Question

Consider a reaction $ A + R \rightarrow Product $. The rate of this reaction is measured to be $ k[A][R] $. At the start of the reaction, the concentration of $ R $, $[R]_0$, is 10-times the concentration of $ A $, $[A]_0$. The reaction can be considered to be a pseudo first order reaction with assumption that $ k[R] = k' $ is constant. Due to this assumption, the relative error (in %) in the rate when this reaction is 40% complete, is ____. [$k$ and $k'$ represent corresponding rate constants]

Hint:

In pseudo-first order reactions, relative error arises because the concentration of the excess reactant changes during the reaction but is approximated as constant.

Answer 1

Step 1: Define variables and assumptions
Given: \[ [R]_0 = 10 [A]_0 \] Assuming \(k[R] = k'\) is constant implies treating the reaction as pseudo-first order with respect to \(A\).
Step 2: Express actual rate and approximate rate
Actual rate: \[ \text{Rate} = k [A][R] \] Since \([R]\) changes during the reaction, actual rate depends on both \([A]\) and \([R]\). Pseudo-first order approximation rate: \[ \text{Rate}_{\text{approx}} = k' [A] = k [R]_0 [A] \] This assumes \([R]\) remains constant at \([R]_0\).
Step 3: Relation between \([R]\) and \([A]\) during reaction
Reaction consumes \(A\) and \(R\) in 1:1 ratio. Let fraction of reaction completion be \(x\), i.e. fraction of \(A\) consumed: \[ [A] = [A]_0 (1 - x) \] \[ [R] = [R]_0 - [A]_0 x = 10 [A]_0 - [A]_0 x = [A]_0 (10 - x) \]
Step 4: Calculate relative error in rate
Actual rate at completion \(x\): \[ \text{Rate} = k [A][R] = k [A]_0 (1-x) \times [A]_0 (10 - x) = k [A]_0^2 (1 - x)(10 - x) \] Approximate rate: \[ \text{Rate}_{\text{approx}} = k [R]_0 [A] = k \times 10 [A]_0 \times [A]_0 (1-x) = k [A]_0^2 10 (1 - x) \] Relative error (in fraction) due to assumption: \[ \epsilon = \frac{\text{Rate}_{\text{approx}} - \text{Rate}}{\text{Rate}} = \frac{k [A]_0^2 10 (1-x) - k [A]_0^2 (1-x)(10 - x)}{k [A]_0^2 (1-x)(10 - x)}\\ = \frac{10(1-x) - (1-x)(10 - x)}{(1-x)(10 - x)} = \frac{10 - 10x - 10 + x}{(1-x)(10 - x)} = \frac{x(1-x)}{(1-x)(10 - x)} = \frac{x}{10 - x} \] Multiply by 100 to get percentage error: \[ \text{Relative error (%)} = \frac{x}{10 - x} \times 100 \] Step 5: Calculate error at 40% completion At \(x = 0.4\): \[ \text{Relative error} = \frac{0.4}{10 - 0.4} \times 100 = \frac{0.4}{9.6} \times 100 \approx 4.17 % \]