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]

Answer 1

4.166

Answer 2

Let the reaction be $A + R \rightarrow Product$. The rate of the reaction is given by $rate = k[A][R]$. At the start of the reaction, $[R]_0 = 10[A]_0$.

The reaction is considered to be a pseudo first order reaction with the assumption that $k[R] = k'$ is constant. This assumption is typically made by assuming that the concentration of the excess reactant R remains approximately constant at its initial value $[R]_0$. So, the pseudo first order rate constant is $k' = k[R]_0$. The pseudo first order rate expression is $rate_{pseudo} = k'[A] = k[R]_0[A]$.

The actual rate expression is $rate_{actual} = k[A][R]$.

The reaction is 40% complete. This means that 40% of the initial concentration of A has reacted. Let $[A]_0$ be the initial concentration of A. The concentration of A at 40% completion is $[A]_t = [A]_0 - 0.40[A]_0 = 0.60[A]_0$.

According to the stoichiometry of the reaction $A + R \rightarrow Product$, when 1 mole of A reacts, 1 mole of R also reacts. The amount of A reacted is $0.40[A]_0$. So, the amount of R reacted is also $0.40[A]_0$. The initial concentration of R is $[R]_0$. The concentration of R at 40% completion is $[R]_t = [R]_0 - 0.40[A]_0$. Given $[R]_0 = 10[A]_0$, we have $[R]_t = 10[A]_0 - 0.40[A]_0 = (10 - 0.40)[A]_0 = 9.60[A]_0$.

Now, let's calculate the actual rate and the pseudo rate at 40% completion.

The actual rate at 40% completion is $rate_{actual} = k[A]_t[R]_t = k(0.60[A]_0)(9.60[A]_0) = k(0.60 \times 9.60)[A]_0^2 = 5.76 k[A]_0^2$.

The pseudo first order rate at 40% completion is $rate_{pseudo} = k[R]_0[A]_t = k(10[A]_0)(0.60[A]_0) = k(10 \times 0.60)[A]_0^2 = 6.00 k[A]_0^2$.

The relative error in the rate is the difference between the pseudo rate (approximate value) and the actual rate (true value), divided by the actual rate (true value).

$Relative Error = \frac{|rate_{pseudo} - rate_{actual}|}{rate_{actual}}$.

$Relative Error = \frac{|6.00 k[A]_0^2 - 5.76 k[A]_0^2|}{5.76 k[A]_0^2} = \frac{|6.00 - 5.76| k[A]_0^2}{5.76 k[A]_0^2} = \frac{0.24}{5.76}$.

To simplify the fraction $\frac{0.24}{5.76}$:

$\frac{0.24}{5.76} = \frac{24}{576}$.

We can divide both the numerator and the denominator by 24.

$24 \div 24 = 1$.

$576 \div 24 = 24$.

So, the relative error is $\frac{1}{24}$.

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

$Percentage Relative Error = Relative Error \times 100\%$.

$Percentage Relative Error = \frac{1}{24} \times 100\% = \frac{100}{24}\%$.

$\frac{100}{24} = \frac{50}{12} = \frac{25}{6}$.

$Percentage Relative Error = \frac{25}{6}\%$.

As a decimal, $\frac{25}{6} = 4.1666...$

The question asks for the relative error (in %), which suggests a numerical value is expected. The value is $\frac{25}{6}$.