Question

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

The following concentrations are given 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

From before, we got the rate law was

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

where `r_i(t)` would denote the initial rate for trial `i`, `k` is the rate constant, and `[" "]` denotes a molar concentration for one of the reactants.

All you have to do at this point is to choose a trial and solve for the rate constant using the rate and concentrations of that trial.

(You should get the same result no matter what trial you choose, since it is the same reaction at the same temperature. I'll prove it by doing it for all four trials.)

TRIAL 1

`color(blue)(k) = (r_1(t))/(["CHCl"_3]_1["Cl"_2]_1^"1/2") = ("0.0035 M/s")/("0.010 M" cdot ("0.010 M")^"1/2")`

`= ulcolor(blue)("3.5 M"^(-"1/2")cdot"s"^(-1))`

TRIAL 2

`color(blue)(k) = (r_2(t))/(["CHCl"_3]_2["Cl"_2]_2^"1/2") = ("0.0069 M/s")/("0.020 M" cdot ("0.010 M")^"1/2")`

`= "3.45 M"^(-"1/2")cdot"s"^(-1) ~~ ulcolor(blue)("3.5 M"^(-"1/2")cdot"s"^(-1))`

TRIAL 3

`color(blue)(k) = (r_3(t))/(["CHCl"_3]_3["Cl"_2]_3^"1/2") = ("0.0098 M/s")/("0.020 M" cdot ("0.020 M")^"1/2")`

`= "3.47 M"^(-"1/2")cdot"s"^(-1) ~~ ulcolor(blue)("3.5 M"^(-"1/2")cdot"s"^(-1))`

TRIAL 4

`color(blue)(k) = (r_4(t))/(["CHCl"_3]_4["Cl"_2]_4^"1/2") = ("0.0270 M/s")/("0.040 M" cdot ("0.040 M")^"1/2")`

`= "3.38 M"^(-"1/2")cdot"s"^(-1) ~~ ulcolor(red)("3.4 M"^(-"1/2")cdot"s"^(-1))`

[Probably a typo in the question. If it was `"0.028 M/s"`, that would make the difference and give us `"3.5 M"^(-"1/2") cdot "s"^(-1)` exactly.]