Question

Ammonia (NH3) reacts with oxygen to form nitrogen monoxide and water. All the materials involved in this reaction are gasses. 0.100 moles of each of the reactants are initially introduced to a 5.0-liter reaction vessel. a. What would be the quantity of each gas in the container upon completion of the reaction? b.What would be the partial pressure of each gas on the reaction vessel upon reaction completion if the temperature of the system is 105 degrees C? c. What is the total pressure of all the gases on the reaction vessel at 105 degrees C?

Answer 1

Answer:

a. 0.02 moles  of NH₃, 0 moles of O₂, 0.08 moles of NO, 0.12 moles of H₂O

b. $P_{NH_3}$ = 12,576.5 Pa, $P_{NO}$ = 50,306.05 Pa, $P_{H_2O}$ =  74,459.1 Pa

c. The total pressure is 138,341.64 Pa

Explanation:

a. NH₃ + O₂  → NO + H₂O

The balanced chemical equation is first found to be

4NH₃ + 5O₂  → 4NO + 6H₂O

Therefore, we have;

4 moles of NH₃ reacts with 5 moles of O₂ to form 4 moles of NO and 6 moles H₂O

Dividing by the reactant with the highest number of moles which is 5 moles of oxygen gives;

4/5 moles  of NH₃ reacts with 5/5 moles of O₂ to form 4/5 moles of NO and 6/5 moles H₂O

Which is the same as 4/5 moles  of NH₃ reacts with 1 mole of O₂ to form 4/5 moles of NO and 6/5 moles H₂O

Multiplying by 0.100 gives;

0.1×4/5 moles  of NH₃ reacts with 0.1 mole of O₂ to form 0.1×4/5 moles of NO and 0.1×6/5 moles H₂O

The quantity of each gas in the container upon completion of the reaction is therefore;

(0.1 - 0.1×4/5) = 0.02 moles  of NH₃

0 moles of O₂

0.08 moles of NO

0.12 moles H₂O

b. Given that the temperature = 105°C, we have;

PV = nRT

P = nRT/V

Where:

n = Total number of moles = 0.02 + 0.08 + 0.12 = 0.22 moles

R = Universal gas constant = 8.3145 J/(mol·K)

T = Temperature = 105°C = 378.15 K

V = Volume = 5 litre = 0.005 m³

P = 0.22×8.3145×378.15/0.005 = 138,341.64 Pa

From Dalton's law of partial pressure, we have;

Partial pressure Pₓ = Xₓ × P

Where:

Xₓ = Mole fraction

Which gives for ammonia NH₃ with 0.02 moles;

Mole fraction = 0.02/0.22 = 1/11

$P_{NH_3}$ = 1/11 × 138,341.64  = 12,576.5 Pa

For the 0.08 moles of NO, we have

Mole fraction = 0.08/0.22 = 4/11

$P_{NO}$ = 4/11 × 138,341.64  = 50,306.05 Pa

For the 0.12 moles H₂O

P = 0.12×8.3145×378.15/0.005 = 74,459.1 Pa

Mole fraction = 0.12/0.22 = 6/11

$P_{H_2O}$ = 6/11 × 138,341.64  = 74,459.1 Pa

c. The total pressure = 12,576.5 Pa + 50,306.05 Pa + 74,459.1 Pa = 138,341.64 Pa.