Question

30.00mg of phosphorus (P4) is added to a 2.00Lcontaining only molecular oxygen, PO2= 352 mmHg, and Argon, PAr= 489 mmH, at 80.2 degrees C.

A) Calculate amount of heat consumed or produced if only Phosphorus and oxygen react according to the following reaction: P4(s)+5O2(g) --->P4O10(s) ΔH= -2940 kJ

B) Mole fractions and partial pressures of remaining gases?

Answer 1

A)

0.7115kJ of heat is produced

Explanation

P4(s) + 5O2(g) --------->P4O10(s) ∆H= -2940KJ

stoichiometrically ,1mole of P4 react with 5 mole of O2

No of mole of P4 = 0.030g/123.892g = 0.000242

No of mole of O2 is obtained from ideal gas law

n = PV/RT

n = 0.463atm× 2L /0.08257(L atm /mol K) × 353.35K

= 0.0319

P4 is limiting

1mole of P4 produce 2940kJ of heat

Therefore,

0.000242 mole of P4 produce = 0.000242 × 2940kJ = 0.7115kJ

B)

mole fraction of Ar = 0.5912

mole fraction of O2 = 0.4088

partial pressire of Ar =0.6434atm

partial pressure of O2 =0.4445atm

Explanation

mole of Ar , n = 0.6434atm × 2L/0.082057(L atm/mol K) × 353.35K

= 0.04438

Remaining mole of O2 = 0.0319 - (5×0.000242) = 0.03069

Total mole gas = 0.04438 + 0.03069 = 0.07507mol

mole fraction of Ar = 0.04438/0.07507 =0.5912

mole fraction of O2 = 0.03069/0.07507 = 0.4088

Total pressure, P = nRT/V

= 0.07507 × 0.082057(L atm / mol K) ×353.35K/2L

= 1.0883atm

Partial pressure of Ar = 0.5912 × 1.0883atm = 0.6434atm

Partial pressure of O2 = 0.4088 ×1.0883atm = 0.4445atm