# In a second order reaction, the concentration of A has dropped to "0.020 M" in one hour. If the initial concentration of A is "0.050 M" and that of B is "0.080 M", what is the rate constant? What then is the half-life, if [B] is held fixed?

Jan 17, 2016

RATE LAW

We can write the rate law as:

$\setminus m a t h b f \left(r \left(t\right) = - \frac{d \left[A\right]}{\mathrm{dt}} = k {\left[A\right]}^{m} {\left[B\right]}^{n}\right)$

In this case, we have:

$\setminus m a t h b f \left(r \left(t\right) = k \left[A\right] \left[B\right]\right)$

Since the initial concentration of $B$ was given, I'll assume NOT that it's zero order with respect to $B$, but that $\left[B\right]$ is fixed.

RATE CONSTANT

To get the rate constant, we need the initial rate, which we don't actually have yet. For now, let's solve for the final equation.

$\textcolor{g r e e n}{k = \frac{{r}_{0} \left(t\right)}{{\left[A\right]}_{0} {\left[B\right]}_{0}}}$

Now, $\left[A\right] = \frac{2}{5} {\left[A\right]}_{0}$ after $\text{1 hour}$ ($\text{3600 s}$), so we can acquire a rate by recognizing that the concentration decreased by $\frac{3}{5} {\left[A\right]}_{0}$ in $\text{3600 s}$:

|r(t)| = |-(d[A])/(dt)| = |(-3/5 [A]_0)/("3600 s")| = 8.bar33xx10^(-6) "M/s"

So, now we get:

color(blue)(k) = (8.bar33xx10^(-6) cancel"M""/s")/(("0.050 M")("0.080" cancel"M"))

$= \textcolor{b l u e}{2.08 \overline{3} \times {10}^{- 3}}$ $\textcolor{b l u e}{\text{1/""M"*"s}}$

THIS REACTION'S HALF-LIFE IS KNOWN IF $\setminus m a t h b f \left(\Delta \left[B\right] = 0\right)$

The half-life is done like so, since apparently, $\left[B\right]$ doesn't change during that hour, which is nice. It simplifies our calculations.

$- \frac{d \left[A\right]}{\mathrm{dt}} = k \left[A\right] \left[B\right]$

Separation of variables gives:

$- \frac{1}{\left[A\right] \left[B\right]} d \left[A\right] = k \mathrm{dt}$

Now we integrate the left side for $\Delta \left[A\right]$ and the right side for $\Delta t$. $\Delta \left[B\right] = 0$, so we can pull that out of the integral.

$- \frac{1}{\left[B\right]} {\int}_{{\left[A\right]}_{0}}^{\left[A\right]} \frac{1}{\left[A\right]} d \left[A\right] = {\int}_{0}^{t} k \mathrm{dt}$

$- \frac{1}{\left[B\right]} \left(\ln \left[A\right] - \ln {\left[A\right]}_{0}\right) = k t$

Now, for the half-life, $\left[A\right] = {\left[A\right]}_{0} / 2$, which simplifies this down to:

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

Use the properties of logarithms to turn their subtraction into the logarithm of a fractional argument:

$- \frac{1}{\left[B\right]} \ln \setminus \frac{\cancel{{\left[A\right]}_{0}}}{2 \cancel{{\left[A\right]}_{0}}} = k {t}_{\text{1/2}}$

Here, $\ln \left(\frac{1}{2}\right) = - \ln 2$, thus cancelling out the negative sign out front.

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

And solving for the half-life, we get:

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

This makes sense, because it looks like the half-life of a first-order reaction involving only $\left[A\right]$, which would have been ${t}_{\text{1/2}} = \frac{\ln 2}{k}$. In fact, here, the units become:

1/"s" = 1/"M" * 1/(1/("M/s"))

$\frac{1}{\text{s" = 1/cancel"M" * cancel"M"/"s}}$

$\textcolor{g r e e n}{\frac{1}{\text{s" = 1/"s}}}$

So, the half-life is therefore:

t_"1/2" = 1/("0.080" cancel"M")(ln2)/(2.08bar3 xx 10^(-3) 1/(cancel"M"*"s")

$=$ $\text{4158.88 s}$

$\approx$ $\textcolor{b l u e}{\text{1.155 hours}}$

Naturally, if it took an hour to get to $\frac{2}{5}$ the concentration of $A$, incorporating $B$ into the reaction in such a way that $\Delta \left[B\right] = 0$ should increase the half-life by some amount because it is relative to both reactants at once.

Thus, it being near an hour makes sense.