Question

The activation energy of a certain reaction is 45.5 kJ/mol . At 21 ∘C , the rate constant is 0.0110s−1 . At what temperature in degrees Celsius would this reaction go twice as fast?

Answer 1

The temperature in degrees Celsius that the reaction will go twice as fast is  32.4 ⁰C.

Temperature when the reaction rate is twice faster

ln(k₂/k₁) = E/R(1/T₁ - 1/T₂)

where;

• T₁ is initial temperature = 21 ⁰C = 294 K
• k₁ is initial rate
• k₂ is final rate
• T₂ is final temperature
• E is activation energy
• R ideal gas constant

when rate is twice faster, k₂ = 2k₁ = 2(0.011) s⁻¹ = 0.022 s⁻¹

ln(0.022/0.011) = (45,500/8.31)(1/294 - 1/T₂)

0.693 = 5475.33(1/294 - 1/T₂)

1.2657 x 10⁻⁴ = 1/294 - 1/T₂

1/T₂ = 1/294 - 1.2657 x 10⁻⁴

1/T₂ = 3.275 x 10⁻³

T₂ = 305.4 K = 32.4 ⁰C

Thus, the temperature in degrees Celsius that the reaction will go twice as fast is  32.4 ⁰C.

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