A sample of milk splits after 60 min. at 300 K and after 40 min. at 400 K when the population of lactobacillus acidophilus in it doubles. The activa tion energy (in kJ/ mol) for this process is closest to__________.
(Given, R = 8.3 J mol –1 K –1 , ln ( 3/2 ) = 0.4, e –3 = 4.0)

Solution

Chemistry ma’am · 4 steps

Milk that sours faster has a bigger rate constant, so the ratio of times gives you k₂/k₁ upside down.

1. Rate from time. The same change, doubling the bacteria, happens at both temperatures. So the rate constant is inversely proportional to the time taken: k ∝ 1/t. That gives k₄₀₀/k₃₀₀ = 60/40 = 3/2.

   ln k₂/k₁ = Eₐ/R(1/T₁-1/T₂)

2. Put in numbers. The left side is ln 3/2 = 0.4. The bracket is 1/300-1/400 = (4-3)/1200 = 1/1200 K⁻¹.
3. Solve for Ea. Eₐ = 0.4 × 8.3 × 1200 = 3984 J mol⁻¹. The value of e⁻³ is not needed.
4. So. Eₐ ≈ 3.98 kJ mol⁻¹, which is closest to 4.

The common slip: 3984. Leaving Eₐ in joules gives 3984, not 4. Divide by 1000 at the end because the answer is asked in kJ/mol.

The idea to keep. When the same amount of change happens in different times, k ∝ 1/t. Put the shorter time in the denominator so the hotter rate constant comes out larger.

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