In the reaction PCl3 + Cl2 ---> PCl5.
If the rate doesn't increase when the concentration of PCl3 is doubled and the rate increases by a factor of four when the concentration of Cl2 is doubled, how would I write the rate law for the reaction?
rate = k(PCl3)^x*(Cl2)^y
would x = 0; y = 2 make sense to you?
I understand why PCl3 is zero order, but I don't get why Cl2 is second order.
rate 1 = k*(PCl3)^0(Cl2)^y
rate 2 = k*(PCl3)^0(Cl2)^y
The problem says rate increases by 4 when we double Cl2. So let's do that. Let's call rate 1 = 1 to make things simple; also, since (PCl3)^0 = 1 we can dispense with that, too. Then we will call (Cl2)= 1 with rate 1, again, to make things simple, then (Cl2) = 2 with rate 2.
rate 1 = 1, (Cl2)^y = (1)^y
rate 2 = 4, (Cl2)^y = (2)^y
Now divide the two equations, we get
1/4 = (1)^y/(2)^y = (1/2)^y so
what must y be to raise 1/2 to that power in order to obtain 1/4. y must = 2. Does this help? It may be easier to see if you divide rate 2 by rate 1, in which case we have
4/1 = (2/1)^y and y must be 2 so that 2^2 = 4.
To determine the rate law for the reaction, you need to establish the relationship between the rate of the reaction and the concentrations of the reactants.
Based on the given information:
1. The rate doesn't increase when the concentration of PCl3 is doubled: This indicates that the concentration of PCl3 does not affect the rate of the reaction. Therefore, the reaction rate is independent of the concentration of PCl3.
2. The rate increases by a factor of four when the concentration of Cl2 is doubled: This tells us that the rate is directly proportional to the square of the Cl2 concentration (c^2).
Based on these observations, we can write the rate law for the reaction as:
rate = k[PCl3]^0[Cl2]^2
Since [PCl3]^0 is equal to 1 (because any value raised to the power of 0 is 1), it can be omitted from the rate law equation.
Therefore, the rate law for the reaction can be written as:
rate = k[Cl2]^2
where k is the rate constant.