Question

From the data, what is the rate law for the reaction? `"CHCl"_3(g) + "Cl"_2(g) -> "CCl"_4(g) + "HCl"(g)`

The following concentrations are in `"M"`.

`ul(["CHCl"_3]" "" "["Cl"_2]" "" "r(t)("M/s"))`
`0.010" "" "" "0.010" "" "0.0035`
`0.020" "" "" "0.010" "" "0.0069`
`0.020" "" "" "0.020" "" "0.0098`
`0.040" "" "" "0.040" "" "0.0270`

Answer 1

The orders are `1` for `"CHCl"_3` and `1/2` for `"Cl"_2`, so the rate law would be:

`r(t) = k["CHCl"_3]["Cl"_2]^"1/2"`

---

You'll always want to write out a general rate law for the problem to keep track of what you're solving for:

`r(t) = k["CHCl"_3]^m["Cl"_2]^n`

where:

• `r(t)` is the initial rate of reaction.
• `k` is the rate constant in the appropriate units.
• `[" "]` is the molar concentration of a reactant.
• `m` and `n` are the reaction orders for only `"CHCl"_3` and only `"Cl"_2`, respectively.

The usual way to find the order for one reactant is to pick two trials where the concentration of the other reactant stays constant.

• If you want the order for `"CHCl"_3`, you'll have to pick trials `1` and `2`, where `["Cl"_2]` stays constant.

• If you want the order for `"Cl"_2`, you'll have to pick trials `2` and `3`, where `["CHCl"_3]` stays constant.

(We will of course, then, label the first row of data as trial `1`, the second row as trial `2`, and so on.)

And you then take the ratio of the rates vs. the ratio of the concentrations.

ORDER FOR `bb("CHCl"_3)`

The rate constant stays the same for the same reaction at the same temperature. Here's how you could do it in general, and we'll simplify the case afterwards.

`(r_2(t))/(r_1(t)) = cancel(k/k) (["CHCl"_3]_2^m)/(["CHCl"_3]_1^m)cancel((("Cl"_2]_2^n)/(["Cl"_2]_1^n))^(1)`

We crossed out the ratio of `"Cl"_2` concentrations, because as we said, `["Cl"_2]_1 = ["Cl"_2]_2 = "0.010 M"`, and `1^n = 1` no matter what finite `n` we have.

It is then convenient to take the `log` of both sides to solve for `m`.

`log((r_2(t))/(r_1(t))) = log((["CHCl"_3]_2)/(["CHCl"_3]_1))^m`

`= mlog((["CHCl"_3]_2)/(["CHCl"_3]_1))`

Therefore, the order for `"CHCl"_3`, for any `m` (decimal or integer), is:

`barul(stackrel(" ")(|" "m = log((r_2(t))/(r_1(t)))/log((["CHCl"_3]_2)/(["CHCl"_3]_1))" ")|)`

The numbers here happen to be nice, and `m` should end up being a whole number. The order for `"CHCl"_3` is:

`bbm = log(0.0069/0.0035)/log(0.020/0.010) = 0.2948/(0.3010) ~~ bb1`

In a case like this with nice numbers, we could have stopped and skipped past using logarithms:

`((r_2(t))/(r_1(t))) = ((["CHCl"_3]_2)/(["CHCl"_3]_1))^m`

`=> "0.0069 M/s"/"0.0035 M/s" = ("0.020 M"/"0.010 M")^m`

`=> 1.971 = 2^m`

And we can then see that `color(blue)ulbb(m ~~ 1)` just from that.

So, the reaction is first order for `"CHCl"_3`.

ORDER FOR `bb("Cl"_2)`

We'll first try it the easy way here, now that you see the more general way to do it. The more rigorous way, which is meant for decimal orders, gives:

`barul(stackrel(" ")(|" "n = log((r_3(t))/(r_2(t)))/log((["Cl"_2]_3)/(["Cl"_2]_2))" ")|)`

We compare trials `2` and `3` to get:

`=> "0.0098 M/s"/"0.0069 M/s" = ("0.020 M"/"0.010 M")^n`

`1.42 = 2^n`

This actually looks close to `1.414`, which is `sqrt2`, so `color(blue)bbul(n ~~ 1/2)` since `x^"1/2" = sqrtx`.

If you don't see that right away, you can do it the more rigorous way:

`n = log(0.0098/0.0069)/log(0.020/0.010) = 0.1524/0.3010`

`= 0.5063 ~~ 1/2`

and you again get that the reaction is half-order for `"Cl"_2`.