Question

Heya☺

Question in attachment

Kinetics(Activation Energy)

Answer plzZ✔✔✔

Answer 1

Hey mate!!!!

We know that..

Arrhenius Equation :-

$= > K = A .{e}^{ - \frac{Ea}{RT} }$
$\\ = > log_{}K = log_{}A - \frac{Ea}{2.303RT}$
For second Reaction :-
$= > logK _{2} = logA - \frac{Ea _{2} }{2.303RT} \: \: \: ...(Eq _{1})$
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For First Reaction :-

$= > logK _{1} = logA - \frac{Ea _{1} }{2.303RT} \: \: \: ..(Eq _{2})$
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Substituting eq(2) from eq(1)

$= > log \frac{K _{2}}{K _{1} } = \frac{Ea _{1} - Ea _{2} }{2.303RT}$
________________

Given :-

$= > Ea _{1} - Ea _{2} = 24.3 \: kj {mol}^{ - 1} \\ = 24900 \: j {mol}^{ - 1}$

=> T = 300 K

$= > R = 8.314 \: j {k}^{ - 1} {mol}^{ - 1}$
___________

$= > log \frac{K _{2} }{K _{1}} = \frac{24900}{2.303 \times 8.314 \times 300} = 4.34$
Then

$= > \frac{K _{2} }{K _{1}} = 21877.6 = 2.19 \times {10}^{4}$
_______________[ANSWER]

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Answer 2

Answer:

satyam sorry mein tumhe unblock nhi kar payi