Question

A flask is filled with 4.0 atm of gaseous H₂ and 3.0 atm of gaseous N₂. When a lit match is thrown into the flask, the gases react and form gaseous NH₃. What is the pressure in the flask after the reaction?

Answer: 4.33 atm.

Please provide step by step on how to solve.

Answer 1

Ans. Part A: Let the volume of the flask be V liters, and the reaction temperature be T kelvin.

Using ideal gas equation Ideal gas Law: pV = nRT           - equation 1

            Where, p = pressure in atm

            V = volume in L

            n = number of moles

            R = universal gas constant= 0.082057338 atm L mol^-1K^-1

            T = absolute temperature

Number of moles of H₂ gas, n1 = pV / RT = (4.0 atm.) V / RT

Number of moles of N₂ gas, n2 = pV / RT = (3.0 atm.) V / RT

Balanced reaction: 3 H₂ + N₂ -------> 2 NH₃

Stoichiometry: 3 moles of H₂ reacts with 1 mol of N₂ to produce 2 moles NH₃ gas.

Number of moles if directly proportional to pressure. 3 moles H₂ is required per 1 mol N₂ (H2 : N2 :: 3:1) ; while the pressure ration is 4:3. So, H₂ is the limiting reactant.

Part B: 3 moles of H₂ produces 2 moles NH₃. So, total moles of NH₃ produced from given amount of H₂ gas (limiting reagent)-

            Moles of NH₃ produced = (2/3) x [(4.0 atm.) V / RT]                   - statement 1

3 moles of H₂ reacts with 1 mol N₂ gas. So, total moles of N₂ gas consumed in the reaction-

            Moles of N₂ consumed = (1/3) x [(4.0 atm.) V / RT]

Remaining moles of N₂ = Initial moles of N₂ – moles of N₂ consumed during reaction

                                    = [(3.0 atm.) V / RT ] - (1/3) x [(4.0 atm.) V / RT]

                                    = (5 atm. V) / 3 RT                                           - Statement 2

Total moles of gases at the end of reaction-

                        Moles of NH₃ produced + moles of N₂ remaining            ; [note: all H₂ consumed]

                        = statement 1 + statement 2

                        = (2/3) x [(4.0 atm.) V / RT] + (5 atm. V) / 3 RT

                        = (8.0 atm. V / 3RT) + (5 atm. V / 3 RT)

                        = (13.0 atm. V / 3RT)                                                      - statement 3

Thus, total number of moles of gases at the end of the reaction = (8.0 atm. P / 3RT) moles

Pressure exerted by (13.0 atm. P / 3RT) moles-

                        P = nRT / V = [(13.0 atm. V / 3RT) x RT] / V

                        Or, P = 13.0 atm. / 3 = 4.33 atm.

Thus, pressure at the end of the reaction = 4.33 atm