Rate constant of a first order reaction is 0.0693 min -1 . If we start with 20 mol L-1, it is reduced to 2.5 mol L-1 in:

Question

Rate constant of a first order reaction is 0.0693 min -1 . If we start with 20 mol L-1, it is reduced to 2.5 mol L-1 in: (A) 10 min (B) 20 min (C) 30 min (D) 40 min

Tardigrade Admin · Accepted Answer

For, 1st order reaction, k=(2.303/t)= log (a/(a-x)) Hence, t=(2.303/0.0693)× log (20/2.5) =(2.303/0.0693)× log (2)3 =(2.303/0.0693)× 3× 0.3010=30 min

Q. Rate constant of a first order reaction is $0.0693 min^{- 1}$ . If we start with $20 m o l L^{- 1},$ it is reduced to $2.5 m o l L^{- 1}$ in:

10 min

20 min

30 min

40 min

For, 1st order reaction, $k = \frac{2.303}{t} = lo g \frac{a}{( a - x )}$ Hence, $t = \frac{2.303}{0.0693} \times lo g \frac{20}{2.5}$ $= \frac{2.303}{0.0693} \times lo g (2)^{3}$ $= \frac{2.303}{0.0693} \times 3 \times 0.3010 = 30 min$

Q. Rate constant of a first order reaction is 0.0693min−1 . If we start with 20molL−1, it is reduced to 2.5molL−1 in: