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Consider the consecutive reaction \mathrm{A \rightarrow B \rightarrow C}. If the initial concentration of \mathrm{A } is \mathrm{2M }. & the rate constants for the first  reaction & second reaction are \mathrm{0.2 s^{-1} } and \mathrm{0.1 s^{-1} } respectively, what is the concentration of \mathrm{B } when the reaction reaches equilibrium ?

Option: 1

0.5 M


Option: 2

1 M


Option: 3

1.5 M


Option: 4

2 M


Answers (1)

best_answer

In the consecutive reaction , the concentration of in intermediate (B) is given as follows:
\mathrm{[B]= [A]_{0}\left ( k_{2}/k_{1} \right )\left ( \frac{k_{2}}{k_{1}-k_{2}} \right )}
\mathrm{[B]=2 \times \left ( \frac{0.1}{0.2} \right )\frac{0.1}{0.2-0.1}}
\mathrm{[B]=2 \times\frac{1}{2}= 1\, M}

Posted by

Deependra Verma

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