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CBSE Class 12 Chemistry 2016 Outside Delhi Set 1 Solved Paper

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Question : 12 of 26
Marks: +1, -0
For the first order thermal decomposition reaction, the following data were obtained :
C2H5Cl(g)→C2H4(g)+HCl(g)\mathrm{C}_2\mathrm{H}_5\mathrm{Cl}(g) \rightarrow \mathrm{C}_2\mathrm{H}_4(g) + \mathrm{HCl}(g)
Time / sTotal pressure / atm00.303000.50\begin{array}{c c} \text{Time / }\mathrm{s} & \text{Total pressure / atm} \\ 0 & 0.30 \\ 300 & 0.50 \end{array}
Calculate the rate constant
(Given : log⁡2=0.3010,log⁡3=0.4771,\log 2 = 0.3010, \log 3 = 0.4771, log⁡4=0.6021)\log 4 = 0.6021)
Solution:  
Given: Initial pressure,\text{Given: Initial pressure,}
P0=0.30 atmP_0 = 0.30\ \mathrm{atm}
Pt=0.50 atmP_t = 0.50\ \mathrm{atm}
t=300 st = 300\ \mathrm{s}
Rate constant, k=2.303tlog⁡P02P0−Pt\text{Rate constant,}\ k = \frac{2.303}{t} \log \frac{P_0}{2P_0 - P_t}
=2.303300 slog⁡0.302×0.30−0.50= \frac{2.303}{300\ \mathrm{s}} \log \frac{0.30}{2 \times 0.30 - 0.50}
=2.303300 slog⁡0.300.60−0.50= \frac{2.303}{300\ \mathrm{s}} \log \frac{0.30}{0.60 - 0.50}
=2.303300 slog⁡0.300.10= \frac{2.303}{300\ \mathrm{s}} \log \frac{0.30}{0.10}
=2.303300 slog⁡3= \frac{2.303}{300\ \mathrm{s}} \log 3
=2.303300 s×0.4771= \frac{2.303}{300\ \mathrm{s}} \times 0.4771
=1.099300 s= \frac{1.099}{300\ \mathrm{s}}
=0.0036 s−1= 0.0036\ \mathrm{s}^{-1}
=3.6×10−3 s−1= 3.6 \times 10^{-3}\ \mathrm{s}^{-1}
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