Question

For the consecutive reaction, \[ \text{X} \xrightarrow{k_X} \text{Y} \xrightarrow{k_Y} \text{Z} \] $C_0$ is the initial concentration of X. The concentrations of X, Y, and Z at time $t$ are $C_X$, $C_Y$, and $C_Z$, respectively. The expression for the concentration of Y at time $t$ is

• A. $\dfrac{k_X C_0}{k_Y - k_X} \left(e^{-k_X t} - e^{-k_Y t}\right)$
• B. $\dfrac{k_Y C_X}{k_Y - k_X} \left(e^{-k_X t} - e^{-k_Y t}\right)$
• C. $\dfrac{k_X C_0}{k_Y - k_X} \left(e^{-k_Y t} - e^{-k_X t}\right)$
• D. $\dfrac{k_Y C_X}{k_Y - k_X} \left(e^{-k_Y t} - e^{-k_X t}\right)$

Hint:

In consecutive first-order reactions, intermediate concentration (Y) first rises and then falls. Use integrating factor method for solving coupled first-order rate equations.

Answer 1

The Correct Option is A

Step 1: Write rate equations for each species.
For the given consecutive first-order reactions: \[ \text{X} \xrightarrow{k_X} \text{Y} \xrightarrow{k_Y} \text{Z} \] Rate equations are: \[ \frac{dC_X}{dt} = -k_X C_X, \quad \frac{dC_Y}{dt} = k_X C_X - k_Y C_Y \] Step 2: Solve for $C_X$.
Integrating the first equation: \[ C_X = C_0 e^{-k_X t} \] Step 3: Substitute $C_X$ into Y’s rate equation.
\[ \frac{dC_Y}{dt} + k_Y C_Y = k_X C_0 e^{-k_X t} \] This is a linear first-order differential equation. Using the integrating factor $e^{k_Y t}$: \[ \frac{d}{dt}(C_Y e^{k_Y t}) = k_X C_0 e^{(k_Y - k_X)t} \] Integrating both sides: \[ C_Y e^{k_Y t} = \frac{k_X C_0}{k_Y - k_X}\left(e^{(k_Y - k_X)t} - 1\right) \] Step 4: Simplify to get $C_Y$.
\[ C_Y = \frac{k_X C_0}{k_Y - k_X} \left(e^{-k_X t} - e^{-k_Y t}\right) \] Step 5: Conclusion.
Hence, the correct expression for the concentration of Y at time $t$ is: \[ C_Y = \frac{k_X C_0}{k_Y - k_X} \left(e^{-k_X t} - e^{-k_Y t}\right) \] which corresponds to **Option (A)**.