The decomposition of ${Cl}_{2} O_{7}$ at 400 K in the gas phase to ${Cl}_{2}$ and $O_{2}$ is a first order reaction. After 55 seconds at 400 K the pressure of ${Cl}_{2} O_{7}$ falls from 0.062 to 0.044 atm. The rate constant and pressure of ${Cl}_{2} O_{7}$ after 100 sec of decomposition at this temperature are

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$5.8 \times 10^{- 3} \sec^{- 1}; 0.44 atm$

$5.2 \times 10^{- 4} \sec^{- 1}; 0.05 atm$

$6.2 \times 10^{- 3} \sec^{- 1}; 0.033 atm$

$4.6 \times 10^{- 3} \sec^{- 1}; 0.005 atm$

answer is B.

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Detailed Solution

The equation for first order reaction is

$K = \frac{2 .303}{t} \log (\frac{a}{a - x})$

(i) Here $a = 0 .062, a - x = 0 .044$

$k = \frac{2 .303}{55} \log \frac{0 .062}{0 .044} = 6 .2 \times 10^{- 3} \sec^{- 1}$

(ii) Here we are to find the value of $a - x$

$\begin{array}{l} 6 .20 \times 10^{- 3} = \frac{2 .303}{100} \log \frac{0 .062}{a - x} \\ \log \frac{0 .062}{a - x} = 2 .69 \times 10^{- 1} \\ \frac{0 .062}{a - x} = 1 .86 \\ a - x = \frac{0 .062}{1 .86} = 0 .033 atm \end{array}$

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