Question

Select the equation for [B] at any time t in the following reaction. Let the initial concentration of A be $\left[A_0\right]$ and B and C be 0.

• A. $\left[B\right]=\left[B{]}_0\left(\frac{k_1\left[A\right]}{k_1+k_2}\right)\right(1-e^{-(k_1+k_2)t})$
• B. $\left[B\right]=\frac{k_1\left[A\right]}{k_1+k_2}(1-e^{(k_1+k_2)t})$
• C. $\left[B\right]=\frac{k_1[A{]}_0}{k_1+k_2}(1-e^{-(k_1+k_2)t})$
• D. $\left[B\right]=\frac{k_1[A{]}_0}{k_1+k_2}(1-e^{(k_1+k_2)t})$

Answer 1

The correct option is C $\left[B\right]=\frac{k_1[A{]}_0}{k_1+k_2}(1-e^{-(k_1+k_2)t})$

The overall net rate of consumption of A or net rate reaction

​$=\frac{-d\left[A\right]}{dt}=-{\left(\frac{d\left[A\right]}{dt}\right)}_1+{\left(\frac{-d\left[A\right]}{dt}\right)}_2\frac{d\left[A\right]}{dt}=k_1\left[A\right]+k_2\left[A\right]\frac{d\left[A\right]}{dt}=[k_1+k_2]\left[A\right]\frac{d\left[A\right]}{dt}=k_{overall}\left[A\right]Where=k_{overall}=k_1+k_2$

Integrating Eq. (i), we get

$\left[A\right]=[A{]}_0{e}^{-(k_1+k_2)t}$

Also, from the rate equation:

$\frac{d\left[B\right]}{dt}=k_1\left[A\right]and\frac{d\left[C\right]}{dt}=k_2\left[A\right]$


Substituting the value of [A] (t) in the differential equation for [B] and [C] and integrating, we get

$\left[B\right]=[B{]}_0+{\int }_0^t{k}_1[A{]}_0{e}^{-(k_1+k_2)t)}dt$

$\left[B\right]=\frac{k_1\left[A\right]}{(k_1+k_2)}(1-e^{-(k_1+k_2)t}),since[B{]}_0=0$