Question

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

`ul("Pressure of B"" "" ""time(min)")`
`"300 Pa"" "" "" "" "10`
`"200 Pa"" "" "" "" ""completed reaction"`

Answer 1

I got `k = 1.44xx10^(-2) "min"^(-1)`.

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RATE LAW OF THIS KIND OF REACTION

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

`\mathbf(2A(g) -> B(g) + C(s))`

is written as

`\mathbf(r(t) = k[A]^m)`

where:

• `r(t)` is the rate of reaction in `"M/min"`.
• `k` is the rate constant in units of `1/("M"^(-1+m)cdot"min")`.
• `[A]` is the concentration of `A` in `"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(t) = kP_A,`

where `P_A` is the total pressure in `"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 `"Pa"/"min"`, so with the units of the total pressure being `"Pa"`, the rate constant has units of `"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 `\mathbf(P_A)` over time (remember to include the stoichiometric coefficient as `1/nu`):

`kP_A = -1/2(dP_A)/(dt)`

`-2kdt = 1/(P_A)dP_A`

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

`-2int_(0)^(t) kdt = int_(P_(A,i))^(P_"tot") 1/(P_A)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_"tot"|` is `y`, `ln|P_(A,i)|` is `b` (the y-intercept), and `-2k` is `m` (the slope). Anyways...

`-2kt = ln|(P_"tot")/(P_(A,i))|`

`color(green)(k = -1/(2t)ln|(P_"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_"tot" = P_B + P_A = "300 Pa"` (where `P_A` and `P_B` are partial pressures), but after completion, `P_"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

`(P_(A,i)cancel(V))/(P_(B,t)cancel(V)) = (n_Acancel(RT))/(n_Bcancel(RT))`

With `(P_(A,i))/P_(B,t) = (P_(A,i))/(P_"tot") = 2/1 = 2` upon reaction completion, `color(green)(P_(A,i) = "400 Pa")` and `P_"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_"tot"` after 10 minutes.

ACQUIRING THE RATE CONSTANT

Hence, the rate constant at `t = "10 min"` should be:

`k = -1/(2*"10 min")ln|(P_"tot"^"10 min")/(P_(A,i))|`

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

`= (ln(4/3))/(2*"10 min")`

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

`color(blue)(k = 1.44xx10^(-2) "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.