JEE Main · 2019 · Shift-IIeasyCK-039

NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation: {2N_2O_5(g) 4NO_2(g)…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

NO2\mathrm{NO_2} required for a reaction is produced by the decomposition of N2O5\mathrm{N_2O_5} in CCl4\mathrm{CCl_4} as per the equation: 2N2O5(g)→4NO2(g)+O2(g)\mathrm{2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)} The initial concentration of N2O5\mathrm{N_2O_5} is 3.00 mol L−13.00\,\mathrm{mol\,L^{-1}} and it is 2.75 mol L−12.75\,\mathrm{mol\,L^{-1}} after 30 minutes. The rate of formation of NO2\mathrm{NO_2} is:

Options
  1. a

    1.667×10−2 mol L−1 min−11.667 \times 10^{-2}\,\mathrm{mol\,L^{-1}\,min^{-1}}

    ✓
  2. b

    8.333×10−3 mol L−1 min−18.333 \times 10^{-3}\,\mathrm{mol\,L^{-1}\,min^{-1}}

  3. c

    4.167×10−3 mol L−1 min−14.167 \times 10^{-3}\,\mathrm{mol\,L^{-1}\,min^{-1}}

  4. d

    2.083×10−3 mol L−1 min−12.083 \times 10^{-3}\,\mathrm{mol\,L^{-1}\,min^{-1}}

Correct Answera

1.667×10−2 mol L−1 min−11.667 \times 10^{-2}\,\mathrm{mol\,L^{-1}\,min^{-1}}

Detailed Solution

Rate of disappearance of N2O5\mathrm{N_2O_5}: −Δ[N2O5]Δt=3.00−2.7530=0.2530=8.33×10−3 mol L−1 min−1-\frac{\Delta[\mathrm{N_2O_5}]}{\Delta t} = \frac{3.00 - 2.75}{30} = \frac{0.25}{30} = 8.33 \times 10^{-3}\,\mathrm{mol\,L^{-1}\,min^{-1}}

From stoichiometry (4 NO24\,\mathrm{NO_2} for every 2 N2O52\,\mathrm{N_2O_5}): d[NO2]dt=2×8.33×10−3=1.667×10−2 mol L−1 min−1\frac{d[\mathrm{NO_2}]}{dt} = 2 \times 8.33 \times 10^{-3} = 1.667 \times 10^{-2}\,\mathrm{mol\,L^{-1}\,min^{-1}}

Answer: Option (1)

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