Question

The first-order reaction A→products has a halflife, t1/2, of 45.8 min at 25∘C and 2.0 min at 102∘C.
A Calculate the activation energy of this reaction.
B At what temperature would the half-life be 12.0 min ?

Answer 1

Answer:

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Answer 2

Given:

Half-life at T1 = 25°C = 298K

t1 = 45.8 min

Half-life at T2 = 102°C = 375K

t1 = 2 min

To Find:

activation energy (E) =?

the temperature(T3) at which half-life would be 12 min =?

Solution:

we know that,

$t_{1/2} = \dfrac{0.693}{k}$

thus, for T1 = 25°C = 298K, on putting values in above equation

$45.8 = \dfrac{0.693}{k1}$

k1 = 0.015

now, for T2 = 102°C = 375K, on putting values in above equation

$2 = \dfrac{0.693}{k2}$

k2 = 0.346

Now, for activation energy

$log\dfrac{k1}{k2} = \dfrac{E}{2.303R} \times \dfrac{T2-T1}{T2 T1}$

$log\dfrac{0.015}{0.346} = \dfrac{E}{2.303\times 8.314} \times \dfrac{375-298}{375\times 298}$

on solving for E

E = 37791.97 J

E = 37.8 KJ

Now, for half life(t3) = 12 min

$t_{1/2} = \dfrac{0.693}{k}$

$12 = \dfrac{0.693}{k3}$

k3 = 0.057

now for temperature

$log\dfrac{k3}{k1} = \dfrac{E}{2.303R} \times \dfrac{T3-T1}{T3 T1}$

$log\dfrac{0.057}{0.015} = \dfrac{37791.9}{2.303\times 8.314} \times \dfrac{T3-298}{T3\times 298}$

On solving for T3

T3 = 326.39K