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Byju's Answer

Standard XII

Chemistry

Ideal Gas Equation

A mixture in ...

Question

A mixture in which the mole ration of $H_{2}$ and $O_{2}$ is 2:1 is used to prepare water by the reaction
${2 H}_{2} (g) + O_{2} (g) ⟶ {2 H}_{2} O (g)$
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 $20^{\circ} C$
initial partial pressure of these gases
pp of $O_{2} = \frac{0.8}{3} = 0.26 a t m$
pp of $H_{2} = 0.26 \times 2 = 0.52$ atm
Given:- yield = 80%
therefore, 80% of original molecules left unreacted
$∴$ Partial pressure of original gases after reaction.
$H_{2} = 0.52 \times 0.20 = 0.104 a t m$
$O_{2} = 0.36 \times 0.20 = 0.052$ atm
Or the Basis of strichiometry, amount of $H_{2} O$ formed will be equal to
the amount of $H_{2}$ that realized therefore,
$0.52 \times 0.80 = 0.416 a t m$
Total pressure at $20^{\circ} C =$
Sum of partial pressure of gases
$P T (20^{\circ} C) = 0.104 + 0.052 + 0.416$
$= 0.572 a t m$
Pressure at $120^{\circ} C : -$
$P (120^{\circ} C) = \frac{0.57 \times 3.93}{293} = 0.76 a t m$

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