Chemistry Daily Challenge 23-July-2015

The rates of the two reactions are as given below:

$\begin{cases} - \frac{dC_A}{dt} = k_1 t &...(1) \\ \frac{dC_R}{dt} = \underbrace{k_1 C_A}_{appearance} - \underbrace{k_2 C_R}_{disappearance} &...(2) \end{cases}$

Let $k = k_1 = k_2$ .

$\begin{aligned} (1): \quad \int \frac{dC_A}{C_A} & = - \int k \space dt \\ \ln C_A & = -kt + \text{constant} \\ \Rightarrow C_A & = C_{A0} e^{-kt} \\ (2): \quad \quad \frac{dC_R}{dt} & = k C_A - k C_R \\ & = kC_{A0} e^{-kt} - kC_R \quad ...(3) \\ \Rightarrow \color{#3D99F6}{C_R} & \color{#3D99F6}{= kC_{A0} t e^{-kt} \quad \quad \quad \quad \small \text{See Note}} \end{aligned}$

For maximum $C_R$ , then:

$\begin{aligned} \Rightarrow \frac{dC_R}{dt} & = 0 \\ (3): \quad \quad kC_{A0} e^{-kt} - kC_R & = 0 \\ kC_{A0} e^{-kt} - k^2C_{A0} t e^{-kt} & = 0 \\ kC_{A0} e^{-kt} & = k^2C_{A0} t e^{-kt} \\ \Rightarrow t & = \frac{1}{k} \\ \Rightarrow C_{R_{max}} & = kC_{A0} \left(\frac{1}{k}\right) e^{-k\left(\frac{1}{k}\right)} \\ & = \boxed{\dfrac{C_{A0}}{e}} \end{aligned}$

$$

$\color{#3D99F6}{\text{Note:}}$ If $\color{#3D99F6}{C_R = kC_{A0} t e^{-kt}}$ , then:

$\begin{aligned} \frac{dC_R}{dt} & = \frac{d}{dt} kC_{A0} t e^{-kt} \\ & = kC_{A0} e^{-kt} - k^2C_{A0} t e^{-kt} \\ & = kC_{A0} e^{-kt} - kC_R \quad ...(3) \end{aligned}$