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Anestetic [448]
3 years ago
5

The reaction system

Chemistry
1 answer:
tankabanditka [31]3 years ago
3 0

Answer:

e. As equilibrium is reestablished, the partial pressure of hydrogen decreases.

Explanation:

Hello,

In this case, for the given chemical reaction:

CS_2(g) + 4H_2(g)\rightleftharpoons CH_4(g) + 2H_2S(g)

Considering the Le Chatelier's principle which states that adding more reactant shifts the reaction towards the products, condition that is related with the doubling of the hydrogen partial pressure, we can notice that e. As equilibrium is reestablished, the partial pressure of hydrogen decreases since the addition of hydrogen will be used to form more methane and hydrogen sulfide, for that reason, its amount is decreased to reattain equilibrium.

Regards.

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12. The vapor pressure of water at 90°C is 0.692 atm. What is the vapor pressure (in atm) of a solution made by dissolving 3.68
luda_lava [24]

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of H_2O = 1.00 kg = 1000 g

Moles of CsF = 3.68 mole

Molar mass of H_2O = 18 g/mole

Vapor pressure of water = 0.692 atm

First we have to calculate the moles of H_2O.

\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1000g}{18g/mole}=55.55mole

Now we have to calculate the mole fraction of H_2O

\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CsF}=\frac{55.55}{55.55+3.68}=0.938

Now we have to partial pressure of solution.

According to the Raoult's law,

P_{Solution}=X_{H_2O}\times P^o_{H_2O}

where,

P_{Solution} = vapor pressure of solution

P^o_{H_2O} = vapor pressure of water = 0.692 atm

X_{H_2O} = mole fraction of water = 0.938

P_{Solution}=X_{H_2O}\times P^o_{H_2O}

P_{Solution}=0.938\times 0.692atm

P_{Solution}=0.649atm

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

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