# Chemistry

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?

1. 👍
2. 👎
3. 👁
1. rate = k(PCl3)^x*(Cl2)^y
would x = 0; y = 2 make sense to you?

1. 👍
2. 👎
2. I understand why PCl3 is zero order, but I don't get why Cl2 is second order.

1. 👍
2. 👎
3. 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.

1. 👍
2. 👎

## Similar Questions

1. ### Chemistry

A 0.60mol sample of PCl3(g) and a 0.70mol sample of Cl2(g) are placed in a previously evacuated 1.0L rigid container, and the reaction represented above takes place. At equilibrium, the concentration of PCl5(g) in the container is

2. ### Chemistry

The equilibrium constant (Kp) for the interconversion of PCl5 and PCl3 is 0.0121: PCl5 (g) ↔ PCl3 (g) + Cl2 (g) A vessel is charged with PCl5, giving an initial pressure of 0.123 atm. At equilibrium, the partial pressure of PCl3

3. ### Chemistry

Consider the following equilibrium: PCl3 (g) + Cl2 (g) yields PCl5 (g) delta H= -92KJ The concentration of PCl3 at equilibrium may be increased by: decreasing the temp (think this is it) adding Cl2 to the system the addition of

4. ### Chemistry

PCl3(g)+Cl2(g)⇄PCl5(g) Kc=0.11 A 0.60mol sample of PCl3(g) and a 0.70mol sample of Cl2(g) are placed in a previously evacuated 1.0L rigid container, and the reaction represented above takes place. At equilibrium, the

2. ### Chem II

The equilibrium constant, Kp for the reaction PCl5 PCl3 + Cl2 is 1.05 at 250 degrees C. The reaction is started with PCl5, PCL3 and Cl2 at 0.177, 0.223, and 0.111 atm at 250 degrees C. When the reaction comes to equilibrium, the

3. ### chemistry

Phosphorus pentachloride decomposes according to the chemical equation PCl5(g) -----> PCl3 (g)+ Cl2(g) Kc= 1.80 at 250 degrees C A 0.463 mol sample of PCl5(g) is injected into an empty 4.80 L reaction vessel held at 250 °C.

4. ### Chemistry

What effect do the following changes have on the position of equilibrium (which way does the equilibrium shift) for the reversible endothermic reaction? PCL5+ heat PCL3(g)+ CL2(g) A) removal of PCL3, as it is formed B)addition of

1. ### chemistry

What is the percent yield for the reaction PCl3(g) + Cl2(g) --> PCl5(g) if 119.3 g of PCl5 are formed when 61.3 g of Cl2 react with excess PCl3? I don't understand how to set this up?

2. ### Chmistry

pcl3(g) cl2(g) ¨

3. ### Chemistry

You wish to know the enthalpy change for the formation of liquid PCl3 from the elements shown below. P4(s) + 6 Cl2(g)--> 4 PCl3(l) ΔH°f = ? The enthalpy change for the formation of PCl5 from the elements can be determined

4. ### AP Chemistry

2 Questions (Multi-choice and T/F, though not sure if they're right, and these 2 questions have no correspondence to each other) 1.) a.) Consider the following system at equilibrium where Kc = 1.20E-2 and H° = 87.9 kJ/mol at 500