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  • For a two step reaction;               <img alt="\mathrm{1^{\text {st }} \text { step: } \quad A \stackrel{\text { ker }}{\rightleftharpoons} 2 B}"

For a two step reaction;

              \mathrm{1^{\text {st }} \text { step: } \quad A \stackrel{\text { ker }}{\rightleftharpoons} 2 B}

              \mathrm{2^{\text {nd }} \text { step: } A+B \stackrel{k}{\longrightarrow} P}

 Assuming equilibrium time to be very small, the rate of formation of P is proportional to

Option: 1

\mathrm{[A]^{1 / 2}}


Option: 2

\mathrm{[A]}


Option: 3

\mathrm{[A]^{2}}


Option: 4

\mathrm{[A]^{3/2}}


Answers (1)

for \mathrm{2^{\text {nd }} step, \, \, \, \frac{d[P]}{d t}=k[A][B]-(1)}

from \mathrm{1 ^{st} \, step, \quad K_{e q}=\frac{[B]^2}{[A]} \Rightarrow[B]=(\operatorname{Keq}[A])^{1 / 2}}

Putin (1), \mathrm{\frac{d[P]}{d t}=K_{e q}^{1 / 2}[A]^{3 / 2}}

                     \mathrm{\frac{d[P]}{d t} \propto[A]^{3 / 2}}

Posted by

Ramraj Saini

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