Question

1. the reaction a + b → c is first - order in both a and b. the reaction is carried out where a_0 = 2.0 m and b_0 = 1.0×10^4 m.

a. illustrate that the reaction is run under flooding (pseudo - order) conditions by calculating the concentration of a when b is depleted.
b. under these conditions, it was determined that the plot of lnb vs t gave a straight line with slope - 0.30. determine the rate constant, k, for the reaction.

Explanation:

Step1: Determine the relationship for pseudo - first - order conditions

In a reaction $A + B
ightarrow C$ that is first - order in both $A$ and $B$, under pseudo - first - order conditions (when $[A]\gg[B]$), we consider the reaction as first - order with respect to $B$. When $B$ is depleted, the change in $[A]$ is negligible. Given $[A]_0 = 2.0\ M$ and $[B]_0=1.0\times10^{-4}\ M$. Since $[A]_0\gg[B]_0$, when $B$ is completely consumed, the concentration of $A$, $[A]\approx[A]_0 = 2.0\ M$.

Step2: Relate the slope of the $\ln[B]$ vs $t$ plot to the rate constant

For a first - order reaction of the form $\text{Rate}=k'[B]$, the integrated rate law is $\ln[B]=\ln[B]_0 - k't$. The slope of the $\ln[B]$ vs $t$ plot is equal to $-k'$. Here, the slope of the $\ln[B]$ vs $t$ plot is $- 0.30\ s^{-1}$. In the pseudo - first - order case where the reaction is considered first - order with respect to $B$ (because $[A]$ is in large excess), the rate constant $k'$ is the pseudo - first - order rate constant. If we consider the overall second - order reaction $A + B
ightarrow C$ (first - order in $A$ and first - order in $B$), and let the overall rate constant be $k$. Under pseudo - first - order conditions ($[A]$ is constant), $k' = k[A]$. But if we just want the pseudo - first - order rate constant for the reaction with respect to $B$ under these conditions, the rate constant $k$ (the pseudo - first - order rate constant in this context) is equal to the absolute value of the slope of the $\ln[B]$ vs $t$ plot. So $k = 0.30\ s^{-1}$.

Answer:

a. $[A]=2.0\ M$
b. $k = 0.30\ s^{-1}$