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(D) The rate law for the reaction $2AB + B 	o A_2B_3$ is given by $r = k[AB]^2[B]^1$.Let the initial moles of $AB$ and $B$ be $n$ each.In vessel $1$

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$8 : 1$

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(D) The rate law for the reaction $2AB + B 	o A_2B_3$ is given by $r = k[AB]^2[B]^1$.Let the initial moles of $AB$ and $B$ be $n$ each.In vessel $1$ $(V_1 = 1 \ dm^3)$,the concentrations are $[AB]_1 = n/1 = n$ and $[B]_1 = n/1 = n$.So,$r_1 = k(n)^2(n) = kn^3$.In vessel $2$ $(V_2 = 2 \ dm^3)$,the concentrations are $[AB]_2 = n/2$ and $[B]_2 = n/2$.So,$r_2 = k(n/2)^2(n/2) = k(n^3/8) = kn^3/8$.The ratio $r_1/r_2 = (kn^3) / (kn^3/8) = 8/1$ or $8 : 1$.

The initial reaction $2AB + B \to A_2B_3$ is carried out in vessels of volume $1 \ dm^3$ and $2 \ dm^3$ respectively,using equal moles of reactants. The ratio of the reaction rates $(r_1/r_2)$ is:

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