5.22 * 10^-4 mole of a gzs containing H2, O2 & N2 exerted a pressure of 67.4 mm in a certain standard volume. The gas was passed over a hot platinum filament which combined H2 & O2 into H2O which was frozen out. When the gas was returnred to the same volume , the pressure was 14.3 mm. Extra oxyzen was added to increase pressure to 44.3. The combustion was repeated, after which the pressure read 32.9 mm. What was the mole fraction of H2, O2 & N2 in the gas sample?

Asked by pspratimasingh588 | 13th Aug, 2010, 10:41: PM

Expert Answer:

Nitrogen is inert in these reactions. After step 1 you're left with an excess of either H2 or O2, with as much H2O created (and removed) as the limiting reagent permitted. In step 2 you add O2, and create more H2O - that means the limiting reagent in step 1 was O2, and there was H2 left over.

In step 2 you consume p = (44.3mm - 32.9mm) = pO2 + pH2'. The O2 reacted with all the H2 left in the box to form water. (If there was still H2 left and all the O2 was consumed, the final pressure should have been less than 32.9mm, right?) So, the partial pressure of H2 after step 1 is
O2 + 2 H2 → 2 H2O
pO2 + pH2' = pH2' (1 mol O2 / 2 mol H2) + pH2' = 1.5 pH2' = 11.4mm
pH2' = 7.6mm

There was no O2 left after step 1, so the partial pressure of N2 must be
pN2 + pH2' = 14.3mm
pN2 = 14.3mm - 7.6mm = 6.7mm

Reduce the original pressure by that of the gases left after step 1, and all you have are the pressures of the reactants in step 1:
pO2 + pH2* = 67.4mm - (pN2 + pH2') = 67.4mm - 14.3mm = 53.1mm
pO2 + pO2 (2 mol H2 / 1 mol O2) = 3 pO2 = 53.1mm
pO2 = 17.7mm

So, the original composition was 17.7mm O2, 6.7mm N2, (2*17.7+7.6)mm H2, and the mole fractions are 17.7/67.4, 6.7/67.4, and 43.0/67.4, respectively

Answered by  | 14th Aug, 2010, 10:05: PM

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