Ask Questions, Get Answers


For the reaction $A\;\;\;\underrightarrow{k_1}\;\;\;B\;\;\;\underrightarrow{k_2}\;\;\;C$, select the correct statement:

(A) $[B]_\text{max}=[A_o]\big(\frac{k_1}{k_2-k_1}\big)^{\large{k_2/k_2+k_1}}$ <\p> (B) $[B] = \large\frac{k_2[A_o]}{k_2-k_1}$$\big(e^{-k_2t}-e^{-k_1t})$ <\p> (C) $[C] = [A]_0$$ \big ( 1 - \large\frac{k_1 e^{-k_2t} - k_2e^{k_1t}}{k_2+k_1}$$\big)$ <\p> (D) $t_\text{max} = \large\frac{1}{k_2-k_1}$$\ln \large\frac{k_1}{k_2}$, the time at which concentration of $[B]$ is maximum

1 Answer

Answer: $t_\text{max} = \large\frac{1}{k_2-k_1}$$\ln \large\frac{k_1}{k_2}$ is correct
For the consecutive reactions $A \rightarrow B \rightarrow C$, at $t=0, [A] = [A]_0, [B] = 0, [C] = 0$, and at all times $[A]+[B]+[C] = [A]_0$
It follows that the rate equations for concentrations of A, B and C are as follows:
$\quad \large\frac{d[A]}{dt}$$ = -k_1[A]$
$\quad \large\frac{d[B]}{dt}$$ = -k_1[A] - k_2[B]$
$\quad \large\frac{d[C]}{dt}$$ = -k_1[B]$
Integrating $\large\frac{d[A]}{dt}$$ = -k_1[A] \rightarrow [A] = [A]_0 e^{-k_1t}$
Integrating $\large\frac{d[B]}{dt}$$ \rightarrow [B] = \large\frac{k_1[A_o]}{k_2-k_1}$$\big(e^{-k_1t}-e^{-k_2t})$
Since $[C] = [A]_0 - [B] - [A]$, we get: $[C] = [A]_0$$ \big ( 1 + \large\frac{k_1 e^{-k_2t} - k_2e^{k_1t}}{k_2-k_1}$$\big)$
To arrive at $[B]_\text{max}, \large\frac{d[B]}{dt}$$ = 0 = \large\frac{k_1[A_o]}{k_2-k_1}$$\big(e^{-k_1t}-e^{-k_2t})$
$\Rightarrow t_\text{max} = \large\frac{1}{k_2-k_1}$$\ln \large\frac{k_1}{k_2}$, which when substituted back in the equation for [B]. we get:
From the above analysis, its clear that the correct statement is $t_\text{max} = \large\frac{1}{k_2-k_1}$$\ln \large\frac{k_1}{k_2}$, the rest of the statements which are values for $[A], [B]$ and $[C]$ as stated in the question are incorrect.
answered Dec 17, 2013 by sreemathi.v
edited Jul 27, 2014 by balaji.thirumalai

Related questions

Download clay6 mobile appDownload clay6 mobile app