Chemistry, 30.03.2021 01:00 kamkam1713

A gas tank contains 2.5 moles of oxygen, 7.0 moles of nitrogen and 0.5 mole of water vapor. The total pressure of the mixture is 852 torr. Calculate the partial pressure of each gas. Show work.

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Answer from: OliTepley8032

The partial pressure of each gas is:

Oxygen= 213 torrNitrogen= 596.4 torrWater Vapor= 42.6 torr

Explanation:

The pressure exerted by a particular gas in a mixture is known as its partial pressure. So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

PT = PA + PB

This relationship is due to the assumption that there are no attractive forces between the gases.

Dalton's partial pressure law can also be expressed in terms of the mole fraction of the gas in the mixture. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of a component to the number of moles of all the components present.

The mole fraction of a gas in a gas mixture is given by:

XA = nA / nT

This fraction will always be less than 1.

So in a mixture of two or more gases, the partial pressure of gas A can be expressed as:

PA = XA * PT

In this case, you know:

number of moles of oxygen= 2.5 molesnumber of moles of nitrogen= 7.0 molesnumber of moles of water vapor= 0.5 moles

So, the number of total moles nT is calculated as:

nT= number of moles of oxygen + number of moles of nitrogen +number of moles of water vapor

Then:

nT= 2.5 moles + 7.0 moles + 0.5 moles

nT= 10

So, the mole fraction of the gas in the mixture is:

mole fraction of oxygen = $\frac{2.5 moles}{10 moles}$ = 0.25mole fraction of nitrogen = $\frac{7 moles}{10 moles}$ = 0.7mole fraction of water vapor = $\frac{0.5 moles}{10 moles}$ = 0.05

Being the total presurre (PT) equal to 852 torr, then:

Poxygen= 0.25*852 torr= 213 torrPnitrogen= 0.7*852 torr= 596.4 torrPwater vapor= 0.05*852 torr= 42.6 torr

The partial pressure of each gas is:

Oxygen= 213 torrNitrogen= 596.4 torrWater Vapor= 42.6 torr

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

Woah that’s a lot to read xd i’ll try though. i’ll get back t you if. i know the andwer

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A gas tank contains 2.5 moles of oxygen, 7.0 moles of nitrogen and 0.5 mole of water vapor. The tota...

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