Hydrogen can be extracted from natural gas according to the following reaction:

CH4 (g) + CO2 (g) 2CO (g) + 2H2 (g)

Kp=4.5*102 at 825 K

An 85.0 L reaction container initially contains 22.3 kg of CH4 and 55.4 kg of CO2 at 825 K.

a) Assuming ideal gas behavior, calculate the mass of H2 (in g) present in the reaction mixture at equilibrium.

b) What is the percent yield of the reaction under these conditions?

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 %