# Consider the reaction 3A + B + C ---> D + E where the rate law is defined as -D[A]/Dt=k[A]^2[B][C]?

## The initial concentration of $B$ and $C$ are both $\text{1.00 M}$. The initial concentration of $A$ is $1.00 \times x {10}^{- 4} \text{M}$, and after 3.00 min, it became $3.26 \times {10}^{- 5} \text{M}$. 1) Find the rate constant for the pseudo-second-order reaction in ${\text{M"^(-3)cdot"s}}^{- 1}$. 2) Find the pseudo-second-order half-life in seconds. 3) Under what assumptions are $\left(1\right)$ and $\left(2\right)$ true? 4) Determine the concentration of $A$ after 10.00 min have passed.

Jun 11, 2017

Alright, first, let's write down what we were given here, and also what we can write out from what we know so far.

$A + B + C \to D + E$

(I assume the 3A is a typo, because the rate given, $- \frac{\Delta \left[A\right]}{\Delta t}$, has no coefficient when it should if the stoichiometric cofficient of $A$ were not $1$.)

• Initial concentration of $B$, ["B"]_0 = ["C"]_0 = "1.00 M"
• Initial concentration of $A$, ${\left[\text{A}\right]}_{0} = 1.00 \times {10}^{- 4}$ $\text{M}$
• Current concentration of $A$, ["A"]_"3.00 min." = 3.26 xx 10^(-5) "M"

$r \left(t\right) = k {\left[A\right]}^{2} \left[B\right] \left[C\right]$

$= - \frac{1}{1} \frac{\Delta \left[A\right]}{\Delta t} = - \frac{1}{1} \frac{\Delta \left[B\right]}{\Delta t} = - \frac{1}{1} \frac{\Delta \left[C\right]}{\Delta t}$

$= + \frac{1}{1} \frac{\Delta \left[D\right]}{\Delta t} = + \frac{1}{1} \frac{\Delta \left[E\right]}{\Delta t}$

1)

The rate constant can be found by using the integrated rate law to track the fraction of just $A$ leftover after some time $t$.

The integrated rate law with respect to ONLY $A$ should be the second-order version as noted from the rate law, and fortunately we have tracked what the concentration of $A$ had become in the context of $B$ and $C$ (which influence $A$!).

This version can be derived (assuming $B$ and $C$ don't interfere), but the derivation will be omitted. If you need to know it, it should be in your textbook, maybe off to the side? It requires a bit of calculus.

$\boldsymbol{{\overbrace{\frac{1}{\left[A\right]}}}^{y} = {\overbrace{k}}^{m} {\overbrace{t}}^{x} + {\overbrace{\frac{1}{{\left[A\right]}_{0}}}}^{b}}$

assuming $A$ has a stoichiometric coefficient of $1$. Get to know this equation, because we'll be using it a lot.

Now, let's plot the two data points we have (and hope that that's enough).

Since the slope is $6891.6$, the rate constant is $k = {\text{6891.6 M"^(-3)"min}}^{- 1}$. The units are this way because the reaction is fourth order overall, and the change in $\left[A\right]$ was in the context of $B$ and $C$:

${\overbrace{\text{M"/"min")^(r(t)) = overbrace(1/("M"^3 cdot "min"))^(k) xx overbrace("M"^2)^([A]^2) xx overbrace("M")^([B]) xx overbrace("M}}}^{\left[C\right]}$

$\textcolor{b l u e}{k} = \text{6891.6 M"^(-3)cdot"min"^(-1) xx "1 min"/"60 s}$

$= \textcolor{b l u e}{{\text{114.86 M"^(-3)cdot"s}}^{- 1}} \approx 115$

2)

Since we know the form of the integrated rate law for just $A$ in the context of $B$ and $C$, we can use the rate constant to find the half-life. The half-life is, again, when $\left[A\right] = \frac{1}{2} {\left[A\right]}_{0}$, so:

$\frac{1}{\frac{1}{2} {\left[A\right]}_{0}} = k {t}_{\text{1/2}} + \frac{1}{{\left[A\right]}_{0}}$

$\frac{1}{{\left[A\right]}_{0}} = k {t}_{\text{1/2}}$

Therefore, the half-life for the reaction based on only the second-order behavior of $A$ in the context of $B$ and $C$ is:

$\textcolor{b l u e}{{t}_{\text{1/2}}} = \frac{1}{k {\left[A\right]}_{0}}$

$= \frac{1}{114.86 \left(1.00 \times {10}^{- 4}\right)} \text{ s}$

$=$ $\textcolor{b l u e}{\text{87.06 s}}$

3)

As it turns out, we have been assuming in $\left(1\right)$ and $\left(2\right)$ to NOT consider the concentrations of $B$ and $C$, and so, the half-life does not seem to be influenced by $B$ or $C$...

That is not strictly true, unless the concentration of $B$ and $C$ do not change, or are held fixed at $\textcolor{b l u e}{\text{1.00 M}}$ each.

4)

Using the second-order integrated rate law once more, denote $\alpha$ as the fraction of $A$ leftover after $t = \text{10.0 min}$. This fraction will be less than $0.5$ because the half-life was about $\text{1.45 min}$.

Then:

$\frac{1}{\left[A\right]} = k t + \frac{1}{{\left[A\right]}_{0}}$

$\frac{1}{\alpha {\left[A\right]}_{0}} - \frac{1}{{\left[A\right]}_{0}} = k t$

$\frac{\frac{1}{\alpha} - 1}{{\left[A\right]}_{0}} = k t$

$\implies k t {\left[A\right]}_{0} + 1 = \frac{1}{\alpha}$

So, the fraction of $\boldsymbol{A}$ leftover is:

$\alpha = \frac{1}{k t {\left[A\right]}_{0} + 1}$

$= \frac{1}{114.86 \cdot 600 \left(1.00 \times {10}^{- 4}\right) + 1}$

$= 0.1267$

Therefore, the amount of $A$ left is:

$\textcolor{b l u e}{{\left[A\right]}_{\text{10.00 min}}} = \alpha {\left[A\right]}_{0} = 0.1267 \times 1.00 \times {10}^{- 4}$

$= \textcolor{b l u e}{1.27 \times {10}^{- 5} \text{M}}$