Chemical Kinetics - Result Question 7
####7. For a reaction scheme, $A \xrightarrow{k _1} B \xrightarrow{k _2} C$, if the rate of formation of $B$ is set to be zero then the concentration of $B$ is given by
(2019 Main, 8 April II)
(a) $k _1 k _2[A]$
(b) $\left(\frac{k _1}{k _2}\right)[A]$
(c) $\left(k _1-k _2\right)[A]$
(d) $\left(k _1+k _2\right)[A]$
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Answer:
Correct Answer: 7. (a)
Solution:
- $A \xrightarrow{K _1} B \xrightarrow{K _2} C$
Rate of formation of $B$ is
$$ \begin{array}{rlr} & \frac{d[B]}{d t}=k _1[A]-k _2[B] & \ \Rightarrow & \quad 0=k _1[A]-k _2[B] \quad\left[\because \text { Given, } \frac{d[B]}{d t}=0\right] \ \Rightarrow & k _2[B]=k _1[A] & \ \Rightarrow & \text { Concentration of } B,[B]=\frac{k _1}{k _2}[A] & \end{array} $$
