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BITSAT 2012
Exam
BITSAT
Year
2012
Subject
Chemistry
Chapter
Chemical Kinetics

BITSAT 2012 · Chemistry · Chemical Kinetics

2 N_ 2 O_ 5 leftharpoons 4 NO_ 2 +O_ 2 If rate and rate constant...

2N2O5⇌4NO2+O22 \mathrm{ N}_{2} \mathrm{O}_{5} \rightleftharpoons 4 \mathrm{NO}_{2}+\mathrm{O}_{2} If rate and rate constant for above reaction are 2.40×10−5molL−1s−12.40 \times 10^{-5} \mathrm{ mol} \mathrm{ L}^{-1} \mathrm{ s}^{-1} and 3×10−5s−13 \times 10^{-5} \mathrm{ s}^{-1} respectively, then calculate the concentration of N2O5\mathrm{N}_{2} \mathrm{O}_{5}

Options

  1. A1.4
  2. B1.2
  3. C0.04
  4. D0.8

Correct answer

D. 0.8

Step-by-step solution

The reaction is of first and for a first order reaction, rate, R=k[N2O5]2.4×10−5=3×10−5×[N2O5][N2O5]=2.4×10−53×10−5=0.8molL−1R=k\left[\mathrm{ N}_{2} \mathrm{O}_{5}\right]2.4 \times 10^{-5}=3 \times 10^{-5} \times\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=\frac{2.4 \times 10^{-5}}{3 \times 10^{-5}}=0.8 \mathrm{ mol} \mathrm{ L}^{-1}
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