Question Video: Calculating the Equilibrium Constant for Partial Pressures Given the Total Pressure and the Percent Dissociation | Nagwa Question Video: Calculating the Equilibrium Constant for Partial Pressures Given the Total Pressure and the Percent Dissociation | Nagwa

# Question Video: Calculating the Equilibrium Constant for Partial Pressures Given the Total Pressure and the Percent Dissociation Chemistry • Third Year of Secondary School

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A sealed flask at a pressure of 1.50 atm contains dinitrogen tetroxide. The gas is found to have dissociated by 20.00% at 298 K, as shown: N₂O₄ ⇌ 2NO₂. What is the value of the equilibrium constant, 𝐾_𝑝, at 298 K? Give your answer to two decimal places.

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### 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.

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