Partial pressure and gas equilibria

In a nutshell

Partial pressure is the pressure each individual gas exerts out of the entire mixture of gases. The total pressure can be calculated by adding all of the partial pressures. Mole fractions can be used to calculate partial pressures. $K_p$ is the equilibrium constant for a gaseous equilibrium.

Equations

word equation	symbol equation
$total\>pressure=sum\>of\>partial\>pressures$	$P=\sum partial \, pressures$
$mole\>fraction\>of\>a\>gas=\dfrac{number \>of\>moles\>of\>gas}{total\>number\>of\>moles\>in\>gas\>mixture}$	$X_A=\dfrac{n_{A}}{n_{total}}$
$partial\>pressure\>of\>a\>gas=mole\>fraction\>of\>a\>gas\>\times\>total\>pressure$	$p_A=X_{A}\times P$
$gaseous\>equilibrium\>constant=\dfrac{(product\>partial\>pressure)^{product\>moles\>ratio}}{(reactant\>partial\>pressure)^{reactant\>moles\>ratio}}$	$K_p=\dfrac{p(products)^n}{p(reactants)^m}$

Variable units

Quantity name	symbol	unit name	unit
$total\>pressure$	$P$	$atmosphere \>or\>pascals$	$atm\>or\>Pa$
$partial \>pressure$	$p_A$	$atmosphere \>or\>pascals$	$atm\>or\>Pa$
$moles$	$n$	$moles$	$mol$
$mole\>fraction$	$X_{i}$	$no\>units$	$no\>units$
$gaseous\>equilibrium\>constant$	$K_p$	$dependent\>on\>reaction$	$dependent\>on\>reaction$

Partial pressure

Partial pressure is the pressure each individual gas exerts out of the entire mixture of gases. The total pressure can be calculated by adding all of the partial pressures.

The partial pressure of an individual gas can be calculated if the total gas pressure and other partial pressures for the other gases are known.

$total\>pressure=sum\>of\>partial\>pressures \\ \ \\ P=\sum partial \, pressures$

Example

$1.50\>mol$ of the gas $PI_5$ is heated, forming $PI_3$ and $I_2$ . The partial pressure of $I_2$ is $0.31\>atm$ in the equilibrium mixture. The total pressure of the gas mixture is $11\>atm$ . What is the partial pressure for $PI_5$ ?

$PI_5(g)⇌PI_3(g)+I_2(g)$

Write the equation for total partial pressure for the given equation:

$P=p(PI_5)+p(PI_3)+p(I_2)$

Work out the partial pressure for $PI_3$ :

$p(PI_3)=p(I_2)=0.31\>atm$

Both have the same number of moles at equilibrium so their partial pressures will also be equal.

Insert known values into total pressure equation:

$11=p(PI_5)+0.31+0.31$ 

Rearrange for $p(PI_5)$ and calculate:

$p(PI_5)=11-0.35-0.35=10.3\>atm$

The partial pressure for $PI_5$ is $\underline{10.3\>atm}$ .

Mole fraction

The mole fraction is the fraction a gas out of the entire gas mixture.

$mole\>fraction\>of\>a\>gas=\dfrac{number \>of\>moles\>of\>gas}{total\>number\>of\>moles\>in\>gas\>mixture} \\ \ \\ X_A=\dfrac{n_{A}}{n_{total}}$

This can be used to calculate the partial pressure of a gas.

$partial\>pressure\>of\>a\>gas=mole\>fraction\>of\>a\>gas\>\times\>total\>pressure \newline p_A=X_{A}\times P$

Calculating the partial pressure of a gas by working out the mole fraction is described below.

Procedure

1.	Calculate the mole fraction of a gas using the equation: $X_A=\dfrac{n_{A}}{n_{total}}$
2.	Insert the value for the mole fraction and total pressure to calculate the partial pressure of the gas: $p_A=X_{A}\times P$

Example

$5.0\>mol$ of $PBr_5$ was heated and decomposed to $PBr_3$ and $Br_2$ . The moles of $PBr_3$ is $1.1\>mol$ at equilibrium. The total pressure of the gas mixture is $10\>atm$ . What is the partial pressure of $PBr_5$ at equilibrium? Round your answer to two significant figures.

$PBr_5(g)⇌PBr_3(g)+Br_2(g)$ 

Work out the moles of $Br_2$ at equilibrium:

$Br_2\>moles=PBr_3\>moles=1.1\>mol$

Their mole ratios are equal so their equilibrium moles will also be equal.

