0.01 mol of sulphuryl chloride $S O_{2} C l_{2} (g)$ is taken in a sealed tube and heated to 400K where it decomposes following a first order kinetics according to the following reaction:
$S O_{2} C l_{2} (g) \to S O_{2} (g) + C l_{2} (g)$
The tube is broken after 4.0 h and has passed through a 15 mL of an acidified 1.0 N iodine solution where all $S O_{2}$ is oxidised to $S O_{4}^{2 -}$ . The resulting solution required 7.0 mL of 1 M hypo solution. Calculate half life ( in hour) of given first order reaction.
[ log 2 = 0.3, log(5/3) = 0.22 ]

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answer is 5.42.

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Detailed Solution

Ans: 5.42 hr to 5.46 hr
SOL- Meq of hypo consumed $= 7 = m e q$ of iodine remaining.
meq of $I_{2}$ reacted with $S O_{2} = 15 - 7 = 8 = m e q$ of $S O_{2}$
$\Rightarrow m m o l$ of $S O_{2}$ produced in 4.0 $h r = 4.00$
Total moles of gases after 4.0 hr
$= (0.01 - 0.004) + 2 \times 004 = 0.014$
$\Rightarrow p = \frac{n R T}{V} = \frac{0.014 \times 0.082 \times 400}{0.2} = 2.3 a t m$
Also $4 k = \ln \frac{0.01}{0.01 - 0.004} = \ln \frac{5}{3}$

$t_{1 / 2} = \frac{\ln 2}{k} = \frac{4 \ln 2}{\ln (5 / 3)} = 5.45 h r$