Question

 2 NH3 (g)  +   2 H2O (g)   =   2 NO (g)   +   5 H2 (g)

When 12.0 grams of ammonia; 16.0 grams of water; and 20 grams of NO (g) are put into a closed, expandable container kept at 160 ^oC, 1.0 atm, final partial pressure of NO (g) is 20 % greater than the final partial pressure of the hydrogen gas. What is the final volume of the expandable container?

(Answer = 98.1 Liters )

Answer 1

Ans. Moles of NH₃ = 12.0 g / (17.03056 g/mol) = 0.704616 mol

Moles of water (vapor) = 16.0 g / (18.01528 g/ mol) = 0.888135 mol

Moles of NO = 20.0 g / (30.00614 g/ mol) = 0.666530 mol

# Total initial moles of gaseous species in vessel =

0.704616 mol + 0.888135 mol + 0.666530 mol

= 2.259281 mol

# Given, Temperature, T = 160.0⁰C = 433.15 K

            Pressure, P = 1.00 atm

# Calculate the initial volume of the reaction vessel using ideal gas equation-

Ideal gas equation: PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Putting the values in equation 1 -

1.00 atm x V = 2.259281 mol x (0.0821 atm L mol^-1K^-1) x 433.15 K

            Hence, V = 80.344 L

Therefore, initial volume of reaction vessel = 80.344 L

# Calculate initial mole fraction and partial pressure of each gas

Now,

pNH₃ = mole fraction of NH₃ x Initial pressure in vessel

                        = (Moles of NH₃ / Total initial moles) x Initial pressure in vessel

                        = (0.704616 mol / 2.259281 mol) x 1.0 atm

                        = 0.311876 atm

            pH₂O(g) = (0.888135 mol / 2.259281 mol) x 1.00 atm = 0.393105 atm

            pNO = (0.666530 / 2.259281 mol) x 1.00 atm = 0.295019 atm

# Given-

I. Final partial pressure of NO(g) is 20% greater than final partial pressure of H₂ gas.

            II. The vessel is closed but expandable. So, the pressure and temperature of the vessel must remain constant at 1.00 atm and 433.15K during the reaction.

Now,

            Let the final partial pressure of H₂ = X atm = Change in partial pressure of H₂ (because there is no H₂ initially present in the vessel)

            So, final partial pressure of NO = X atm + 20.0 % of X atm = 1.2X atm

# In the balanced reaction, formation of 5 mol H₂ is associated with formation of 2 mol NO. Moles of a gas is directly proportional to its partial pressure.

Now,

Change in pNO = (2/5) x change in partial pressure of H₂ = (2/5) X atm= 0.4X atm

            Final pNO = Initial pNO + Change pNO = 0.295019 atm + 0.4X atm

So far we have, final partial pressure of NO = 1.2X atm

                        Final pNO = 0.295019 atm + 0.4X atm

Since both the values are final partial pressure of same NO, they must be equal.

            Or, 1.2X atm = 0.295019 atm + 0.4X atm

            Or, 1.2X – 0.4X = 0.295019

            Or, X = 0.295019 / 0.8

            Hence, X = 0.36877375

Therefore, change in pH₂ = X atm = 0.36877375 atm

# Calculate final partial pressure of all gases –

Final pH₂ = 0.36877375 atm

Final pNO = 1.2X atm = 1.2 x (0.36877375) atm = 0.4425285 atm

Final pNH₃ = Initial pNH₃ – Change pNH₃

                        = Initial pNH₃ – (2/5) x change pH₂

                        = 0.311876 atm – 0.4 (0.36877375 atm)

                        = 0.1643665 atm

Final pH₂O = 0.393105 atm – 0.4 (0.36877375 atm) = 0.2455955 atm

# Calculate theoretical total pressure in vessel after reaction (assuming volume has not expanded yet):

Total pressure in vessel = 0.36877375 atm + 0.4425285 atm + 0.1643665 atm + 0.2455955 atm = 1.22126425 atm

# So far, we have-

            Initial pressure = 1.00 atm                          ; Initial volume = 80.344 L

            Final pressure = 1.22126425 atm             ; final volume = ?

Given that the vessel is expandable, the pressure remains constant at 1.0 atm that causes volume expansion.

Now (noting that volume is directly proportional to total pressure)

            Final volume = (Initial volume / Initial pressure) x Final Pressure

                                    = (80.344 L / 1.00 atm) x 1.22126425 atm           

                                    = 98.121 L

Therefore, final volume of expandable reaction vessel = 98.1 L