Work out the moles of $PBr_5$ at equilibrium:

$equilibrium\>moles\>of\>PBr_5=initial\>moles\>-\>moles\>reacted\>=5-1.1=3.9\>mol$

Write the mole fraction equation:

 $X_A=\dfrac{n_{A}}{n_{total}}$

Add all the moles together to calculate the total moles of gas in the gas mixture:

$n_{total}=3.9+1.1+1.1=6.1\>mol$

Write the mole fraction equation for $PBr_5$ and insert known values to calculate it:

$X_{PBr_5}=\dfrac{n(PBr_5)}{n_{total}}=\dfrac{3.9\>mol}{6.1\>mol}=\dfrac{39}{61}$

Write the equation linking partial pressure and mole fraction:

$p_A=X_{A}\times P$

Insert known values for $PBr_5$ :

$p(PBr_5)=X_{PBr_5}\times P=(\dfrac{39}{61})\times(10\>atm)=6.3934\>atm$

Round to two significant figures:

$p(PBr_5)=6.4\>atm$

The partial pressure of $PBr_5$ is $\underline{6.4\>atm}$ .

Equilibrium constant, K_p

K_pexpression

The equilibrium constant, $K_p$ , is an equilibrium constant for a gaseous equilibrium. The working out is the same as equilibrium constant, $K_c$ . The only difference is that normal brackets are used instead of square brackets as pressure is being used instead of concentration.

This is the expression for $K_p$ :

$K_p=\dfrac{p(products)^n}{p(reactants)^m}$

$n= number\>of\>product\>moles$

$m=number\>of\>reactant\>moles$

You will need to write expressions for $K_p$ for any given reaction.

Procedure

1.	In the numerator, write the products separately in brackets after the letter p.
2.	Write the number of moles each product has in the power form for each product.
3.	In the denominator, write the reactants separately in brackets after the letter p.
4.	Write the number of moles each reactant has in the power form for each reactant.

Example

Write the expression for $K_p$ for the production of nitrogen dioxide from nitrogen and oxygen.

Write the balanced equation:

$N_2(g)+2O_2(g)⇌2NO_2(g)$

Write the $K_p$ expression:

$\underline{K_p=\dfrac{p(NO_2)^2}{p(N_2)\>p(O_2)^2}}$

Note: Normal brackets are used instead of square brackets.

K_p units

The units for the gaseous equilibrium constant, $K_p$ , can vary depending on the reaction. You need to figure out the units every single time you calculate $K_p$ .

procedure

1.	In the numerator, write the pressure units all to the power of the total product moles.
2.	In the denominator, write the pressure units all to the power of the total reactant moles.
3.	Simplify it fully.

Example

What are the units for $K_p$ for the given reaction? The unit for pressure is $atm$ .

$N_2(g)+2O_2(g)⇌2NO_2(g)$

Write the $K_p$ expression:

$K_p=\dfrac{p(NO_2)^2}{p(N_2)\>p(O_2)^2}$

Insert the pressure units into the equation:

$K_p=\dfrac{(atm)^2}{(atm)(atm)^2}$

Simplify fully:

$K_p=\dfrac{(atm)^2}{(atm)(atm)^2}=\dfrac{(atm)^2}{(atm)^3}=\dfrac{1}{atm}=atm^{-1}$

The units for $K_p$ are $\underline{atm^{-1}}$ .

Calculating K_p

The value of the gaseous equilibrium constant can be calculated with known partial pressures at equilibrium. Sometimes you have to calculate the partial pressures at equilibrium first.

procedure

1.	Write the balanced chemical equation.
2.	If the partial pressure isn't given, calculate the partial pressure for each species.
3.	Write down the $K_p$ expression.
4.	Insert partial pressure values into the $K_p$ equation to calculate $K_p$ .

Example

Calculate the value of $K_p$ when gaseous $PBr_5$ dissociates into the gases, $PBr_3$ and $Br_2$ at $400\>K$ . The partial pressure for $PBr_5$ is $2.6\>atm$ , for $PBr_3$ is $2.0\>atm$ and for $Br_2$ is $1.8\>atm$ .

Write the balanced equation:

$PBr_5(g)⇌PBr_3(g)+Br_2(g)$ 

Write the $K_p$ expression:

$K_p=\dfrac{p(PBr_3)\>p(Br_2)}{p(PBr_5)}$ 

Insert known partial pressure values to calculate $K_p$ :

$K_p=\dfrac{(2.0\>atm)\times(1.8\>atm)}{(2.6\>atm)}=1.38\>atm$

The value of $K_p$ is $\underline{1.38\>atm}$ .

Finding partial pressure with a known K_p

When a value of $K_p$ is given, you can use the equation to solve for an unknown partial pressure.

procedure

1.	Write the $K_p$ expression.
2.	Insert known values.
3.	Rearrange and solve for unknown partial pressure.

K_p and heterogeneous equilibrium

Heterogeneous equilibrium is when all the species aren't in the same physical state. Don't include non-gaseous species in the expression for $K_p$ .

Example

Sulfur and oxygen react to form sulfur dioxide. The chemical equation is given below. Write the expression for $K_p$ .

$S(s)+O_2(g)⇌SO_2(g)$

Write the $K_p$ expression:

$\underline{K_p=\dfrac{p(SO_2)}{p(O_2)}}$

Sulfur isn't included because it's a solid.