Asked by maura
Kp=1.6
Kp=1.0*10^-4
A<->2B
Find the equilibrium partial pressures of A and B
B=1.0 atm
A=0.0 atm
This is what I have so far
A 2B
0 1.0
+p -p
p 1-p
[B]^2/A (1-p)^2/p = 1.6
then got a quadratic equation of p^2+1.6p-1 to get x=.48
Kp=1.0*10^-4
A<->2B
Find the equilibrium partial pressures of A and B
B=1.0 atm
A=0.0 atm
This is what I have so far
A 2B
0 1.0
+p -p
p 1-p
[B]^2/A (1-p)^2/p = 1.6
then got a quadratic equation of p^2+1.6p-1 to get x=.48
Answers
Answered by
DrBob222
Look at your algebra. I did that and obtained p^2 - 3.6p + 1.0 = 0
Answered by
maura
Not sure what I did there? my x value came to be .3035. Should I plug this number into the (1-p)^2/p
Answered by
maura
if that is correct then i got a value of 1.6 after plugging it in. This is the step that Im not sure what to do.
Answered by
DrBob222
I get 0.48 if I solve p^2 + 1.6p -1 = 0
If I solve p^2 -3.6p + 1 = 0 I get 0.303
If I solve p^2 -3.6p + 1 = 0 I get 0.303
Answered by
maura
I got the same answers for both, but I don't know how to get two pressures from these numbers
Answered by
DrBob222
p (your x) = 0.303 = partial pressure of A.
1-0.303 = ? = partial pressure of B.
That is if you are going with 0.303 for p.
1-0.303 = ? = partial pressure of B.
That is if you are going with 0.303 for p.
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