Reaction is CO2(g) + C(s) <=> 2CO(g) Kp = 5.7, 1200K Calcualte the total pressure at equilibrium when 4.45g CO2 is introduced into a 10.0L container and heated to 1200K in the presence of 2.00g of graphite
What effect does the mass of graphite have on this?
first step find the pressure exerted pv=nRT use this relation and find the pressure at initial state
so u get the patial pressure of CO2 and graphite so this is the pressure at initial state at equilibrium, co2+ c------2co pco2-k pc-k 2k Kc=2k^2/pco2^2 *pc^2
How can a solid have partial pressure?
very good question!................is the graphite in pure form?
Yes
yes it should be then concentration=moles/volume=density/molar mass density and molar mass being constant graphite has concentration=1 therefore Pc=1
I do not understand what effect does the mass of graphite have on this question. The question after this asks me to do this entire process again but for mass of graphite = 0.50g...
Would not the graphite burn as well?
yes but not contribute to kc which we are bothered abt here and to equilibrium concentration
So you are saying that in both cases, the answer is the same? (ie. when it is 0.50g graphite, or the mass given in this question)...
yeah it souind vague but yeah thats the thing i am sure about it it is only when u have 2gms of c that the raection proceeds if it were less then half the reaction will take place wat i mewan by this is for 1 mol of co2 1 mol of c is to be present to give 2 mol of CO so thats basically the idea
n1 =moles of CO2 (find them) n2 = moles of C (find them) CO2(g) + C(s) <=> 2CO(g) n1 n2 0 -x -x +2x n1-x n2-x 2x Kp=P^2 (CO)/Pco2 ..........use PV=nRT Kp=(2x)^2/n1-x *RT/V solve to x then P(total)=n(total)*R*T/V......Ptot=....
Salini, you might want to review. Both answers being the same is incorrect.
Mos, the Carbon solid is a constant. So it does not take place in an equilibrium ICE table.
yes you are right i should put n(total) as n (gas total)=n(co2)+n(co)
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