The progress of the reaction `A hArr nB` with time is persented 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

a. `A hArr nB`
From the graph, it is clear that in `4` h, decrease in concentration of `A=0.2 "mol" L^(-1)`, increase in concentration of `B=0.4 "mol" L^(-1)`, increase in concentration of B is double as compared to decrease in concentration of A.
Therefore, `n=2`
b. `A hArr 2B`
At equilibrium,
`[A]=0.3`
At equilibrium,
`[B]=0.6`
`K_(c)=([B]^(2))/([A])=0.6^(2)/0.3=1.2`
c. For the first half hour, change in concentration of A is `0.6-0.5=0.1 "mol" L^(-1)`
Initial rate of conversion of A into B is
`(Delta[A])/(Deltat)=0.1/1=0.1 "mol" L^(-1) h^(-1)`