The reaction `2AX(g)+2B_(2)(g)rarr A_(2)(g)+2B_(2)X(g)` has been studied kinetically and on the baiss of the rate law following mechanism has been proposed.
I. `2A X hArr A_(2)X_(2) " " ("fast and reverse")`
II. `A_(2)X_(2)+B_(2)rarrA_(2)X+B_(2)X`
III. `A_(2)X+B_(2)rarrA_(2)+B_(2)X`
where all the reaction intermediates are gases under ordinary condition.
form the above mechanism in which the steps (elementary) differ conisderably in their rates, the rate law is derived uisng the principle that the slowest step is the rate-determining step (RDS) and the rate of any step varies as the Product of the molar concentrations of each reacting speacting species raised to the power equal to their respective stoichiometric coefficients (law of mass action). If a reacting species is solid or pure liquid, its active mass, i.e., molar concentration is taken to be unity, the standard state. In qrder to find out the final rate law of the reaction, the concentration of any intermediate appearing in the rate law of the RDS is substituted in terms of the concentration of the reactant(s) by means of the law of mass action applied on equilibrium step.
Let the equilibrium constant of Step I be `2xx10^(-3) mol^(-1) L` and the rate constants for the formation of `A_(2)X` and `A_(2)` in Step II and III are `3.0xx10^(-2) mol^(-1) L min^(-1)` and `1xx10^(3) mol^(-1) L min^(-1)` (all data at `25^(@)C)`, then what is the overall rate constant `(mol^(-2) L^(2) min^(-1))` of the consumption of `B_(2)`?

Options

A. `6xx10^(-5)`
B. `1.2xx10^(-4)`
C. `3xx10^(-5)`
D. `1.5xx10^(-5)`

Answer

B

Explanation

Rate of overall reacation `=`Rate of step II
`(+(d[N_(2)O])/(dt))=k_(II)[A_(2)X_(2)][B_(2)]`
form equilibrium step, i.e., step I
`k_(c )=([A_(2)X_(2)])/([AX]^(2)), k_(c )=` equilibrium constant of step I
`[A_(2)X_(2)]=k_(c )[AX]^(2)`
(`k_(II)=` rate of constant of formation of `A_(2)X)`
`=k[AX]^(2)[B_(2)]`
`k=`Rate constant of overall reacation
Since for the rate of `A_(2)X` consumed `2mol` of `B_(2)` are formed so rate constant for the consmption of `B_(2)` will twice that of overall rate constant, i.e., `2k_(II)k_(c )`
`2xx3xx10^(-2)xx2xx10^(-3)= 1.2xx10^(-4)mol^(-2)L^(2)min^(-1)`