Step-by-Step Solution

1. Define Initial Conditions and Concentrations at 40% Completion

The reaction is given as $A + R \rightarrow$ Product, with the rate law: Rate $= k[A][R]$.

Initial concentrations are given as $[R]_0 = 10[A]_0$. Let's denote $[A]_0 = a$. Then, $[R]_0 = 10a$. Since the concentration of A is much smaller than R, A is the limiting reactant.

The completion of the reaction is based on the limiting reactant, A. When the reaction is 40% complete, 40% of A has been consumed.

• Amount of A reacted = $0.40 \times [A]_0 = 0.4a$.
• Concentration of A at 40% completion: $[A]_t = [A]_0 - 0.4a = a - 0.4a = 0.6a$.

According to the stoichiometry of the reaction ($A+R$), for every mole of A that reacts, one mole of R also reacts.

• Amount of R reacted = Amount of A reacted = $0.4a$.
• Concentration of R at 40% completion: $[R]_t = [R]_0 - 0.4a = 10a - 0.4a = 9.6a$.

2. Calculate the True Rate of Reaction

The true rate of reaction at any given time depends on the instantaneous concentrations of the reactants. The rate law is $Rate_{true} = k[A][R]$.

At 40% completion, the true rate is: $Rate_{true} = k[A]_t[R]_t = k(0.6a)(9.6a) = 5.76ka^2$

3. Calculate the Approximated Rate (Pseudo-First-Order)

The pseudo-first-order approximation assumes that the concentration of the reactant in excess (R) remains constant at its initial value throughout the reaction.

• Assumed concentration of R: $[R] \approx [R]_0 = 10a$.

The approximated rate law becomes $Rate_{approx} = k[A][R]_0$. This is often written as $Rate_{approx} = k'[A]$, where the pseudo-first-order rate constant is $k' = k[R]_0$.

At 40% completion, the concentration of A is $[A]_t = 0.6a$. The approximated rate at this point is: $Rate_{approx} = k[A]_t[R]_0 = k(0.6a)(10a) = 6.0ka^2$

4. Calculate the Relative Error in Percentage

The relative error is defined as the absolute difference between the approximated value and the true value, divided by the true value.

• Absolute Error = $|Rate_{approx} - Rate_{true}| = |6.0ka^2 - 5.76ka^2| = 0.24ka^2$.

• Relative Error = $\frac{|Rate_{approx} - Rate_{true}|}{Rate_{true}} = \frac{0.24ka^2}{5.76ka^2}$

• Relative Error = $\frac{0.24}{5.76} = \frac{24}{576}$

To simplify the fraction, we can note that $576 = 24^2$.

• Relative Error = $\frac{24}{24 \times 24} = \frac{1}{24}$.

The question asks for the relative error in percentage (%):

• % Relative Error = Relative Error $\times 100 = \frac{1}{24} \times 100 = \frac{100}{24} = \frac{25}{6}$

• % Relative Error $= 4.1666... \% \approx 4.17\%$

This value lies within the given answer range of 4 to 4.25.