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CBSE Class 12 Chemistry 2014 Outside Delhi Set 1 Solved Paper

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Question : 20 of 30
Marks: +1, -0
The following data were obtained during the first order thermal decomposition of SO2Cl2\mathrm{SO}_2\mathrm{Cl}_2 at a constant volume:
SO2Cl2(g)→SO2(g)+Cl2(g)\mathrm{SO}_2\mathrm{Cl}_2(\mathrm{g}) \rightarrow \mathrm{SO}_2(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})
 Experiment  Time s−1\mathrm{s}^{-1}  Total pressure/atm
 1  0  0.4
 2  100  0.7
Calculate the rate constant. (Given: log⁡4=0.6021\log 4 = 0.6021, log⁡2=0.3010)\log 2 = 0.3010)
Solution:  
SO2Cl2(g)→SO2(g)+Cl2(g)\mathrm{SO}_2\mathrm{Cl}_2(\mathrm{g}) \rightarrow \mathrm{SO}_2(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g})
At t=0P000At t=100 secP0−PPP\begin{array}{lccc} \text{At } t=0 & P^0 & 0 & 0 \\ \text{At } t=100 \text{ sec} & P^0-P & P & P \end{array}
P0=0.4  atmP^0 = 0.4\;\mathrm{atm}
P0−P+P+P=0.7  atmP^0 - P + P + P = 0.7\;\mathrm{atm}
P0+P=0.7  atmP^0 + P = 0.7\;\mathrm{atm}
P=0.3  atmP = 0.3\;\mathrm{atm}
It's a first order reaction.
  K=  2.303tlog⁡  P0P0−P\;K = \;\frac{2.303}{t} \log \;\frac{P^0}{P^0-P}
  K=  2.303tlog⁡  0.40.1\;K = \;\frac{2.303}{t} \log \;\frac{0.4}{0.1}
So, the rate constant, K=1.39×10−2  sec−1K = 1.39 \times 10^{-2}\;\mathrm{sec}^{-1}
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