First slide

Chemical Equilibrium

Question

The composition of the equilibrium mixture $(C l_{2} \overset{}{⇌} 2 C l),$ which is attained at 1200°C, is determined by measuring the rate of effusion through a pinhole. It is observed that at 1.80 mm Hg pressure, the mixture effuses 1.2 times as fast as krypton effuses under the same conditions. Calculate the fraction of chlorine molecules dissociated into atoms. (Atomic mass of Kr = 84, Cl = 35.5)

Moderate

Solution

$\frac{r_{m i x}}{r_{K r}} = \sqrt{(\frac{M_{K r}}{M_{M i x}})} or 1.2 = \sqrt{(\frac{84}{M})}$
$∴ M = \frac{700}{12}$
For
$\begin{array}{l} C l_{2} ⇌_{}^{} 2 C l \\ 1 0 \\ (1 - α) 2 α \end{array}$
$\frac{N o r m a l m o l a r m a s s}{E x p . m o l a r m a s s} = 1 + α$
$\frac{71}{(\frac{700}{12})} = 1 + α$
$\begin{array}{l} ∴ & α = 0.217 & or & \approx 0.22 \end{array}$

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