JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Chemical Kinetics

For a certain first order reaction

32%

of the reactant is left after 570 s . The rate constant of this reaction is ____

\times 1 0^{- 3} s^{- 1}

. (Round off to the nearest integer).
[Given :

lo g_{10} 2 = 0.301, ln 10 = 2.303

]

Answer: 2

Solution

For a first-order reaction

k = \frac{2.303}{t} lo g \frac{[ A ] _{0}}{[ A ]}

With

32%

of the reactant left,

\frac{[ A ] _{0}}{[ A ]} = \frac{100}{32}

and

lo g \frac{100}{32} = 2 - 5 lo g 2 = 2 - 5 (0.301) = 0.495

. Substituting,

k = \frac{2.303 \times 0.495}{570} = 2.0 \times 1 0^{- 3} s^{- 1}

so the rate constant is

2 \times 1 0^{- 3}

^{- 1}

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Integer answer