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Question

The decomposition of N2O5 according to the equation, N2O5(g)4NO2(g)+O2(g)
is a first order reaction. After 30 minutes from the start of the decomposition in a closed vessel, the total pressure developed is found to be 280 mm Hg and on completion, the total pressure is 580 mm Hg. The rate constant of the reaction is x×103 min1. Find the value of x. Take: log (1.16)=0.065

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Solution

2N2O54NO2+O2
On decomposition of 2 moles of N2O5, 4 moles of NO2 and 1 mole of O2 are produced.

2N2O54NO2+O2
At t=0: P0 0 0
At t=30: P02P 4P P
At t=: 0 2P0 P0/2
At t=
Total pressure =2P0+P0/2=52P0
Initial pressure of N2O5,P0=25×580=232mmHg

At t=30 min
Total pressure =P02P+4P+P
P0+3P=280
3P=280232=48mmHg
P=483=16mmHg
Pressure of N2O5 after 30 minutes =232(2×16)
=200mmHg
k=2.30330log10232200=5×103min1

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