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 $373 K$ ?
$(1985,4 M)$
Show Answer
Solution:
- $A _2(g)+B _2(g) \rightleftharpoons 2 A B(g)$
$$ \begin{array}{rc} & K=\frac{[A B]^{2}}{\left[A _2\right]\left[B _2\right]}=\frac{\left(n _{A B}\right)^{2}}{n _{A _2} \cdot n _{B _2}}=\frac{(2 x)^{2}}{(1-x)(2-x)} \ \Rightarrow \quad & 50=\frac{4 x^{2}}{x^{2}-3 x+2} \Rightarrow 23 x^{2}-75 x+50=0 \ \Rightarrow & x=\frac{75 \pm \sqrt{75^{2}-4 \times 23 \times 50}}{46}=0.93,2.32 \end{array} $$
