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.
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?
Choose one answer.
a. 3.4
b. 0.19
c. 0.16
d. 0.13
e. 0.20
3 answers
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?
Choose one answer.
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
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?
Choose one answer.
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
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.