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Question

A mixture in which the mole ration of H2 and O2 is 2:1 is used to prepare water by the reaction
2H2(g)+O2(g)2H2O(g)
The total pressure of the container is 0.8 atm at 20 before the reaction. Determine the final pressure at 120 after reaction assuming 80% yield of water.

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Solution

Concept :
The total pressure and partial pressure are proportional to the number
of moles in the system.
Explanation : Total pressure = 0.8 atm at 20C
initial partial pressure of these gases
pp of O2=0.83=0.26atm
pp of H2=0.26×2=0.52 atm
Given:- yield = 80%
therefore, 80% of original molecules left unreacted
Partial pressure of original gases after reaction.
H2=0.52×0.20=0.104atm
O2=0.36×0.20=0.052 atm
Or the Basis of strichiometry, amount of H2O formed will be equal to
the amount of H2 that realized therefore,
0.52×0.80=0.416atm
Total pressure at 20C=
Sum of partial pressure of gases
PT(20C)=0.104+0.052+0.416
=0.572atm
Pressure at 120C:
P(120C)=0.57×3.93293=0.76atm


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