CHM 152

posted by .

Consider the following equilibrium:

PCl3 + Cl2 = PCl5 Kc=26 @ 250 degree Celcius

1.0 mol of PCl3 , 1.0 mol of Cl2 , and 5.0 mol of PCl5 are placed in a 10.0L flask at 250 C and allowed to come to equilibrium. What is the concentration, in M, of at equilibrium?

a. 3.4

b. 0.19

c. 0.16

d. 0.13

e. 0.20

• CHM 152 -

1. You didn't finish the problem What is the concn of ??????at equilibrium?
2. Note the correct spelling of celsius.

Kc = (PCl5)/(PCl3)(Cl2)
Convert moles to M. Set up an ICE chart and solve.

• CHM 152 -

Consider the following equilibrium:

PCl3 + Cl2 = PCl5 Kc=26 @ 250 degree Celcius

1.0 mol of PCl3 , 1.0 mol of Cl2 , and 5.0 mol of PCl5 are placed in a 10.0L flask at 250 C and allowed to come to equilibrium. What is the concentration, in M, of Cl2 at equilibrium?

a. 3.4

b. 0.19

c. 0.16

d. 0.13

e. 0.20

I set up ICE:
PCl3 + Cl2 = PCl5
I 1.0/10.0 1/10.0 5/10.0

C -x -x +x

E 0.1-x 0.1-x 0.5+x

26= 0.5+x/(0.1-x)^2

so I then take the sqaure root of both sides to get rid of the ^2 then multiply 26(0.1-x)=0.5+x then take the square root of both sides

-0.5*5.099(0.1)=x2

but then I think I am confused in my math because I am thinking set up a quadratic equation formula for this problem?????Please Help

• CHM 152 -

You have it going the wrong way. See my comments at your later post. Also, you can't solve by taking the square root of both side BECAUSE there is no squared term on top (just on the bottom). You must solve the quadratic equation.

Similar Questions

1. chemistry

Given the equilibrium system PCl5(g) <---> PCl3(g) + Cl2(g) K = 12.5 at 60 degrees Celsius. A 1.0-L reaction vessel is analyzed and found to contain 3.2 mol Cl2(g), 1.5 mol PCL3(g) and 3.0 mol PCl5(g). Show that the reaction …
2. Chemistry

Initially, 1.68 mol of PCl5(g) and 0.36 mol of PCl3(g) are in mixed in a 2.00 l container. It is later found that 1.44 mol of PCl5 are present when the system has reached equillibrium. Calculate the value of the equllibrium. This is …
3. chm152

PCl3 + Cl2 = PCl5 Kc=26 @ 250 degree Celcius 1.0 mol of PCl3 , 1.0 mol of Cl2 , and 5.0 mol of PCl5 are placed in a 10.0L flask at 250 C and allowed to come to equilibrium. What is the concentration, in M,of cl2 of at equilibrium?
4. Chemistry

When a sample of PCl5(g) (0.02087 mol/L) is placed in 83.00 L reaction vessel at 491.0 °C and allowed to come to equilibrium the mixture contains 103.0 grams of PCl3(g). What is the equilibrium concentration (mol/L) of Cl2(g)?
5. chemistry

A reaction starts with 1.00 mol each of PCl3 and Cl2 in a 1.00-L flask. When equilibrium is established at 250 degree Celsius in the reaction PCl3(g)+ Cl2(g)--> PCl5(g), the amount of PCl5 present is 0.82 mol. What is Kc for this …
6. Chemistry

Phosphorus pentachloride decomposes according to the chemical equation PCl5(g) <-> PCl3(g)+Cl2(g) Kc=1.80 at 250 degrees Celsius A 0.222 mol sample of PCl5(g) is injected into an empty 3.25 L reaction vessel held at 250 °C. …
7. Chemistry

Phosphorus pentachloride decomposes according to the chemical equation: PCl5(g)<--> PCl3(g)+Cl2(g) Kc = 1.80 at 250 degrees Celsius A 0.352 mol sample of PCl5(g) is injected into an empty 4.45 L reaction vessel held at 250 °C. …
8. Chemistry

The initial concentration for the compounds involved in the reaction displayed were determined to be [PCl5(g)] = 0.4107 mol/L, [PCl3(g)] = 0.3463 mol/L, [Cl2(g)] = 0.04085 mol/L. Calculate the value of the equilibrium constant (Kc) …
9. Chemistry

At 25C, an equilibrium mixture of gases con- tains 0.00680 mol/L PCl3, 0.0290 mol/L Cl2, and 0.00500 mol/L PCl5. PCl5(g) ⇀ ↽ PCl3(g) + Cl2(g) What is the equilibrium constant for the reaction?