form the gaseous reaction
`2A + B_(2) rarr 2AB`, the following rate data were obtained at `300 K`.

Calculate the rate constant for the reaction and the rate of formation of `AB` when `[A]` is `0.02` and `[B_(2)]` is `0.04 mol L^(-1)` at `300K`.

Explanation

`2A +B_(2) rarr 2AB`
Let the rate of reaction is
`r = k[A]^(alpha)[B_(2)]^(beta)`
`r_(1) = 1.8 xx 10^(-3) = k[0.015]^(alpha) . [0.15]^(beta)` …(i)
`r_(2) = 1.08 xx 10^(-2) = k[0.09]^(alpha).[0.15]^(beta)` …(ii)
`r_(3) = 5.4 xx 10^(-3) = k[0.015]^(alpha).[0.45]^(beta)` ...(iii)
Divide Eq. (ii) by Eq. (i).
`(1.08 xx 10^(-2))/(1.8 xx 10^(-3)) = (k[0.09]^(alpha)[0.15]^(beta))/(k[0.015]^(alpha)[0.15]^(beta)) rArr (6)^(1) = (6)^(alpha) rArr alpha = 1`
Divide Eq. (iii) by Eq.(i),
`(5.4 xx 10^(-3))/(1.8 xx 10^(-3)) = (k[0.015]^(alpha)[0.45]^(beta))/(k[0.015]^(alpha)[0.15]^(beta)) rArr (3)^(1) = (3)^(beta) rArr beta = 1`
`:. r = k[A]^(1)[B_(2)]^(1)`
Substitute the value of `r`, `[A]`, and `[B_(2)]` in Eq. (i) and calculate `k`.
`1.8 xx 10^(-3) = k[0.015]^(1)[0.15]^(1)`
`:. k = 0.8 L mol^(-1) min^(-1)` (unit of second order reaction)
`2A + B_(2) rarr 2AB`
`(-d[A])/(2 dt) = (-d[B_(2)])/(dt) = (+d[AB])/(2 dt)`
`(d[AB])/(dt) = 2 xx (-[B_(2)])/(dt)`
Rate `= (-[B_(2)])/(dt) = k[A]^(1)[B_(2)]^(1) = 0.8 xx (0.02)^(1)(0.04)`
`= 0.64 xx 10^(-3) mol L^(-1) min^(-1)`
`:. (d[AB])/(dt) = 2 xx 0.64 xx 10^(-3)`
`= 1.28 xx 10^(-3) mol L^(-1) min^(-1)`