The progress of the reaction, A ⇌ nB with time, is presented in figure given below. Determine.

(i) The value of n,
(ii) The equilibrium constant K, and
(iii) the initial rate of conversion of A.

Explanation

(i) 2 (ii) 1.2 (iii) `0.1" mole"^(–1) hr^(–1)`
(i) According to the figure, in the given time of 4 hours (1 to 5) concentration of A falls from 0.5 to 0.3 M, while in the same time concentration of B increases from 0.2 M to 0.6 M.
Decrease in concentration of A in 4 hours
`=0.5-0.3=0.2 M`
Increase in concentration of B in 4 hours
`=0.6-0.2=0.4 M`
Thus increase in concentration of B in a given time is twice then the decrease in concentration of A. Thus n 2.
(ii) `K=([B^(2)]_(eq))/([A]_(eq))=((0.6)^(2))/(0.3) =1.2 M`
(iii) Initial rate of conversion of A
= Change in conc. of A during 1 h
`=(0.6-0.5)/(1) =0.1" mole"^(-1) "hr"^(-1)`