Consider the reversible reaction of carbon monoxide and water vapor yielding car

Question

Consider the reversible reaction of carbon monoxide and water vapor yielding carbon dioxide and hydrogen. We have 1 mole each of the starting reactants in a 50 L reaction vessel. The equilibrium constant K is 0.58 at 1000 C. How many moles of each substance are in the equilibrium mixture? Also, calculate Kp for this reaction.

I already have the moles of each substance in the equilibrium mixture. I'm having trouble figuring out the Kp for this reaction.

This is what I got so far, but I'm not sure if it is raised to the 0 power or 2nd power in order to find delta n.

Kp = 0.58[(0.082 L*atm/K*mol)(1273 K)]^0 = 60.62

or

Kp = 0.58[(0.082 L*atm/K*mol)(1273 K)]^2 = 3.67x10^3

Thank you in advance for any help!

Explanation / Answer

molar concentration = number of moles/volume in litre

CO(g) + H2O(g) <====> CO2(g) + H2(g)

initially (1/50)M (1/50)M 0 0

at equilibrium (0.02-x) (0.02-x) x x

therefore, Kc = x^2/(0.02-x)^2

or, 0.58 = x^2/(0.02-x)^2

or, x= 0.00865M

therefore moles of CO2 at equilibrium = 50x = 0.4325

moles of H2 at equilibrium = 50x = 0.4325

moles of CO at equilibrium = 50*(0.02-x) = 0.5675

moles of H2O at equilibrium = 50*(0.02-x) = 0.5675

now, Kp = Kc*(RT)^n ; where n= number of moles of gaseous products - number of moles of gaseous reactants in the balanced equation

in this reaction n= 0

therefore Kp =Kc= 0.58   

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects