For a reaction, given below is the graph of ln k vs 1/T. The activation energy for the reaction is equal to ____________ cal mol⁻¹. (nearest integer)
(Given : R=2 cal K⁻¹ mol⁻¹ )

Solution

Chemistry ma’am · 4 steps

On an ln k vs 1/T graph, the slope is -Eₐ/R, so read the slope off two clean points.

1. Straight-line form. Take the natural log of k = A e^(-Eₐ/RT). It becomes the equation of a straight line in ln k and 1/T.

   ln k = ln A - Eₐ/R·1/T

2. Read the slope. The line passes through (0, 20) and (5 K⁻¹, 0). So the slope is (0 - 20)/(5 - 0) = -4 K.
3. Find Ea. Set -Eₐ/R = -4 K. Then Eₐ = 4 K × 2 cal K⁻¹ mol⁻¹ = 8 cal mol⁻¹.
4. So. The activation energy is 8 cal mol⁻¹.

The common slip: 18. Using the log₁₀ form, slope = -Eₐ/2.303R, on an ln k graph gives 2.303 × 2 × 4 ≈ 18, not 8.

The idea to keep. Check the y-axis first. With ln k, the slope is -Eₐ/R. With log k, the slope is -Eₐ/2.303R. The intercept gives ln A, never Eₐ.

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