Answer:
Here's what I get
Explanation:
[tex]\rm A \xrightarrow{k_{1}}I \xrightarrow{k_{2}} P[/tex]
(a) Plot I against t
I assume that Equation 22.40 is something like
[tex]\text{[I]} = \dfrac{k_{1}\text{[A]}_{0}}{k_{2} - k_{1}} \left(e^{-k_{1}t} - e^{-k_{2}t} \right )[/tex]
If the initial conditions are
[A]₀ = 1.0 mol·dm⁻³; k₁ = 10 s⁻¹; k₂ = 1 s⁻¹
The equation becomes
[tex]\text{[I]} = \dfrac{10}{-9} \left(e^{-10t} - e^{-t} \right )[/tex]
The graph of [I] vs t is shown in Fig 1.
(b) Increasing k₂/k₁ ratio
In Fig. 2, I added the same plots, but with k₂ = 3, 11, and 51 s⁻¹ (black, green and purple).
The graphs show that, as k₂ becomes increasingly greater than k₁, the maximum concentration of I becomes smaller and the graph becomes (except for the very beginning) a flat line.
Thus, the approximation that
[tex]\mathbf{\dfrac{\text{d[I]}}{\text{d}t}=0}[/tex]
becomes increasingly valid.
