States of Matter 1 Question 76
76. The composition of the equilibrium mixture $\left(\mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{Cl}\right)$ which is attained at $1200^{\circ} \mathrm{C}$, is determined by measuring the rate of effusion through a pin-hole. It is observed that at $1.80 \mathrm{~mm} \mathrm{Hg}$ 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 $\mathrm{Kr}=84)$
$(1995,4 \mathrm{M})$
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Answer:
Correct Answer: 76. $\left(407 \mathrm{~ms}^{-1}\right)$
Solution:
- If ’ $\alpha$ ’ is the degree of dissociation, then at equilibrium
$$ \begin{array}{llll} & \mathrm{Cl}_{2} \rightleftharpoons 2 \mathrm{Cl} & \ \text { Moles } & 1-\alpha & 2 \alpha \end{array} \quad \text { Total }=1+\alpha $$
From diffusion information
$$ \begin{aligned} & & \frac{r_{(\text {mix })}}{r_{(\mathrm{Kr})}} & =1.16=\sqrt{\frac{84}{M(\mathrm{mix})}} \ \Rightarrow & & M_{(\text {mix })} & =62.4 \ \Rightarrow & & M_{(\text {mix })} & =\frac{71}{1+\alpha}=62.4 \ \Rightarrow & & \alpha & =0.14 \end{aligned} $$
