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MHT CET 202626 AprilMorning Shift
Exam
MHT CET
Year
2026
Date
26 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Kinetics

MHT CET 2026 · 26 April · Morning Shift · Chemistry · Chemical Kinetics

The decomposition of N_2O_ 5(g) occurs as: 2\ N_2O_ 5(g) arrow 4\...

The decomposition of N2O5(g)\text{N}_2\text{O}_{5(g)} occurs as: 2 N2O5(g)→4 NO2(g)+O2(g)2\ \text{N}_2\text{O}_{5(g)} \rightarrow 4\ \text{NO}_{2(g)} + \text{O}_{2(g)}.
If initial concentration of N2O5(g)\text{N}_2\text{O}_{5(g)} is 1.63×10−2 mol L−11.63 \times 10^{-2}\ \text{mol L}^{-1} and after 400400 seconds it is 1.36×10−2 mol L−11.36 \times 10^{-2}\ \text{mol L}^{-1}, calculate the rate of reaction.

Options

  1. A3.38×10−6 mol L−1 \texts−13.38 \times 10^{-6}\ \text{mol L}^{-1}\ \texts^{-1}
  2. B2.25×10−6 mol L−1 \texts−12.25 \times 10^{-6}\ \text{mol L}^{-1}\ \texts^{-1}
  3. C1.78×10−6 mol L−1 \texts−11.78 \times 10^{-6}\ \text{mol L}^{-1}\ \texts^{-1}
  4. D3.0×10−6 mol L−1 \texts−13.0 \times 10^{-6}\ \text{mol L}^{-1}\ \texts^{-1}

Correct answer

A. 3.38×10−6 mol L−1 \texts−13.38 \times 10^{-6}\ \text{mol L}^{-1}\ \texts^{-1}

Step-by-step solution

The rate of reaction is given by the expression:

Rate=−12Δ[N2O5]Δt\text{Rate} = -\dfrac{1}{2} \dfrac{\Delta [\text{N}_2\text{O}_5]}{\Delta t}

Δ[N2O5]=1.36×10−2−1.63×10−2=−0.27×10−2 mol L−1\Delta [\text{N}_2\text{O}_5] = 1.36 \times 10^{-2} - 1.63 \times 10^{-2} = -0.27 \times 10^{-2}\ \text{mol L}^{-1}

Δt=400 s\Delta t = 400\ \text{s}

Rate=−12(−0.27×10−2400)<br><br>Rate=0.27×10−2800=3.375×10−6 mol L−1 \texts−1\text{Rate} = -\dfrac{1}{2} \left( \dfrac{-0.27 \times 10^{-2}}{400} \right)<br><br>\text{Rate} = \dfrac{0.27 \times 10^{-2}}{800} =3.375 \times10^{-6}\ \text{mol L}^{-1}\ \texts^{-1}

Rounding off, the rate of reaction is 3.38×10−6 mol L−1 \texts−13.38 \times10^{-6}\ \text{mol L}^{-1}\ \texts^{-1}.

Answer: 3.38×10−6 mol L−1 \texts−13.38 \times10^{-6}\ \text{mol L}^{-1}\ \texts^{-1}
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