The progress of the reaction `A hArr nB`, with times is presented in the figure below.

Determine:
(a) The value of `n` , (b) `K_(eq)`
( c) Initial rate of converison of `A`

Explanation

In the given time of `4 hr` (`1` to `5`), concentration of `A` falls form `0.5` to `0.3 M`, while in the same time concentration of `B` increases form `0.2` to `0.6 M`. The increase in the concentration of `B` in a given time s twice the decrease in the concentration of `A`.
(a) Thus, `n = 2`
(b) `([B_(eq)]^(2))/([A]_(eq)) = (0.6)^(2)/(0.3) = 1.2`
( c) Initial rate of converison of `A = (0.6-0.5)/(1-0) = 0.1 M hr^(-1)`