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  • As per first order rate law, the following gas phase decomposition takes place where <img alt="\mathrm{\begin{array}{rl} k=7.6 \times 10^{-4} \mathrm{~s}^{-1} & \\ 2 \mathrm{~N}_2 \ma

As per first order rate law, the following gas phase decomposition takes place where

\mathrm{\begin{array}{rl} k=7.6 \times 10^{-4} \mathrm{~s}^{-1} & \\ 2 \mathrm{~N}_2 \mathrm{O}_5 & \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2 \end{array}}

What will be the decomposition time of \mathrm{\mathrm{N}_2 \mathrm{O}_5} so that the pressure becomes 0.25 atm where initially it was 0,5 atm ?

Option: 1

200 s


Option: 2

385 s


Option: 3

408 s


Option: 4

0.5 s


Answers (1)

best_answer

\mathrm{2 \mathrm{~N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2+\mathrm{O}_2}

\mathrm{\begin{array}{lll} 0.50 \mathrm{tm} & - \\ 0.5-x & 2 x & \frac{x}{2} \end{array}}

Total Pressure  \mathrm{ =0.5-x+2 x+\frac{x}{2}}

\mathrm{\begin{aligned} 0.25 & =0.5+\frac{3 x}{2} \\ 0.20 & =\frac{3 x}{2} \\ x & =\frac{0.40}{3}=0.133 \\ t & =\frac{2.303}{k} \log \frac{P_0}{\rho} \\ & =\frac{2.303}{7.6 \times 10^{-4}} \log =\frac{0.5}{0.5-x} \\ & =\frac{2.303}{7.6 \times 10^{-4}} \log \frac{0.5}{0.5-\frac{0.4}{3}} \\ t & =408 \mathrm{~s} \end{aligned}}

 

Posted by

shivangi.bhatnagar

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