calculation of molar fraction and pressure

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anto007
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calculation of molar fraction and pressure

Post by anto007 »

Hi everyone, i need your help again. I have this exercise: "In a closed reactor of 8.00 dm ^ 3, thermostated at a certain temperature, 2.50 dm ^ 3 of methane, CH4, measured at the pressure of 1.45 * 10 ^ 5 Pa, are mixed with 4.50 dm ^ 3 of oxygen, measured at 2.70 * 10 ^ 5 Pa and at the same temperature. A spark ignites the combustion of methane:
CH4 (g) + 2O2 (g) -> CO2 (g) + 2H2O (g)
calculate a) the total pressure and partial pressures of the gases before combustion; b) the molar fractions and partial pressures after combustion, knowing that at this temperature the water is in the form of vapor. "

I solved point a) like this:
I apply Boyle's isothermal law P1V1=P2V2
P1(CH4)=1.45 * 10 ^ 5 Pa
V1(CH4)=2.50 dm^3
V2= 8.00 dm^3
P2(CH4)=(P1(CH4)*V1(CH4))/V2=(1.45 * 10 ^ 5 Pa*2.50 dm^3)/8.00 dm^3= 4.53*10^4 Pa

P1(O2)=2.70 * 10 ^ 5 Pa
V1(O2)= 4.50 dm^3
V2= 8.00 dm^3
P2(O2)=(P1(O2)*V1(O2))/V2= (2.70 * 10 ^ 5 Pa* 4.50 dm^3)/8.00dm^3= 1.52*10^5 Pa

P(tot)= P2(CH4)+P2(O2)=(4.53*10^4 Pa)+(1.52*10^5 Pa)= 1.973*10^5 Pa

b)
Now to calculate the molar fractions I need to know the total moles and the moles of the individual gases of the products, but I can't figure out how to get them from the information I have. I know the water vapor temperature is 100 ° C (373.15 F). Could you help me understand how to extrapolate this information? Thank you very much
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ChenBeier
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Re: calculation of molar fraction and pressure

Post by ChenBeier »

Consider the vessel is 8 dm^3. In this is filled 2,5 dm^3 CH4 and 4.5 dm^3 O2 . In total 7 dm^3. 1 dm^3 is missing so we have some expansion.
of the gases. The two pressures are also different.

p1V1 + p2V2 = ptotal* Vvessel

pV ~ n. RT is constant.

From this you have to get total pressure and the partial pressures.
After burning you get new products, but what is really different , compare the moles.
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anto007
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Re: calculation of molar fraction and pressure

Post by anto007 »

Hello, sorry if I answer you late, but there was a fire emergency, here the woods have been burning for days.
Here is my development:
P(CH4)=1.45*10^5 Pa = 1.43 atm n(CH4)= (1.43 atm*2.50 L)/(0.08206*T) = 43.6/T mol
P(O2)= 2.7.*10^5 Pa = 2.66 atm n(O2) = (2.66 atm*4.5 L)/(0.08206*T) = 146.1/T mol
n tot = 189.6 mol
CH4 (g) + 2O2 (g) -> CO2 (g) + 2H2O (g)
43.6/T 146.1/T 0 0
-43.6/T -87.2/T +43.6/T +87.2/T
0 58.9/T 43.6/T 87.2/T
n tot =189.6/T
X(O2) = (58.9/T)/(189.6/T) = 0.311
X(CO2) = (43.6/T)/(189.6/T) = 0.230
X(H2O) = (87.2/T)/(189.6/T) = 0.460

Ptot= (189.6*0.08206*T)/(T*8) = 1.95 atm = 1.97*10^5 Pa
P(O2) = 0.311*1.97*10^5 Pa = 6.13*10^4 Pa
P(CO2) = 0.230*1.97*10^5 Pa = 4.53*10^4 Pa
P(H2O) = 0.460*1.97*10^5 Pa = 9.06*10^4 Pa
The mind is like a parachute, it only works if it opens. -A. Einstein
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ChenBeier
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Re: calculation of molar fraction and pressure

Post by ChenBeier »

I am really sorry to hear that a fire is close to your place. This planet is dying.

Your results are correct.
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