The rate of a reaction depends upon the temperature and is quantitatively expressed as
k = A e![]()
i) If a graph is plotted between log k and 1/T, write the expression for the slope of the reaction?
ii) If at under different conditions Ea1 and Ea2 are the activation energy of two reactions. If Ea1 = 40 J / mol and Ea2 = 80 J / mol. Which of the two has a larger value of the rate constant?
Given, rate of a reaction depends upon the temperature and is expressed as
k = A e![]()
Taking logarithm on both sides, we get
![]()
The equation is like
. If the graph is plotted between y (
) and x (
), the slope (m) can be calculated.

(i) So, slope is = ![]()
(ii) As the value of activation energy, EA increases, the value of rate constant, k decreases.
So if, EA1 (40 J / mol ) < EA2 (80 J / mol) then
k1>k2 respectively.
Couldn't generate an explanation.
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