JEE Main 31-Aug-2021 Shift 1 Solved Paper

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

2.303

t

‌log‌

a

a−x

where, a = initial concentration of reactant and
a - x − = concentration of reactant after time t.
For 50% completion,
k=

2.303

t50%

‌log‌

100

100−50

...(i)
For 75% completion,
k=

2.303

t75%

‌log‌

100

100−75

...(ii)
From Eqs. (i) and (ii),

2.303

t50%

‌log‌

100

50

=

2.303

t75%

‌log‌

100

25

0.3010

t50%

=

0.6020

t75%

t75%=2t50%
∴

t75%

t50%

=

2

1