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Heptane gas reacts with oxygen gas to give carbon dioxide gas and water vapor (gas). If you mix heptane and oxygen in the correct stoichiometric ratio, and if the total pressure of the mixture is 300 mm Hg, what are the partial pressures of heptane ( 25 mmHg) and oxygen ( 275 mm Hg)? If the temperature and volume do not change, what is the pressure of the water vapor ( 200 mm Hg) after reaction?

Answer :

Answer:

Explanation:

Total pressure of the mixture = 300 mm Hg

equation of reaction

C₇ H₁₆(g) + 11 O₂ (g) →  7 CO₂(g) + 8 H₂O(g)

partial pressure of heptane = mole fraction heptane × total pressure = 1 / 12 × 300 mm Hg = 25 mm Hg

partial pressure of oxygen = mole fraction oxygen × total pressure = 11 / 12 × 300 mm Hg = 275 mm Hg

After the reaction

total number of mole before the reaction = 12

total number of mole after the reaction = 15

temperature and volume did not change

if 12  to 300 mm Hg

15 will be 15 × 300 / 12 = 375 mm Hg

partial pressure of water vapor = mole fraction of water vapor × 375 mm Hg = 8 / 15 × 375 mm Hg = 200 mm Hg

The pressure of the water vapor on the solution is 200mm Hg. The pressure exerted on the solution by the vapor is known as vapor pressure.

What is vapor pressure?

The pressure exerted on the solution by the vapor is known as vapor pressure.

[tex]{P_{H_2O} = X_{H_2O}\times P_{sol}[/tex]

Where,

[tex]{P_{H_2O}[/tex] - Vapour pressure

[tex]X_{H_2O}[/tex] - Mole fraction of water = 8/15

[tex]P_{sol}[/tex] - Vapour Pressure of solution = 375 mm Hg

Put the values in the formula,

[tex]{P_{H_2O} = \dfrac 8 { 15} \times 375 { \rm \ mm Hg}\\{P_{H_2O} = 200 { \rm \ mm Hg}[/tex]

Therefore, the pressure of the water vapor on the solution is 200mm Hg.

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