The activation energy of one of the reactions in a biochemical process is 532611 J mol⁻¹. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k₃₀₀ = x × 10⁻³ k₃₁₀. The value of x is ____________.
[Given : ln 10 = 2.3, R = 8.3 J K⁻¹ mol⁻¹]

Solution

Chemistry ma’am · 4 steps

Use the two-temperature Arrhenius form; the numbers are chosen so the log comes out as exactly 3 ln 10.

1. Two-temperature form. Write k = A e^(-Eₐ/RT) at 310 K and at 300 K and divide one by the other. A cancels, so only Eₐ, R and the two temperatures remain.

   ln k₃₁₀/k₃₀₀ = Eₐ/R(1/300-1/310)

2. Put in numbers. Eₐ/R = 532611/8.3 = 64170 K, and the bracket is 10/93000 K⁻¹. So ln k₃₁₀/k₃₀₀ = 641700/93000 = 6.9.
3. Undo the log. 6.9 = 3 × 2.3 = 3 ln 10 = ln 10³. So k₃₁₀/k₃₀₀ = 10³, which means k₃₀₀ = 10⁻³ k₃₁₀.
4. So. Comparing with k₃₀₀ = x × 10⁻³ k₃₁₀ gives x = 1.

The common slip: 1000. Reporting the ratio k₃₁₀/k₃₀₀ = 1000 as x gives 1000. The question writes k₃₀₀ as x × 10⁻³ k₃₁₀, so the 10⁻³ is already there and x is 1.

The idea to keep. For a rate-constant ratio between two temperatures, use ln k₂/k₁ = Eₐ/R(1/T₁-1/T₂). Then check whether the log is a neat multiple of ln 10.

Answer from the NTA final answer key. Solution by PrepWiser.