Question

Solve this
(Question type : Image)

Answer 1

We have to find out the value of pre-exponential factor and Activation energy.

Arrhenius equation is given as :-

$\boxed{ \rm \large K = A {e}^{ \frac{ - Ea }{RT} } }$

Where :- A = frequency or pre-exponential factor

Ea = Activation energy

R = gas constant

T = temperature

$\rm \longrightarrow\large K = A {e}^{ \frac{ - Ea }{RT} } \\ \\ \\ \rm \longrightarrow\large \: log K =log( A {e}^{ \frac{ - Ea }{RT} } ) \\ \\ \\ \rm \longrightarrow\large \: log K =log A + log \: {e}^{ \frac{ - Ea }{RT} } \\ \\ \\ \rm \longrightarrow\large \: log K =log A - { \dfrac{ Ea }{2.303 \: RT} } .....(1)$

In the question , it is given that :-

$\rm \longrightarrow\large \: log K = - (2000) { \dfrac{1}{T} } + 6.0 \\ \\ \\ \rm \longrightarrow\large \: log K = 6.0 - (2000) { \dfrac{1}{T} } .....(2)$

Now , compare (1) and (2)

➡️log A = 6

➡️ A = 10⁶

$\rm \longrightarrow\large \:{ - \dfrac{ Ea }{2.303 \: RT} } = \:( - 2000){ \dfrac{ 1}{T} } \\ \\ \\ \rm \longrightarrow\large \:{ \dfrac{ Ea }{2.303 \: RT} } = \:( 2000){ \dfrac{ 1}{T} } \\ \\ \\ \rm \longrightarrow\large \:{ \dfrac{ Ea }{2.303 \times 8.314} } = \:( 2000)\\ \\ \\ \rm \longrightarrow\large \:{ Ea } = \:2000 \times 2.303 \times 8.314\\ \\ \\ \rm \longrightarrow\large \:{ Ea } = 38294.2 \: J$

Now , we will convert 38294.2 J/mol into Kilo joule/mol.

➡️38. 2942 KJ/mol

Answer :-

pre-exponential factor (A) = 10⁶

Activation energy = 38. 2942 KJ/mol

Therefore, option (D) is correct