Respuesta :
Answer:
There will be produced 379 grams of H2
The percent yield under these conditions is 7.45 %
Explanation:
Step 1: The balanced equation
CH4(g)+CO2(g) → 2CO(g)+2H2(g)
Step 2: Given data
Kp = 4.5*10²
Temperature = 825 Kelvin
Volume = 85.0 L
Mass of CH4 = 22.3 Kg
Mass of CO2 = 55.4 Kg
Molar mass of CH4 = 16.04 g/mole
Molar mass of CO2 = 44.01 g/mole
Step 3: Calculate moles of CH4
moles of CH4 = mass of CH4 / Molar mass of CH4
moles of CH4 = 22300 / 16.04 = 1390.3
Step 4: Calculate moles of CO2
moles of CO2 = mass of CO2 / Molar mass of CO2
moles of CO2 = 55400 / 44.01 = 1258.80 moles
Step 5: Calculate moles of H2
For 1 mole of CH4 we need 1 mole of CO2 to produce 2 moles of H2
so there will be produced 2*1258.8 = 2517.6 moles of H2
Step 6: Calculate theoretical mass of H2
mass of H2 = Number of moles of H2 * Molar mass of H2
mass of H2 = 2517.6 moles *2.02 g/mole = 5085.552 grams
Step 7: Calculate pressure of CH4
P*V = n*R*T
P =(n*R*T) / V
P = 1390.3*.0821*825/85 = 1107.86atm
Step 8: Calculate pressure of CO2
P*V = n*R*T
P =(n*R*T) / V
P = 1258.80*.0821*825/85 = 1003.08atm
Step 9
Kp= 4.5 *100 =[P(H2)^2 * P(CO)^2]/[P(CO2) * P(CH4)]
450=16X^4/[(1107.86-X)(1003.08-X)]
450=16X^4/[(1107.86)(1003.08)]
(1107.86)(1003.08)*450=16X^4
X = 74.77
We plug this value in for "X" in the ICE chart. Only H2 since thats what the problem wants.
for H2 (2 moles) so 2X = 2*74.77 = 149.54
Step 10: Calculate number of moles of H2
n = P*V/ R*T
n = (149.54 *85 )/(0.0821 * 825) = 187.66 moles H2
Step 11: Calculate mass of H2
mass of H2 = Number of moles H2 * Molar mass of H2
mass of H2 = 187.66 * 2.02 g/moles = 379.07 grams H2 ≈ 379 grams of H2
Step 12: Calculate the yield
(379 grams of H2 / 5085.552 grams of H2 ) * 100 % = 7.45 %
The percent yield under these conditions is 7.45 %
Based on the data provided, the mass of H2 at equilibrium is 379 grams and percent yield is 7.45 %.
What is the mass of hydrogen gas at equilibrium?
The theoretical mass of the hydrogen gas at equilibrium is determined from the equation of the reaction.
The balanced equation of the reaction is as follows:
- CH4 (g) + CO2 (g) → 2 CO (g) +2 H2 (g)
The moles of CH4 is first determined:
- moles of CH4 = mass of CH4 / Molar mass of CH4
Mass of CH4 = 22.3 Kg = 22300 g
Molar mass of CH4 = 16.04 g/mole
moles of CH4 = 22300 g / 16.04 g= 1390.3 moles
The moles of CO2 is then determined:
moles of CO2 = mass of CO2 / Molar mass of CO2
Mass of CO2 = 55.4 Kg = 55400 g
Molar mass of CO2 = 44.01 g/mole
moles of CO2 = 55400 / 44.01 = 1258.80 moles
Moles of H2 produced is calculated from the molar equation
mole ratio of CO2 to H2 is 1:2
moles of H2 produced 2 * 1258.8 = 2517.6 moles of H2
Then;
mass of H2 = 2517.6 moles *2.02 g/mole = 5085.552 grams
The actual yield of H2 is determined from the partial pressures of each of the gases thus:
Prssure of CH4
Using the ideal gas equation,
- PV = nRT
where;
- R = 0.0821 atm.L/K.mol
- T = 825 K
- V = 85.0 L
P = nRT/V
P = 1390.3*.0821*825/85
PCH4 = 1107.86atm
Pressure of CO2
P = nRT/V
P = 1258.80*.0821*825/85
PCO2= 1003.08atm
From Kp = 4.5 × 10² = 450
Kp = 450 = [P(H2)^2 × P(CO)^2] / [P(CO2) * P(CH4)]
450 = 16X^4 / [(1107.86-X)(1003.08-X)]
450=16X^4/[(1107.86)(1003.08)]
(1107.86)(1003.08) × 450 = 16X^4
X = 74.77
Substituting X for H2 in the ICE chart:
H2 = 2 moles
Therefore,
2X = 2 × 74.77 = 149.54
Calculating number of moles of H2 using ideal gas equation:
n = PV/RT
n = (149.54 *85 )/(0.0821 × 825)
n = 187.66 moles H2
mass of H2 = Number of moles × Molar mass
mass of H2 = 187.66 × 2.02 g/moles
mass of H2 = 379 grams
Percent yield = actual yield/theoretical yield × 100%
Percent yield = (379 g / 5085.552 g) * 100 % Percent yield = 7.45 %
Therefore, the mass of H2 at equilibrium is 379 grams and percent yield is 7.45 %
Learn more about mass and percent yield at: https://brainly.com/question/8638404
