The equilibrium constant Kc for the following reaction equals 49 at 230°C.
PCl3(g) + Cl2(g) <--> PCl5(g)
If 0.313 mol each of phosphorus trichloride and chlorine are added to a 3.8-L reaction vessel, what is the equilibrium composition of the mixture at 230°C?
0.313mol/3.8L = 0.0824M
I PCl3=Cl2= 0.0824M; PCl5= 0
C PCl3=Cl2= -x; PCl5= +x
E PCl3=Cl2= 0.0824 - x; PCl5= x
kc = [PCl5]/[PCl3][Cl2]
49 = x/(0.0824-x)(0.0824-x)
49 = x/0.00679 - 0.0824x - 0.0824x + x^2
49 = x/0.00679 - 0.1648x +x^2
x=49x^2 - 8.0752x + 0.3325
subtract x to both sides
so 49x^2 - 9.0752x + 0.33271 = 0
a = 49
b = -9.0752
c = 0.33271
x=(-b±?(b^2-4ac))/2a
x=(9.0752±?82.27-65.17)/98
X=9.0752±4.15
SO THIS WHEN THE PROBLEM OCCURS, THE ANSWER HAS 2 POSSITIVE ANSWERS, WHAT CAN I DO NOW???
Please help me!!!!! i will be really appreciated. Thank you for your time
SO THE QUESTION IS --> WHAT IS THE EQUILIBRIUM OF PCl3, Cl2, PCl5?
PCl3=?M
PCl5=?M
Cl2=?M
PCl3(g) + Cl2(g) <--> PCl5(g)
If 0.313 mol each of phosphorus trichloride and chlorine are added to a 3.8-L reaction vessel, what is the equilibrium composition of the mixture at 230°C?
0.313mol/3.8L = 0.0824M
I PCl3=Cl2= 0.0824M; PCl5= 0
C PCl3=Cl2= -x; PCl5= +x
E PCl3=Cl2= 0.0824 - x; PCl5= x
kc = [PCl5]/[PCl3][Cl2]
49 = x/(0.0824-x)(0.0824-x)
49 = x/0.00679 - 0.0824x - 0.0824x + x^2
49 = x/0.00679 - 0.1648x +x^2
x=49x^2 - 8.0752x + 0.3325
subtract x to both sides
so 49x^2 - 9.0752x + 0.33271 = 0
a = 49
b = -9.0752
c = 0.33271
x=(-b±?(b^2-4ac))/2a
x=(9.0752±?82.27-65.17)/98
X=9.0752±4.15
SO THIS WHEN THE PROBLEM OCCURS, THE ANSWER HAS 2 POSSITIVE ANSWERS, WHAT CAN I DO NOW???
Please help me!!!!! i will be really appreciated. Thank you for your time
SO THE QUESTION IS --> WHAT IS THE EQUILIBRIUM OF PCl3, Cl2, PCl5?
PCl3=?M
PCl5=?M
Cl2=?M