The progress of the reaction `A hArr nB` with time is presented in the figure given below:

Determine
a. The value of n.
b. The equilibrium constant K.
c. The initial rate of concentration of A.

Explanation

Loss in concentration of A in `I` hour `=(0.6-0.5)/(1)=0.1`
Gain in concentration of `B` in I hour `=(0.2-0)/(1)=0.2`
(i) `:' 0.1` mole of `A` changes to `0.2` mole of `B` in a given time and thus, `n=2`
(ii) Equilibrium constant,
`K=([B]^(n))/([A])=([0.6]^(2))/(0.3)=1.2 mol litre^(-1)`
( `:'` Equilibrium is attained after `5` hr, where `[B]=0.6` and `[A]=0.3`)
(iii) Initial rate of conversion of A = changes in conc. of `A` during `I` hour
`=0.1/1=0.1 mol litre^(-1) hour^(-1)`