The rate constant (k) of a reaction is measured at differenct temperatures (T), and the data are plotted in the given figure. The activation energy of the reaction in kJ mol –1 is :
(R is gas constant)

Solution

Chemistry ma’am · 4 steps

The slope of a ln k versus 1/T line is -Eₐ/R, so read the slope and watch the 10³ on the x-axis.

1. Straight-line form. Take the natural log of k = A e^(-Eₐ/RT). Plotted against 1/T, ln k is a straight line with slope -Eₐ/R.

   ln k = ln A - Eₐ/R·1/T

2. Read the slope. The line meets the y-axis at ln k = 10 and the x-axis at 10³/T = 5. So ln k falls by 2 for every 1 unit of 10³/T.
3. Undo the axis scale. One unit of 10³/T is 10⁻³ K⁻¹ of 1/T. The true slope is -2/10⁻³ = -2000 K, so Eₐ/R = 2000 K.
4. So. Eₐ = 2000R J mol⁻¹ = 2R kJ mol⁻¹, which is option (b).

The idea to keep. For a ln k versus 1/T plot, slope = -Eₐ/R. Always check how the x-axis is scaled. A 10³ there turns the slope from J into kJ for free.

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