$\text{Consider a reaction }A+R⟶\text{Product}.\text{The rate of this reaction is measured to be}\mathrm{ k}\left[A\right]\left[R\right].\text{At the start of}$

$\text{the reaction, the concentration of}\mathrm{ R},[R{]}_0,\text{is}10\text{-times the concentration of}\mathrm{ A},[A{]}_0.\text{The reaction can be}$

$\text{considered to be a pseudo first order reaction with assumption that}\mathrm{ k}\left[R\right]=k^{\prime }\text{is constant.}\text{ Due to this assumption,}$

$\text{the relative error (in \%) in the rate when this reaction is}\mathrm{ 40}\%\text{complete, is}\text{\_\_\_\_\_\_}$
$\left[k\text{ and}{\mathrm{ k}}^{\prime }\text{represent corresponding rate constants}\right]$

The correct answer is 4.17.

Let the initial concentration of reactant $A$ be $[A{]}_0=a$.
Since the initial concentration of $R$ is 10 times that of $A$, we have:
$[R{]}_0=10a$

When the reaction is 40% complete, the amount of $A$ consumed is:
$x=0.4a$

Therefore, the concentrations of $A$ and $R$ at 40% completion are:
$\left[A\right]=a-0.4a=0.6a$
$\left[R\right]=10a-0.4a=9.6a$

Now, let's find the actual rate of reaction at this point:
${\text{Rate}}_{\text{actual}}=k\left[A\right]\left[R\right]=k\left(0.6a\right)\left(9.6a\right)=5.76k{a}^2$

Under the pseudo-first-order assumption, the rate constant $k^{\prime }=k[R{]}_0$ is assumed to remain constant throughout the reaction:
${\text{Rate}}_{\text{pseudo}}=k^{\prime }\left[A\right]=k\left[R{]}_0\right[A]=k(10a\left)\right(0.6a)=6.00k{a}^2$

The relative error (in %) in the rate due to this assumption is given by:
$\text{Relative Error (\%)}=\frac{{\text{Rate}}_{\text{pseudo}}-{\text{Rate}}_{\text{actual}}}{{\text{Rate}}_{\text{actual}}}\times 100$

$\text{Relative Error (\%)}=\frac{6.00k{a}^2-5.76k{a}^2}{5.76k{a}^2}\times 100$

$\text{Relative Error (\%)}=\frac{0.24}{5.76}\times 100=\frac{1}{24}\times 100\approx 4.17\%$