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Question

Chemistry Question on Chemical Kinetics

N2O5N_2O_5 decomposes to NO2NO_2 and O2O_2 and follows first order kinetics. After 5050 minutes, the pressure inside the vessel increases from 50mmHg50\, mm\, Hg to 87.5mmHg.87.5\, mm\, Hg. The pressure of the gaseous mixture after 100100 minute at constant temperature will be :

A

175.0mmHg175.0 \,mm\,Hg

B

116.25mmHg116.25 \,mm\,Hg

C

136.25mmHg136.25 \,mm\,Hg

D

106.25mmHg106.25\, mm\,Hg

Answer

106.25mmHg106.25\, mm\,Hg

Explanation

Solution

The decomposition reaction is We know a=50mmHga=50\, mm\, Hg At t=t50min .ax+2x+12x=87.5t=t_{\text {50\,min }}. a-x+2 x+\frac{1}{2} x =87.5 a+32x=87.5a+\frac{3}{2} x =87.5 32x=87.550=37.5\frac{3}{2} x =87.5-50=37.5 x=37.5×23=25\Rightarrow x= \frac{37.5 \times 2}{3}=25 For first order reaction, kt=2.303log(aax)k t=2.303 \log \left(\frac{a}{a-x}\right) At 50min,kt=2.303log(505025)50\, min,\, k t=2.303 \log \left(\frac{50}{50-25}\right) kt=2.303log2k t=2.303 \log 2 k=2.303×0.301050\Rightarrow k=\frac{2.303 \times 0.3010}{50} At 100minkt=2.303log(aay)100\, min\, k t=2.303 \log \left(\frac{a}{a-y}\right) 100×2.303×0.301050100 \times \frac{2.303 \times 0.3010}{50} =2.303log(50ay)=2.303 \log \left(\frac{50}{a-y}\right) 2×0.3010=log(50ay)2 \times 0.3010=\log \left(\frac{50}{a-y}\right) 50ay=4\frac{50}{a-y}=4 ay=504=12.5a-y=\frac{50}{4}=12.5 50y=12.5y=37.550-y=12.5 \Rightarrow y=37.5 Therefore, total pressure at 100min100 min can be calculated as Total pressure =ay+2y+12y=a-y+2 y+\frac{1}{2} y =a+32y= a+\frac{3}{2} y =50+32×37.5=106.25mmHg= 50+\frac{3}{2} \times 37.5=106.25\, mm\, Hg