Question

The percentage of C that has reacted after 70 minutes.

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%.