Sample 1Medium
A plot of ln k (y-axis) against 1/T (x-axis) for a reaction is a straight line with a slope of −5.0 × 10³ K. Taking R = 8.314 J K⁻¹ mol⁻¹, what is the activation energy of the reaction?
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The answer is C.
From ln k = ln A − Ea/RT, the slope of ln k against 1/T is −Ea/R. So Ea = 5.0 × 10³ K × 8.314 J K⁻¹ mol⁻¹ = 4.16 × 10⁴ J mol⁻¹ = 41.6 kJ mol⁻¹.
- A.
This divides the slope by R instead of multiplying by it.
- B.
Activation energy is always positive; the negative sign belongs to the slope, −Ea/R.
- D.
The factor 2.303 applies only when log k (base 10) is plotted; here the plot uses ln k.
NCERT: Class 12 Chemistry, Chapter 3