A. Consider the following neutral electron configurations in which n has a constant value. Which configuration would belong to t
1 answer:
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<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar
<u>Explanation:</u>
To calculate the number of moles, we use the equation given by ideal gas which follows:
where,
P = pressure of the gas = 5.57 bar
V = Volume of the gas = 2.19 L
T = Temperature of the gas = 298 K
R = Gas constant =
n = Total number of moles = ?
Putting values in above equation, we get:
To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:
........(1)
where,
= partial pressure of carbon dioxide = 0.318 bar
= total pressure = 5.57 bar
= mole fraction of carbon dioxide = ?
Putting values in above equation, we get:
- Mole fraction of a substance is given by:
We are given:
Moles of nitrogen gas = 0.221 moles
Mole fraction of nitrogen gas,
Calculating the partial pressure of oxygen gas by using equation 1, we get:
Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949
Total pressure of the system = 5.57 bar
Putting values in equation 1, we get:
Hence, the partial pressure of oxygen gas is 2.76 bar
Explanation:
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
For the given equation:
A. C is added to the reaction mixture.
If the concentration of reactant specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of reactant specie occurs. So, on increasing C ,equilibrium will shift in right direction.
B. is condensed and removed from the reaction mixture.
If the concentration of reactant specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of reactant specie occurs. So, on removing water vapor ,the equilibrium will shift in left direction.
C. CO is added to the reaction mixture.
If the concentration of product specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of product specie occurs. So, on increasing CO, the equilibrium will shift in left direction.
D. is removed from the reaction mixture.
If the concentration of product specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of product specie occurs. So, on removing hydrogen , the equilibrium will shift in right direction.