I dont understand this chemistry alevel equilibria question
A 0.230 mol sample of carbon monoxide is mixed with hydrogen in a 1:2
mol ratio and allowed to reach equilibrium in a sealed flask at temperature
T.
At equilibrium the mixture contains 0.120 mol of carbon monoxide.
The total pressure of this mixture is 1.04 × 10^4 kPa
Calculate the partial pressure, in kPa, of hydrogen in the equilibrium
mixture.
Partial pressure of hydrogen _______________ kPa
i thought i would do the rice table and found that H2 is 0.24 mol at equilibrium then the mol fraction would be 0.24/0.24 + 0.12(mol of CO at equilibrium) but checking the mark scheme it seems to be (0.24/0.24+0.23(mol of CO before equilibrium)) please help me understand this
mol ratio and allowed to reach equilibrium in a sealed flask at temperature
T.
At equilibrium the mixture contains 0.120 mol of carbon monoxide.
The total pressure of this mixture is 1.04 × 10^4 kPa
Calculate the partial pressure, in kPa, of hydrogen in the equilibrium
mixture.
Partial pressure of hydrogen _______________ kPa
i thought i would do the rice table and found that H2 is 0.24 mol at equilibrium then the mol fraction would be 0.24/0.24 + 0.12(mol of CO at equilibrium) but checking the mark scheme it seems to be (0.24/0.24+0.23(mol of CO before equilibrium)) please help me understand this
(edited 1 year ago)
Carbon monoxide and hydrogen make methanol, and so the equilibrium moles of athanol needs to also be considered when calculating total moles.
CO + 2H2 --> CH3OH
I 0.230 0.460 0
C -0.11 -0.22 +0.11
E 0.120 0.240 0.11
Total moles = 0.120 +0.240+0.11 = 0.47
Mole fraction of H2 = 0.24/0.47
Partial pressure = (0.24/0.47) x 1.04 x 10^4 kPa
CO + 2H2 --> CH3OH
I 0.230 0.460 0
C -0.11 -0.22 +0.11
E 0.120 0.240 0.11
Total moles = 0.120 +0.240+0.11 = 0.47
Mole fraction of H2 = 0.24/0.47
Partial pressure = (0.24/0.47) x 1.04 x 10^4 kPa
Reply 2
Original post by kngikraam
A 0.230 mol sample of carbon monoxide is mixed with hydrogen in a 1:2
mol ratio and allowed to reach equilibrium in a sealed flask at temperature
T.
At equilibrium the mixture contains 0.120 mol of carbon monoxide.
The total pressure of this mixture is 1.04 × 10^4 kPa
Calculate the partial pressure, in kPa, of hydrogen in the equilibrium
mixture.
Partial pressure of hydrogen _______________ kPa
i thought i would do the rice table and found that H2 is 0.24 mol at equilibrium then the mol fraction would be 0.24/0.24 + 0.12(mol of CO at equilibrium) but checking the mark scheme it seems to be (0.24/0.24+0.23(mol of CO before equilibrium)) please help me understand this
mol ratio and allowed to reach equilibrium in a sealed flask at temperature
T.
At equilibrium the mixture contains 0.120 mol of carbon monoxide.
The total pressure of this mixture is 1.04 × 10^4 kPa
Calculate the partial pressure, in kPa, of hydrogen in the equilibrium
mixture.
Partial pressure of hydrogen _______________ kPa
i thought i would do the rice table and found that H2 is 0.24 mol at equilibrium then the mol fraction would be 0.24/0.24 + 0.12(mol of CO at equilibrium) but checking the mark scheme it seems to be (0.24/0.24+0.23(mol of CO before equilibrium)) please help me understand this
Write out the equation.
You know how much CO has reacted, therefore you know how much of the other two components there are.
(This, I imagine is your rice table)
The sum of the partial pressures = total pressure
The partial pressure of each gas = mole fraction x total pressure.
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