Posted by Raj on Wednesday, February 27, 2008 at 5:33pm.
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.
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