Question

A gas cylinder is filled with 211 torr of N2 and 615 torr of F2 and the resulting reaction forms NF3 at constant temperature and volume. N2(g) + 3 F2(g) ⎯→ 2 NF3 (g). What is the total pressure, and what are the Mole Fractions of each gas, after the reaction? (Hint: think ICE) Total Pressure at the end of Rxn: _______________ Mole Fraction of N2 = ___________ F2 = __________ NF3 = ____________

Answer 1

At constant T and V, assuming ideal gas behavior: PV= nRT; Pressure (P_i) is directly proportional to number of moles (n_i), for each gas.

N₂(g) + 3 F₂(g) 2 NF₃ (g)

As per the stoichiometry of the reaction, 1 mole of N₂ require 3 mole F₂ to produce 2 mole of NF3.

Or, for Pressure P of N₂, require 3P pressure of F₂ gas is required.

Given pressure of N₂ = 211 torr, so 3 x 211 torr = 633 torr pressure of F₂ gas is required.

However, only 615 torr is present. It implies F₂ gas is the limiting reagent. It would react completely with (615/3) = 205 torr of N₂ gas and the product would be formed according to it.

Pressure of NF₃ formed = (2/3) x 615 torr = 410 torr

(Since 1 mole F₂ produces 2 mole NF₃ or 1 mole produces (2/3 ) mole of NF₃ )

Pressure of N₂ gas remaining = (211 -205) torr = 6 torr

So, according to Dalton's law of partial pressures: P_total = P_N2 + P_{F2 +} P_NF3

                                                                                                               = (6 + 0 + 410 ) torr = 416 Torr

Mole fraction of gas () = Partial pressure of gas (P_i) / P_total

So, = 410 / 416 = 0.986

      = 6/ 416 = 0.014

         = 0

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