Question 12

Let initial concentrations be $[A]_0 = a$ and $[R]_0 = 10a$.

The actual rate is $r_{act} = k[A][R]$.

The pseudo-first-order rate is assumed as $r_{ps} = k'[A] = k[R]_0 [A]$.

When the reaction is $40\%$ complete, amount of $A$ reacted $x = 0.4a$.

$[A] = a - 0.4a = 0.6a$.

$[R] = 10a - 0.4a = 9.6a$.

Actual rate at this point: $r_{act} = k(0.6a)(9.6a) = 5.76 k a^2$.

Pseudo-rate at this point: $r_{ps} = k(10a)(0.6a) = 6.0 k a^2$.

Relative error $\% = \frac{r_{ps} - r_{act}}{r_{act}} \times 100$

Relative error $\% = \frac{6.0 k a^2 - 5.76 k a^2}{5.76 k a^2} \times 100 = \frac{0.24}{5.76} \times 100 = \frac{1}{24} \times 100 \approx 4.166...$

Rounding to two decimal places, we get $4.17$.