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Ideal Gas Equation

A mixture in ...

Question

A mixture in which the mole ratio of $H_{2}$ and $O_{2}$ is :1 is used to prepare water by the reaction,
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g)$
The total pressure in the container is 0.8 atm at $20^{o} C$ before the reaction. Determine the final pressure at $120^{o} C$ after reaction assuming 80% yield of water.

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Solution

$P_{H_{2}} = \frac{2}{3} \times 0.8 = 0.533 a t m$

$P_{O_{2}} = \frac{1}{3} \times 0.8 = 0.244 a t m$

$2 H_{2} + O_{2} \to 2 H_{2} O$
$i = 0$ 0.533 0.266 0

After the reaction $\frac{0.533 \times 20}{100} \frac{0.266 \times 20}{100} \frac{0.533 \times 80}{100}$

Total pressure $= 0.1066 + 0.0533 + 0.4264 = 0.5863$ atm

Using Gay-Lussac's law, $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
$\frac{0.5863}{293} = \frac{P_{2}}{393}$

$P_{2} = 0.7864 a t m$

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H_{2}

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{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.

Q. A mixture in which the mole ratio of

H_{2}

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2 H_{2} (g) + O_{2} (g) \to 2 H_{2} (g) O

The total pressure in the container is

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