M PCl5 = 0.463 mols/4.80L = 0.09646 which is more places than allowed but I like to keep one extra place, go through the calculations and round in the last step.
..........PCl5 ==>PCl3 + Cl2
initial.0.09646......0......0
change...-x.........x.......x
equil..0.09646-x.....x......x
Kc = 1.80 = (PCl3)(Cl2)/(PCl5)
Substitute from the ICE chart into the Kc expression and solve for x, then for 0.09646
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. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.
PLEASE HELP!!!! Can you confirm the correct answer as well. I have one more attempt. Thanks
9 answers
What answers did u get for PCl5 and PCl3...I want to verify them please.
I didn't work it. Give me your answers and I will check it.
I got x=0.091766....now what do i do? I'm confused. Also can check if this is right for x?
I obtained 0.0918M for x and that is (PCl3). The problem also asked for (PCl5) and that is 0.09646-x = ? and round to 3 s.f. I obtained 0.00466 for that.
yes it is correct! thank you :) I really appreciate it!
How do you solve for x with the equation 1.8 = x^2 / .0964 - x? i don't understand what to do isolate x
I want to know that too ^
That always stumps me on the hw!
That always stumps me on the hw!
you cannot isolate X by its self without using the quadratic equation:
x = -B +or- square-root(B^2 - 4AC)
------------------------------
2A
x = -B +or- square-root(B^2 - 4AC)
------------------------------
2A
1 answer