Question

The slope of the line in a graph of logk versus 1/T for a reaction is -5841.calculate the activation energy of the reaction.

Answer 1

Answer : The activation energy of the reaction is, 111.838 kJ/mole

Explanation :

The Arrhenius equation is written as:

$K=A\times e^{\frac{-Ea}{RT}}$

Taking logarithm on both the sides, we get:

$\ln k=-\frac{Ea}{RT}+\ln A$

or,

$\log k=-\frac{Ea}{2.303RT}+\log A$       ............(1)

where,

k = rate constant

Ea = activation energy

T = temperature

R = gas constant  = 8.314 J/K.mole

A = pre-exponential factor

The equation (1) is of the form of , y = mx + c i.e, the equation of a straight line.

Thus, if we plot a graph of $\log k$ vs $\frac{1}{T}$ then the graph shows a straight line with negative slope. That means,

Slope of the line = $-\frac{Ea}{2.303R}$

As we are given that:

Slope of the line = -5841

Now we have to calculate the activation energy of the reaction by using above relation.

Slope of the line = $-\frac{Ea}{2.303R}$

$-5841=-\frac{Ea}{2.303\times (8.314 J/K.mole)}$

$Ea=111838.4564J/mole=111.838kJ/mole$

Therefore, the activation energy of the reaction is, 111.838 kJ/mole

Answer 2

Answer:

Activation Energy = 111.8 KJ/mole

Explanation:

just see the uploaded photo