# For the reaction 2A(g) -> B(g) + C(s), find the rate constant at constant temperature and volume?

## $\underline{\text{Pressure of B"" "" ""time(min)}}$ $\text{300 Pa"" "" "" "" } 10$ $\text{200 Pa"" "" "" "" ""completed reaction}$

May 25, 2016

I got $k = 1.44 \times {10}^{- 2} {\text{min}}^{- 1}$.

RATE LAW OF THIS KIND OF REACTION

Recall that the rate law for the one-reactant, first-order reaction

$\setminus m a t h b f \left(2 A \left(g\right) \to B \left(g\right) + C \left(s\right)\right)$

is written as

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

where:

• $r \left(t\right)$ is the rate of reaction in $\text{M/min}$.
• $k$ is the rate constant in units of $\frac{1}{\text{M"^(-1+m)cdot"min}}$.
• $\left[A\right]$ is the concentration of $A$ in $\text{M}$.
• $m = 1$ is the order of the reactant $A$ in the first-order reaction.

RATE LAW FOR THIS PARTICULAR REACTION

This can be rewritten as

$r \left(t\right) = k {P}_{A} ,$

where ${P}_{A}$ is the total pressure in $\text{Pa}$ due to $A$. Remember that since $A$ is the only reactant, its order is equal to the order of the reaction.

This now means the rate has units of $\text{Pa"/"min}$, so with the units of the total pressure being $\text{Pa}$, the rate constant has units of ${\text{min}}^{- 1}$, like stated in the problem.

THE INTEGRATED RATE LAW

Next, we can further set the right-half of the rate law equal to the rate of change of $\setminus m a t h b f \left({P}_{A}\right)$ over time (remember to include the stoichiometric coefficient as $\frac{1}{\nu}$):

$k {P}_{A} = - \frac{1}{2} \frac{{\mathrm{dP}}_{A}}{\mathrm{dt}}$

$- 2 k \mathrm{dt} = \frac{1}{{P}_{A}} {\mathrm{dP}}_{A}$

Now, let's integrate this and see where this lands.

$- 2 {\int}_{0}^{t} k \mathrm{dt} = {\int}_{{P}_{A , i}}^{{P}_{\text{tot}}} \frac{1}{{P}_{A}} {\mathrm{dP}}_{A}$

The total final pressure (upon completion) is due to only $B$, because $C$ is a SOLID! Hence it does not contribute to the total pressure. $A$ contributes at 10 minutes, though, since $A$ is not totally consumed at that time.

stackrel("Incompletely-Derived Integrated Rate Law")overbrace(-2k(t - 0) = ln|P_"tot"| - ln|P_(A,i)|)

The Integrated Rate Law would have been acquired by simply recognizing that $\ln | {P}_{\text{tot}} |$ is $y$, $\ln | {P}_{A , i} |$ is $b$ (the y-intercept), and $- 2 k$ is $m$ (the slope). Anyways...

$- 2 k t = \ln | \frac{{P}_{\text{tot}}}{{P}_{A , i}} |$

$\textcolor{g r e e n}{k = - \frac{1}{2 t} \ln | \frac{{P}_{\text{tot}}}{{P}_{A , i}} |}$

DETERMINING THE TOTAL PRESSURE DUE TO A, AND THEN B

After 10 minutes, $A$ is not totally consumed yet. So, ${P}_{\text{tot" = P_B + P_A = "300 Pa}}$ (where ${P}_{A}$ and ${P}_{B}$ are partial pressures), but after completion, ${P}_{\text{tot" = P_(B,t) = "200 Pa}}$ for the total pressure.

Fortunately, since we were given that the reaction is at constant temperature and volume, that means we now can use the mole ratio of $B$ to $A$ to our advantage.

From the ideal gas law, we have

$\frac{{P}_{A , i} \cancel{V}}{{P}_{B , t} \cancel{V}} = \frac{{n}_{A} \cancel{R T}}{{n}_{B} \cancel{R T}}$

With $\frac{{P}_{A , i}}{P} _ \left(B , t\right) = \frac{{P}_{A , i}}{{P}_{\text{tot}}} = \frac{2}{1} = 2$ upon reaction completion, $\textcolor{g r e e n}{{P}_{A , i} = \text{400 Pa}}$ and ${P}_{\text{tot" = "200 Pa}}$. Now that we know ${P}_{A , i}$, we can move forward.

We only have a known time for the 10-minute trial. So, we must use the value for ${P}_{\text{tot}}$ after 10 minutes.

ACQUIRING THE RATE CONSTANT

Hence, the rate constant at $t = \text{10 min}$ should be:

$k = - \frac{1}{2 \cdot \text{10 min")ln|(P_"tot"^"10 min}} / \left({P}_{A , i}\right) |$

= -1/(2*"10 min")ln|"300 Pa"/"400 Pa"|

$= \frac{\ln \left(\frac{4}{3}\right)}{2 \cdot \text{10 min}}$

Assuming 3 sig figs (instead of 1) for a reasonable answer, we get:

$\textcolor{b l u e}{k = 1.44 \times {10}^{- 2} {\text{min}}^{- 1}}$

Just goes to show you... use every piece of information from the problem if you have to. In this case, we had to.