Question

For the system CaO(s) + CO2(g) = CaCO3(s), I added 1.00 mol of CaO(s) to 1.00L of 0.500M CO2(g) at 200 oC. At equilibrium the [CO2] = 0.150M. What is the value of Kp for this reaction?

Kp would = CaCO3/(CaO)(CO2)

I do not know what to do with the 0.500M and 1.00mol.



Chemistry(Please help) - DrBob222, Wednesday, March 14, 2012 at 10:32pm
First you need to work on the Kp expression. SOLIDS aren't used (pure liquids aren't either) so the expression is 1/pCO2 = Kp.
The problem tells you that AT EQUILIBRIUM [CO2] = 0.150M.
I would calculate Kc from Kc = 1/([CO2]
then convert to Kp by Kp = Kc(RT)Dn



So I did 1/0.150 and got 6.66 so now I use this with RT. I am not sure what R is? Kp=6.66(200C + 273)(R)??

Answer 1

I thought one of the first constants one learned in chemistry from the PV = nRT equation was R. R = 0.08206 L*atm/mol*K.

Answer 2

ok so 6.66(473K)(0.08206) = 258.5 which would be 2.59 which is one of the answer choices. Does this seem correct??

Answer 3

the R will be molecular mass of the equation of the reaction

Answer 4

I don't think so.
Kp = Kc(RT)^Dn.
You have Kc, R, T but you don't have delta n anywhere in the equation. Also, note that it is NOT R*T^{delta n} but (RT)^{delta n}

Answer 5

Hannah, I understand that you are having trouble with the homework but academic dishonesty is not the answer. I will be flagging your academic transcripts by the end of today, please come speak with me as soon as you can, maybe we can work something out.