Consider a reaction $$A + R \to Product$$. 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]

The elementary rate law for the reaction is
$$\text{rate}_{\text{true}} = k[A][R] \;.$$

At $$t = 0$$:
$$[A]_0 = a ,\qquad [R]_0 = 10a \quad(\text{given})$$

Let the extent of reaction be such that the reaction is 40 % complete with respect to $$A$$.
Fraction reacted $$x = 0.40$$, so the amount of $$A$$ consumed is $$0.40a$$.

Concentrations when the reaction is 40 % complete:
$$[A] = a - 0.40a = 0.60a$$
Because the stoichiometry is 1 : 1, the same amount of $$R$$ is consumed:
$$[R] = 10a - 0.40a = 9.60a$$

True rate at this instant:
$$\text{rate}_{\text{true}} = k(0.60a)(9.60a) = k\,(0.60 \times 9.60)\,a^{2} = 5.76\,k\,a^{2}$$

Pseudo-first-order approximation:
We assume $$[R] \approx [R]_0 = 10a$$ is constant, so we replace $$k[R]$$ by
$$k' = k[\,R]_0 = k(10a) \;.$$
Then
$$\text{rate}_{\text{pseudo}} = k'[A] = k(10a)(0.60a) = 6.00\,k\,a^{2}$$

Absolute error in rate:
$$\Delta = \text{rate}_{\text{pseudo}} - \text{rate}_{\text{true}} = (6.00 - 5.76)\,k\,a^{2} = 0.24\,k\,a^{2}$$

Relative error (percentage):
$$\%\text{ error} = \frac{\Delta}{\text{rate}_{\text{true}}}\times 100 = \frac{0.24\,k\,a^{2}}{5.76\,k\,a^{2}}\times 100 = \frac{0.24}{5.76}\times 100 = 4.1667\% \approx 4.17\%$$

Therefore, the relative error in the rate due to treating the reaction as pseudo first order when it is 40 % complete is about $$4.17\%$$, which lies in the range $$4.00 - 4.25\%$$.