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

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#.