Question

For a reaction scheme $\ce{A ->[k_1] B}\ce{ ->[k_2] C}$ , if the rate of formation of $B$ is set to be zero then the concentration of $B$ is given by :

• A. $\left(\frac{k_{1}}{k_{2}}\right)\left[A\right]$
• B. $k_{1}k_{2}[A]$
• C. $(k_{1}+k_{2}) [A]$
• D. $\left(\frac{k_{1}}{k_{2}}\right)[ B ]$

Answer 1

Answer: (A)

Solution:- (A) $\left(\frac{ k _{1}}{ k _{2}}\right)[ A ]$
$A \stackrel{ k _{1}}{\longrightarrow} B \stackrel{ k _{2}}{\longrightarrow} C$
According to given question,
$ \frac{ d [ B ]}{ dt }=0= K _{1}[ A ]- K _{2}[ B ] $
$ \Rightarrow[ B ]=\frac{ K _{1}}{ K _{2}}[ A ] $