The decomposition of a gas on a metal surface is zero order with k = 1.5x10^-3 mol L^-1 min^-1. The time for the concentration to fall from 0.10 to 0.04 mol L^-1 is

JEE Main levelSingle correct optionDifficulty 2 of 3ck037

The decomposition of a gas on a metal surface is zero order with $k = 1.5 \times 1 0^{- 3} mol L^{- 1} mi n^{- 1}$ . The time for the concentration to fall from $0.10$ to $0.04 mol L^{- 1}$ is

For zero order,

[A]_{0} - [A] = k t

. So

t = (0.10 - 0.04) /1.5 \times 1 0^{- 3} = 40 min

. The value 611 min would result from wrongly using the first order formula.