Question

(i) Write the mathematical expression relating the variation of rate constant of a reaction with temperature.
(ii) How can you graphically find the activation energy of the recation from the Arrhenius expression?
(iii) The slope of the line in the graph of log k(k=rate constant) versus $\frac{1}{T}$ is $-5841$. Calculate the activation energy of the reaction.

Answer 1

(i) Mathematical expression is Arrhenius equation:
$k=A{e}^{-Ea/RT}$
where k is rate constant of the given reaction.
A is another constant called frequency factor.
$E_a$ is activation energy
R is gas constant and
T is temperature in Kelvin.

(ii) Finding activation energy graphically:
Log form of Arrhenius equation is $lo{g}_{e}k=lo{g}_{e}A-\frac{E_a}{RT}$
or $lo{g}_{10}k=lo{g}_{10}A-\frac{E_a}{2.303RT}$
It is a straight line equation $(y=mx+c)$ when different values of rate constant k are calculated at different temperatures for a given reaction and if a graph is plotted between $lo{g}_{10}k$ against $\frac{1}{T}$ then slope of the graph line should be equal to $-\frac{E_a}{2.303R}$
Using $\frac{dy}{dx}=-\frac{E_a}{2.303R}$, value of $E_a$ can be calculated.

(iii) Slope$=-\frac{E_a}{2.303R}$
or $-E_a=-5841\times 2.303\times 8.314$
$E_a=111838.45$J $mo{l}^{-1}$
$=111.838$kJ $mo{l}^{-1}$.