x+2} \Rightarrow 23 x^{2}-75 x+50=0 \ \Rightarrow & x=\frac{75 \pm \sqrt{75^{2}-4 \times 23 \times 50}}{46}=0.93,2.32 \end{array} $$
