JEE Main · 2019 · Shift-IImediumCK-108

For the reaction of H2 and I2, the rate constant is 2.5 10-4\,dm3\,mol-1\,s-1 at 327°C and 1.0\,dm3\,mol-1\,s-1 at…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

For the reaction of H2\mathrm{H_2} and I2\mathrm{I_2}, the rate constant is 2.5×10−4 dm3 mol−1 s−12.5 \times 10^{-4}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}} at 327°C327°\mathrm{C} and 1.0 dm3 mol−1 s−11.0\,\mathrm{dm^3\,mol^{-1}\,s^{-1}} at 527°C527°\mathrm{C}. The activation energy for the reaction, in kJ mol−1\mathrm{kJ\,mol^{-1}} is:

(R=8.314 J K−1 mol−1)(R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}})

Options
  1. a

    166

    ✓
  2. b

    59

  3. c

    72

  4. d

    150

Correct Answera

166

Detailed Solution

T1=600 KT_1 = 600\,\mathrm{K}, k1=2.5×10−4k_1 = 2.5 \times 10^{-4}; T2=800 KT_2 = 800\,\mathrm{K}, k2=1.0k_2 = 1.0

ln⁡k2k1=EaR(1T1−1T2)\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

ln⁡1.02.5×10−4=ln⁡4000=ln⁡(4×103)=ln⁡4+3ln⁡10=1.386+6.909=8.295\ln\frac{1.0}{2.5 \times 10^{-4}} = \ln 4000 = \ln(4 \times 10^3) = \ln 4 + 3\ln 10 = 1.386 + 6.909 = 8.295

8.295=Ea8.314(1600−1800)=Ea8.314×800−600600×8008.295 = \frac{E_a}{8.314}\left(\frac{1}{600} - \frac{1}{800}\right) = \frac{E_a}{8.314} \times \frac{800-600}{600 \times 800}

=Ea8.314×200480000=Ea8.314×4.167×10−4= \frac{E_a}{8.314} \times \frac{200}{480000} = \frac{E_a}{8.314} \times 4.167 \times 10^{-4}

Ea=8.295×8.3144.167×10−4=68.974.167×10−4=165500 J/mol≈166 kJ/molE_a = \frac{8.295 \times 8.314}{4.167 \times 10^{-4}} = \frac{68.97}{4.167 \times 10^{-4}} = 165500\,\mathrm{J/mol} \approx 166\,\mathrm{kJ/mol}

Answer: Option (1) — 166 kJ/mol

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For the reaction of H2 and I2, the rate constant is 2.5 10-4\,dm3\,mol-1\,s-1 at 327°C and… (JEE Main 2019) | Canvas Classes