The rate law of a reaction is rate = k[A]^n. When [A] is raised from 0.20 to 0.80 M the initial rate rises from 1.6x10^-3 to 1.28x10^-2 M s^-1. Find k in units of 10^-2 M^-1/2 s^-1 (two decimals; take 0.20^1.5 = 0.08944).

JEE Main levelNumerical answerDifficulty 2 of 3ck018

The rate law of a reaction is rate $= k [A]^{n}$ . When $[A]$ is raised from $0.20$ to $0.80 M$ the initial rate rises from $1.6 \times 1 0^{- 3}$ to $1.28 \times 1 0^{- 2} M s^{- 1}$ . Find $k$ in units of $1 0^{- 2} M^{- 1/2} s^{- 1}$ (two decimals; take $0.2 0^{1.5} = 0.08944$ ).

Numerical answer type, answer in 10^-2 M^-1/2 s^-1, tolerance 0.02.

Rate ratio

= 8.00

for a concentration ratio of 4, so

4^{n} = 8.00

gives

n = ln 8.00/ ln 4 = 1.5

. Then

k = 1.6 \times 1 0^{- 3} /0.08944 = 0.01789

, i.e.

1.79 \times 1 0^{- 2}