For the reaction, aA + bB → cC + dD, the plot of log k vs 1/T is given below :
The temperature at which the rate constant of the reaction is 10⁻⁴ s⁻¹ is _________ K. (Rounded off to the nearest integer)
[Given : The rate constant of the reaction is 10⁻⁵ s⁻¹ at 500 K.]

Solution

Chemistry ma’am · 4 steps

The slope of the log k vs 1/T line already contains Ea, so you do not need to work Ea out at all.

1. Read the slope. From log k = log A - Eₐ/2.303R·1/T, the slope of the graph is -Eₐ/2.303R = -10000 K. So Eₐ/2.303R = 10000 K.

   log k₂/k₁ = Eₐ/2.303R(1/T₁-1/T₂)

2. Put in numbers. k₂/k₁ = 10⁻⁴/10⁻⁵ = 10, so the left side is log 10 = 1. Then 1 = 10000(1/500-1/T₂), which gives 1/500-1/T₂ = 1 × 10⁻⁴ K⁻¹.
3. Solve for T2. 1/T₂ = 0.002 - 0.0001 = 0.0019 K⁻¹, so T₂ = 1/0.0019 = 526.3 K. The rate constant went up, so the temperature must go up from 500 K, and it does.
4. So. The rate constant is 10⁻⁴ s⁻¹ at about 526.3 K, which rounds to 526.

The common slip: 476. Getting the sign of the bracket wrong gives 1/T₂ = 0.0021 K⁻¹ and 476 K. A larger k cannot come with a lower temperature.

The idea to keep. On a log k vs 1/T plot, the slope is -Eₐ/2.303R. Use that whole value directly in the two-temperature equation, and check that a larger k comes with a higher T.

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