Can someone help me with this question... PLZ I realy need your help...?

Question

The equilibrium constant K for the reaction C (s) + CO_2(g) <--> 2CO (g) is 1.9 at 1000 K and 0.133 at 298 K.

1. If excess C is allowed to react with 21.0 g of CO_2 in a 3.00-L flask, how many grams of CO are produced at 298 K? At 1000k?

2. What are the partial pressures of each gas at 298 K? At 1000k?

3. Would you expect K to increase or decrease if the pressure were increased at constant temperature and volume?

Answer 1

I will help you to answer at 1000K and you do your practice at 298K.
The worst thing is that you forget to provide the unit for K. I do not know if the unit is atm, or mol/L, or some thing else.
Assume the unit of K is atm.
Since the molar mass of CO2 is 44.0g/mol, its initial pressure must be:
P = nRT/V = (21.0/44.0)*0.08206*1000/3.00 = 13.1 (atm)
Let X atm be the partial pressure of CO at equilibrium:
C(s) + CO2(g) <--> 2CO(g), K = 1.9 atm
..........13.1-0.5X.........X
K = 1.9 = X^2 /(13.1 - 0.5X)
X^2 + 0.95X - 1.9*13.1 = 0
The positive solution of the equation is: X = 4.54
Hence at 1000k, CO produced is:
28n = 28*PV/RT = 28*4.54*3.00/82.06 = 4.6 (grams)
The partial pressure of CO is 4.5 atm, and of CO2 is 10.8 atm. If pressure increase, K should decrease.

Assume the unit of K is mol/L.
Since the molar mass of CO2 is 44.0g/mol, its initial concentration must be:
n/V = (21.0/44.0)/3 = 0.159 mol/L
Let X mol/L be the concentration of CO at equilibrium:
C(s) + CO2(g) <--> 2CO(g), K = 1.9 mol/L
..........0.159-0.5X.........X
K = 1.9 = X^2 /(0.159 - 0.5X)
X^2 + 0.95X - 1.9*0.159 = 0
The positive solution of the equation is: X = 0.251
Hence at 1000k, CO produced is:
28n = 28*0.251*3 = 21.1 (grams)
The partial pressure of CO is 0.251*82.06 = 20.6 atm, and of CO2 is 0.0335*82.06 = 2.75 atm. If pressure increase, K should decrease.
Now you see how important the unit of K is.

Answer 2

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