States of Matter - Result Question 76
####76. The composition of the equilibrium mixture $\left(Cl _2 \rightleftharpoons 2 Cl\right)$ which is attained at $1200^{\circ} C$, is determined by measuring the rate of effusion through a pin-hole. It is observed that at $1.80 mm Hg$ pressure, the mixture effuses 1.16 times as fast as krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms (atomic weight of $Kr=84)$
$(1995,4 M)$
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Answer:
Correct Answer: 76. $\left(407 ms^{-1}\right)$
Solution:
- If ’ $\alpha$ ’ is the degree of dissociation, then at equilibrium
$$ \begin{array}{llll} & Cl _2 \rightleftharpoons 2 Cl & \ \text { Moles } & 1-\alpha & 2 \alpha \end{array} \quad \text { Total }=1+\alpha $$
From diffusion information
$$ \begin{aligned} & & \frac{r _{(\text {mix })}}{r _{(Kr)}} & =1.16=\sqrt{\frac{84}{M(mix)}} \ \Rightarrow & & M _{(\text {mix })} & =62.4 \ \Rightarrow & & M _{(\text {mix })} & =\frac{71}{1+\alpha}=62.4 \ \Rightarrow & & \alpha & =0.14 \end{aligned} $$
