The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700 - 1000 K. The data has been analysed by plotting ln k vs 10³/T graph. The value of activation energy for the reaction is ___________ kJ mol⁻¹. (Nearest integer)
(Given : R = 8.31 J K⁻¹ mol⁻¹)

Solution

Chemistry ma’am · 3 steps

Read the slope, but watch the x-axis: it is 10³/T, not 1/T, so a factor of 1000 hides in the slope.

1. Arrhenius as a line. Take the natural log of k = A e^(-Eₐ/RT) to get ln k = ln A - Eₐ/R·1/T. Since the graph's x-axis is 10³/T, write 1/T = 1/10³·10³/T.

   ln k = ln A - Eₐ/(10³ R)(10³/T) ⇒ slope = -Eₐ/(10³ R)

2. Use the slope. The graph gives a slope of -18.5, so Eₐ/(10³ R) = 18.5. Then Eₐ = 18.5 × 10³ × 8.31 = 153 735 J mol⁻¹.
3. So. Eₐ ≈ 153.7 kJ mol⁻¹, which rounds to 154.

The common slip: 354. Treating the graph as log k and multiplying by 2.303 gives about 354, but the y-axis here is ln k, so no 2.303 is needed.

The idea to keep. For a ln k vs 1/T plot, slope = -Eₐ/R. Use -Eₐ/(2.303R) only when the y-axis is log k, and always check whether the x-axis is scaled by 10³.

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