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Chemical Kinetics and Nuclear Chemistry question

2019 · 10 Apr · Shift 2 · Q15
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Chemical Kinetics and Nuclear Chemistry question

2019 · 10 Apr · Shift 2 · Q15

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
For the reaction of H2H_2H2​ with I2I_2I2​, the rate constant is 2.5 × 10–4 dm3 mol–1s–1 at 327°C and 1.0 dm3 mol–1 at 527°C. The activation energy for the reaction, in kJ mole–1 is : (R = 8.314 JK–1 mol–1 )
  1. A
    59
  2. B
    166
  3. C
    72
  4. D
    150
View written solutionFree

Correct answer: B

  1. Use the Arrhenius equation in two-temperature form

For two temperatures T1T_1T1​ and T2T_2T2​:

ln⁡(k2k1)=EaR(1T1−1T2)\ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)ln(k1​k2​​)=REa​​(T1​1​−T2​1​)

Given:

k1=2.5×10−4 dm3 mol−1s−1atT1=327∘C=600 Kk_1 = 2.5 \times 10^{-4}\ \text{dm}^3\text{ mol}^{-1}\text{s}^{-1} \quad \text{at} \quad T_1 = 327^\circ C = 600\ \text{K}k1​=2.5×10−4 dm3 mol−1s−1atT1​=327∘C=600 K

k2=1.0 dm3 mol−1s−1atT2=527∘C=800 Kk_2 = 1.0\ \text{dm}^3\text{ mol}^{-1}\text{s}^{-1} \quad \text{at} \quad T_2 = 527^\circ C = 800\ \text{K}k2​=1.0 dm3 mol−1s−1atT2​=527∘C=800 K

R=8.314 J K−1 mol−1R = 8.314\ \text{J K}^{-1}\text{ mol}^{-1}R=8.314 J K−1 mol−1

  1. Compute the ratio k2/k1k_2/k_1k2​/k1​
k2k1=1.02.5×10−4=4000\frac{k_2}{k_1} = \frac{1.0}{2.5 \times 10^{-4}} = 4000k1​k2​​=2.5×10−41.0​=4000

So,

ln⁡(k2k1)=ln⁡(4000)\ln\left(\frac{k_2}{k_1}\right) = \ln(4000)ln(k1​k2​​)=ln(4000) ln⁡(4000)=ln⁡(4×103)=ln⁡4+3ln⁡10\ln(4000)=\ln(4 \times 10^3)=\ln 4 + 3\ln 10ln(4000)=ln(4×103)=ln4+3ln10 =1.3863+3(2.3026)=1.3863+6.9078=8.2941=1.3863 + 3(2.3026)=1.3863+6.9078=8.2941=1.3863+3(2.3026)=1.3863+6.9078=8.2941
  1. Compute the temperature factor
1T1−1T2=1600−1800\frac{1}{T_1}-\frac{1}{T_2} = \frac{1}{600}-\frac{1}{800}T1​1​−T2​1​=6001​−8001​

Taking LCM:

=4−32400=12400 K−1= \frac{4-3}{2400}=\frac{1}{2400}\ \text{K}^{-1}=24004−3​=24001​ K−1
  1. Substitute into Arrhenius equation
8.2941=Ea8.314(12400)8.2941 = \frac{E_a}{8.314}\left(\frac{1}{2400}\right)8.2941=8.314Ea​​(24001​)

Therefore,

Ea=8.2941×8.314×2400E_a = 8.2941 \times 8.314 \times 2400Ea​=8.2941×8.314×2400

First,

8.2941×8.314≈68.968.2941 \times 8.314 \approx 68.968.2941×8.314≈68.96

Then,

Ea≈68.96×2400=165504 J mol−1E_a \approx 68.96 \times 2400 = 165504\ \text{J mol}^{-1}Ea​≈68.96×2400=165504 J mol−1 Ea≈165.5 kJ mol−1E_a \approx 165.5\ \text{kJ mol}^{-1}Ea​≈165.5 kJ mol−1
  1. Choose the nearest option
Ea≈166 kJ mol−1E_a \approx 166\ \text{kJ mol}^{-1}Ea​≈166 kJ mol−1

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

Derived answer: B

They match.

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