Question

a first order reaction is 30% completed in 50 minutes at 300k calculate half life ?​

Answer 1

Answer:

For the first-order reaction, the relationship between the half-life period and the rate constant is

k=

t

1/2

0.693

=

30

0.693

=0.0231min

−1

.

The rate law expression is k=

t

2.303

log

(a−x)

a

.

Substitute values in the above expression.

0.0231=

70

2.303

log

(a−x)

a

log

(a−x)

a

=0.7021

(a−x)

a

=5.036

a

a−x

=

5.036

1

=0.2

Thus, the percentage of the reactant remaining after 70 min will be 20%.

Explanation:

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Answer 2

Answer:

$k=t1/20.693=300.693=0.0231min−1.</p><p>The rate law expression is k=t2.303log(a−x)a.</p><p>Substitute values in the above expression.</p><p></p><p>0.0231=702.303log(a−x)a</p><p></p><p>log(a−x)a=0.7021</p><p></p><p>(a−x)a=5.036</p><p></p><p>aa−x=5.0361=0.2</p><p></p><p>$

Thus, the percentage of the reactant remaining after 70 min will be 20%.