Question

For the following sequential reaction:

$A\stackrel{{\lambda }_1}{\to }B\stackrel{{\lambda }_2}{\to }C$,

If the ${\lambda }_1=1.8\times {10}^{-5}{s}^{-1}$ and ${\lambda }_2=1.1\times {10}^{-2}{s}^{-1}$ and initial molar concentration of $A$ is 1.8 $M$, the concentration of $C$ at time $t=1day$ is :

• A. $1.20M$
• B. $1.42M$
• C. $4.12M$
• D. None of these

Answer 1

Answer: (B)

$1.42M$

We know, $A\stackrel{{\lambda }_1}{\to }B\stackrel{{\lambda }_2}{\to }C$

Given, $[A{]}_0=1.8M,{\lambda }_1=1.8\times {10}^{-5}{s}^{-1}and{\lambda }_2=1.1\times {10}^{-2}{s}^{-1}$.

Since, ${\lambda }_1<<{\lambda }_2$, thus, $B$ will be converted to $C$ at higher rate than $A$ is converted to $B$. Thus, sequential reaction may therefore, be written as:

$A\stackrel{{\lambda }_1}{\to }C$

$\therefore [A{]}_0=[A{]}_t+[C{]}_t$ ... (i)

For $1^{st}$ order reaction, rate expression in integrated form gives:
$[A{]}_t=[A{]}_0{e}^{-{\lambda }_{1}t}$ ... (ii)

Thus, by eqs. (i) and (ii),
$[A{]}_0=[A{]}_0{e}^{-{\lambda }_{1}t}+[C{]}_t$

or $[C{]}_t=[A{]}_0[1-e^{-{\lambda }_{1}t}]$

$=1.8[1-e^{-1.8\times {10}^{-5}\times 24\times 3600}]$

$[C{]}_t=1.42M$