JEE Main · 2020 · Shift-IhardCK-032

For the following reactions: {A} {700\,{K}} {Product} {A} [{catalyst}]{500\,{K}} {Product} It was found that the Ea is…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

For the following reactions: A→700 KProduct\mathrm{A} \xrightarrow{700\,\mathrm{K}} \text{Product} A→catalyst500 KProduct\mathrm{A} \xrightarrow[\text{catalyst}]{500\,\mathrm{K}} \text{Product} It was found that the EaE_a is decreased by 30 kJ/mol30\,\mathrm{kJ/mol} in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre-exponential factor is same)

Options
  1. a

    75 kJ/mol75\,\mathrm{kJ/mol}

    ✓
  2. b

    105 kJ/mol105\,\mathrm{kJ/mol}

  3. c

    135 kJ/mol135\,\mathrm{kJ/mol}

  4. d

    198 kJ/mol198\,\mathrm{kJ/mol}

Correct Answera

75 kJ/mol75\,\mathrm{kJ/mol}

Detailed Solution

Since rate remains unchanged with catalyst at lower temperature: k1=k2 ⇒ Ae−Ea/(R×700)=Ae−(Ea−30000)/(R×500)k_1 = k_2 \ \Rightarrow\ Ae^{-E_a/(R \times 700)} = Ae^{-(E_a - 30000)/(R \times 500)}

Ea700=Ea−30000500\frac{E_a}{700} = \frac{E_a - 30000}{500}

500Ea=700(Ea−30000)500 E_a = 700(E_a - 30000)

500Ea=700Ea−21000000500 E_a = 700 E_a - 21000000

200Ea=21000000200 E_a = 21000000

Ea=105000 J/mol=105 kJ/molE_a = 105000\,\mathrm{J/mol} = 105\,\mathrm{kJ/mol}

Activation energy for catalysed reaction =105−30=75 kJ/mol= 105 - 30 = 75\,\mathrm{kJ/mol}

Answer: Option (1) — 75 kJ/mol

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