May 20, 2019, 04:56:53 PM
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Topic: Partial Pressures at Equilibrium Problem  (Read 503 times)

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Offline holydog23

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Partial Pressures at Equilibrium Problem
« on: May 03, 2019, 08:03:20 PM »
Given the water shift reaction:

CO(g)+H2O(g)⇌CO2(g)+H2(g) such that Kp = 4.0 at 500 °C

and the initial concentrations of 0.100 M CO and 0.100 M H2O, we should be able to find the partial pressures of each component at equilibrium

I wanted to check if my logic was correct in attempting this problem: I assumed the reaction took place in a 1 L container, yielding 0.100 mols of CO and H2O. I then used the ideal gas law to determine the partial pressures of CO and H2O to both be about 6.34 atm.

Then I can use the ICE chart to determine the equillibrium pressures to be 6.34 - x for CO and H2O and just x for the products.

Using Kp,

4.0 = (x^2)/(6.34-x)^2

by the quadratic equation, x = 4.22 or x ≈ 12

The x=4.2 solution is the only one that makes sense because x = 12 would yield a negative answer for pressure which is impossible in this context

Thus, the partial pressures are

P(CO) = P(H2O) = 6.34 - x
                         =  2.1  atm
P(H2) = P(CO2) = x
                        = 4.2 atm

Is this process of doing the problem correctly? My professor won't release the answer key until after the homework set is due and I couldn't find this in any textbook

Offline AWK

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Re: Partial Pressures at Equilibrium Problem
« Reply #1 on: May 04, 2019, 02:34:32 AM »
Once you find a solution, you can always insert the obtained values into the expression for Kp. A result very close to the Kp value shows the correct work. A small error may result from the rounding of numbers. in this case, the check can be done even without a calculator and the answer in the textbook is unnecessary.
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Offline AWK

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Re: Partial Pressures at Equilibrium Problem
« Reply #3 on: May 06, 2019, 12:24:07 PM »
No confusion!
Kp = Kc for this reaction. Instead of partial pressures or concentrations, we can simply use moles with the same dimensionless equilibrium constant. The equation in moles or in concentrations has even the exact solution in fractions, namely 1/15 mol and 1/10-1/15=1/30 mol (or M)
AWK

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