CGPET 2021 Solved Paper

For first order reaction, rate constant is given as
k=

2.303

t

‌log‌

a

(a−x)

.........(i)
(a= amount of reactant used and a−x= amount of reactant left after time t )
Given, t

1

2

( half-life )=14.5‌min
∴ t

1

2

=

0.6932

k

(for first order reaction)
14.5&=

0.6932

k

⇒k=0.0478min−1
From (i),
0.0478min−1=

2.303

24min

‌log‌

100%

a−x

0.0478×24

2.303

=log‌100−log(a−x)[log‌

a

b

=log‌a−log‌b]
0.4982=2−log(a−x)
1.5=log(a−x)
antilog(1.5)=(a−x)
31.8%=(a−x)
∴ 31.8% of reactant will remain after 24 min.