Question

How many times the rate of formation of $A_2$ will increase of concentrations of AX is doubled and that of $B_2$ is increased three fold ?

• A. 36
• B. 12
• C. 6
• D. 8

Answer 1

The correct option is B 12

Rate of overall reaction = Rate of step II
$(+\frac{d\left[N_{2}O\right]}{dt})=k_{II}\left[A_2{X}_2\right]B_2$
From equilibrium step, i.e., step I :
$k_c=\frac{\left[A_2{X}_2\right]}{[AX{]}^2}{k}_c$ = equilibrium constant of step I
$\left[A_2{X}_2\right]=k_c[AX{]}^2$
Thus, rate of overall reaction
$=k_{II}{k}_c\left[AX{]}^2\right[B_2]$
($k_{II}$ = rate constant of formation of $A_{2}X$)
$=k\left[AX{]}^2\right[B_2]$
so it will increase by 12 times.