- Joined
- Sep 20, 2009
- Messages
- 105
- Reaction score
- 1
- Points
- 4,551
- Pre-Dental
Advertisement - Members don't see this ad
3A+2B->8C
at equilibrium, total pressure is 1. Kp = ?
initial conditions, .5 atm = A, .2 atm = B.
i am aware that you can solve this by doing it the other way which involves
A B C
.5 .2 0
.5-3x .2-2x 0+8x
1= .5-3x+.2-2x+8x
then solving for X.
My question however, is:
is there a problem with using mole fractions to solve for concentration of each to then solve for Kp?
for example mole fraction of A = 3/(3+5+8)
partial pressure = mole fraction * total pressure (1)
total moles = partial pressure / moles of A
which then you can plug into Kp = (c)^8/ ((a)^3+(b)^5))
at equilibrium, total pressure is 1. Kp = ?
initial conditions, .5 atm = A, .2 atm = B.
i am aware that you can solve this by doing it the other way which involves
A B C
.5 .2 0
.5-3x .2-2x 0+8x
1= .5-3x+.2-2x+8x
then solving for X.
My question however, is:
is there a problem with using mole fractions to solve for concentration of each to then solve for Kp?
for example mole fraction of A = 3/(3+5+8)
partial pressure = mole fraction * total pressure (1)
total moles = partial pressure / moles of A
which then you can plug into Kp = (c)^8/ ((a)^3+(b)^5))