Video Transcript
A sealed flask at a pressure of
1.50 atmospheres contains dinitrogen tetroxide. The gas is found to have
dissociated by 20.00 percent at 298 kelvin, as shown: N2O4 is in equilibrium with
two NO2. What is the value of the
equilibrium constant, 𝐾 𝑝, at 298 kelvin? Give your answer to two decimal
places.
𝐾 𝑝 is the equilibrium constant
for partial pressures. The equilibrium constant for
partial pressures is the ratio between the partial pressures of the products and
reactants at equilibrium.
Let’s consider a generic reaction
equation where the lowercase letters represent stoichiometric coefficients and the
uppercase letters represent chemical formulas. The 𝐾 𝑝 of this reaction can be
calculated by dividing the partial pressures of the products by the partial
pressures of the reactants, where each individual partial pressure is raised to the
power of the respective stoichiometric coefficient. We can apply this understanding to
the provided chemical reaction to construct an equation for the equilibrium constant
for partial pressures. We find that 𝐾 𝑝 is equal to the
partial pressure of NO2 squared divided by the partial pressure of N2O4.
We now have an equation for the
equilibrium constant. But the question does not provide
us with the partial pressures, so we’ll need to calculate these first.
Partial pressures are the pressures
of individual gas components in a mixture. To calculate the partial pressure
of a gas, we first need to know the mole fraction. The mole fraction is the amount of
substance in moles divided by the total amount of all substances in the mixture also
in moles. In this question, we haven’t been
given the amount of moles of each species either. So before we can calculate the mole
fractions and subsequently the partial pressures, we’ll need to calculate the number
of moles of each species at equilibrium.
To help us, we can use the ICE
method. In this method, we’ll record the
initial number moles of each species involved in the reaction, the change in the
number of moles, and the number of moles at equilibrium. Initially, we know that there will
be zero moles of the product. But we don’t know how many moles of
dinitrogen tetroxide were initially added to the flask. So let’s assume that we initially
have one mole of reactant. In order to reach equilibrium, the
number of moles of reactant will decrease by some amount that we can denote as
𝑥. Simultaneously, the number of moles
of product will increase. We know from the balanced reaction
equation that one mole of reactant will produce two moles of product. So the amount of product should
increase by two times 𝑥.
We are told in the question that
the dinitrogen tetroxide is found to have dissociated by 20 percent. 20 percent of one mole is 0.2
moles. This is the value of 𝑥. This means that the amount of
product will increase by two times 0.2 moles, which is 0.4 moles. If we add the change in amount to
the initial amount, we can calculate the equilibrium amount. So at equilibrium, there are 0.8
moles of N2O4 and 0.4 moles of NO2 for a total of 1.2 moles of gas in the flask.
Now we can calculate the mole
fraction of each species. The mole fraction of N2O4 is equal
to the number of moles of N2O4 divided by the total number of moles. So the mole fraction of dinitrogen
tetroxide is equal to 0.8 moles divided by 1.2 moles or 0.667. Repeating this same process for
NO2, we get a mole fraction of 0.333. Mole fractions are related to
partial pressures in that the partial pressure of a gas is equal to the mole
fraction of that gas times the total pressure of the system.
We are told in the question that
the total pressure of the flask is 1.50 atmospheres. So the partial pressure of N2O4 is
equal to the mole fraction of N2O4 times 1.5 atmospheres. This gives us a partial pressure of
one atmosphere. Repeating this process for NO2, we
find that the partial pressure of NO2 is 0.5 atmospheres.
Now that we know the partial
pressures, we can substitute them into the equation for 𝐾 𝑝. 0.5 atmospheres squared is equal to
0.25 squared atmospheres. One of the atmosphere units in the
numerator and the atmosphere unit in the denominator will cancel. Completing the calculation, we find
that 𝐾 𝑝 is equal to 0.25 atmospheres. This value is already given to two
decimal places.
So, we have determined that the
equilibrium constant for partial pressures for the given reaction is 0.25
atmospheres.
