Skip to main content

Equilibrium- Heterogeneous Reaction & Kp

Submitted by toinfinity on Sat, 03/06/2010 - 20:12

I'm having some trouble understanding the calculation of Kp for heterogeneous reactions and finding the number of moles of certain products/reactants in equilibrium..

250.0 grams of solid copper (II) nitrate is placed in an empty 4.0-L flask. Upon heating the flask to 250°C, some of the solid decomposes into solid copper (II) oxide, gaseous nitrogen (IV) oxide, and oxygen gas. At equilibrium, the pressure is measured and found to be 5.50 atmospheres.

a) Write the balanced equation for the reaction.
For my answer, I got 2Cu(NO3)2 (s) → 2CuO (s) + 4NO2 (g) + O2 (g)

b) Calculate the number of moles of oxygen gas present in the flask at equilibrium.
How do you find the number of moles..? Does it have to do with the Initial/Change/Equilibrium table? and if so, that means converting Kp into Kc then, right? Then in the equation Kp=Kc(RT)Δn, how do you find the Δn? Do you include the number of moles for the solids also, or no since solids aren't included in the calculations of Kp?

c) Calculate the number of grams of solid copper (II) nitrate that remained in the flask at equilibrium.
I thought solids weren't included in calculating the Kp, so how do we find the grams..? Or do you convert value of O2 from b into moles of copper (II) nitrate?

d) Write the equilibrium expression for Kp and calculate the values of the equilibrium constant.
Same as in b.. how do you calculate the Δn when there are solids? Do you in/exclude the moles of solids?

e) If 420.0 grams of the copper (II) nitrate had been placed into the empty flask at 250°C, what would the total pressure have been at equilibrium?
It's a solid, so does it even change the equilibrium pressure?..

I know I have a lot of questions, but if you could provide any help that'd be great. Thanks in advance.

Equation is fine
2Cu(NO3)2 (s) → 2CuO (s) + 4NO2 (g) + O2 (g)

Partial pressure = mole fraction x total pressure.
As you have to get NO2 and O2 in the molar ratio 4:1, then mole fraction of NO2 = moles NO2/total moles = 4/5 = 0.8 and O2 = 1/5 = 0.2
So PP NO2 = 0.8 x 5.50 = 4.40 atm and PP O2 = 0.2 x 5.50 = 1.10 atmos

Use PV = nRT to find the number of moles of each gas using the PP's calculated.

Once you know moles of O2, use the balanced equation to find the mass solid that remained
2Cu(NO3)2 (s) → 2CuO (s) + 4NO2 (g) + O2 (g)
2 mole                                              1 mol
So you can find the number of moles of solid that reacted. Find the mass of this using the molar mass.
Subtract this from the starting mass to give the mass unreacted

Kp = PP NO24 x PP O2  (solids do not appear in the Kp expression and don't affect the pressure)

In the last part, use the Kp calculated to find the PP's of the gases made, knowing that the ratio has to be 4:1. You could say that PP O2 = w and so PP NO2 = 4w. Plug into Kp to find w and add w to 4w to get the total pressure (I think)