Lesson Video: The Equilibrium Constant for Partial Pressures | Nagwa Lesson Video: The Equilibrium Constant for Partial Pressures | Nagwa

# Lesson Video: The Equilibrium Constant for Partial Pressures Chemistry • Third Year of Secondary School

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In this video, we will learn how to construct and calculate the equilibrium constant for partial pressures.

18:05

### Video Transcript

In this video, we will learn how to construct and calculate the equilibrium constant for partial pressures.

Letβs start by briefly recapping equilibrium constants. When a reaction, such as this one, is reversible, in equilibrium, in a closed system, then the rate of the forward reaction is the same as the rate of the reverse reaction. If these four substances are solutions or gases in equilibrium, then we can calculate πΎ c, the equilibrium constant.

πΎ c is equal to the concentration of the products multiplied with each other in the numerator divided by the concentrations of the reactants multiplied with each other in the denominator. And the concentration of each species is raised to the power of the stoichiometric coefficient in front of it in the balanced equation. Sometimes, we will see πΎ c written as πΎ or πΎ eq.

Square brackets refer to molar concentration or molarity, which is moles per liter or moles per decimeter cubed. We are perhaps more familiar with this unit used for solution concentrations. But we can use molar concentration for gases as long as we have the moles of each gas and the volume or capacity of the vessel.

Besides molar concentration, there is another unit we can use instead for gases in an equilibrium constant expression. We know that in a closed system all the gases in a gaseous reaction mixture exert pressure on the walls of the container. We say each gas exerts a partial pressure towards the total pressure within the vessel. The particles of gas are in constant motion, and each gas exerts a fraction of the total pressure.

We can therefore express the equilibrium constant in terms of partial pressures, πΎ π. And in the numerator, we have the partial pressure of gas C raised to the power π multiplied by the partial pressure exerted by gas D raised to the power small π divided by, in the denominator, the partial pressure of gas A raised to the power small π multiplied by the partial pressure of gas B raised to the power π.

We still have all the products in the numerator and all the reactants in the denominator. But this time, they are expressed with units for pressure. There are many possible units for gas pressure. But two common ones are atmospheres, atm, and kilopascals, kPa

We now know how to write the πΎ π expression for a gas reaction mixture in equilibrium. This is the general expression. Letβs look at a real-life example now.

Here is an example of a gaseous reaction at equilibrium. 2SO2 gas plus O2 gas reacting reversibly to give 2SO3 gas. The equilibrium constant for partial pressures for this equilibrium would be πΎ π is equal to the partial pressure of the product SO3 raised to the power of two, because there is a coefficient of two in front of SO3, divided by the partial pressures of the reactants SO2 and O2. And the partial pressure of SO2 will be raised to the power of two because SO2 has a coefficient of two in the balanced equation.

We could raise the partial pressure of O2 to the power of one if we choose because its stoichiometric coefficient is one. Or we can just leave that power out. And therefore, for oxygen, it is not necessary to show the parentheses. So I will erase these too. This is the πΎ π expression for this gas equilibrium reaction.

Note that sometimes partial pressure is written with a capital P with the subscript indicating the substance. But sometimes, we will find partial pressure written with a lowercase p. If we knew the partial pressures of these substances in terms of atmosphere units, we would have atmosphere squared divided by atmosphere squared timesed by atmospheres, which would give us a final answer with a unit of per atmospheres.

So, depending on the reaction equation and the stoichiometric coefficients, we might get a different unit for πΎ p. However, πΎ π is usually considered to be dimensionless or unitless, just like other equilibrium constants, which means that in a calculation we can express the numerical answer without units. The size or magnitude of a πΎ π value for a particular reaction at equilibrium has meaning.

We can interpret the magnitude of πΎ π. If πΎ π is larger than one, we know that the numerator value in the πΎ π expression is larger than the denominator, telling us that the products are favored at equilibrium, or the equilibrium sits to the right. If πΎ π is less than one, we know that the numerator value in the πΎ π expression is small relative to the denominator value, telling us that the reactants are favored. In other words, the equilibrium lies to the left-hand side.

Like other equilibrium constants, πΎ π is temperature dependent. Its magnitude is affected by or influenced by temperature. Heating or cooling will shift the equilibrium to the right or to the left, depending on whether the forward reaction is endothermic or exothermic. So, for a specific temperature, we can calculate a πΎ π value using the partial pressures of all the species in the system. However, sometimes we are not given the partial pressures of each gas, but rather the number of moles of each gas at equilibrium. But we can convert from moles to partial pressure and then calculate πΎ π.

If we know the number of moles of each gas of a reaction mixture present at equilibrium in a closed system, then we can work out the total number of moles of gas particles by summing their values, in a similar way, and we have seen before, that the total pressure is equal to the sum of the partial pressures. From these two expressions, we could derive two handy formulas.

We wonβt do the derivations. But the first expression is mole fraction, given the Greek symbol π, of substance A is equal to the number of moles of substance A divided by the total number of moles. And the other expression is the partial pressure of substance A is equal to the mole fraction of A timesed by the total pressure. In other words, the pressure exerted by gas A is a fraction of the total pressure. We could do the same calculation to work out the partial pressure of gas B, C, and D.

In this first formula, because the units of the numerator and denominator are both moles, they will cancel, giving a unitless or dimensionless answer. This answer will be a fraction less than one. These two equations are based on two assumptions that all the particles are the same size and all the particles behave in the same manner.

Now itβs time to practice using these two equations.

6.00 moles of N2 gas and 20.00 moles of H2 were allowed to react at 650 kelvin and 50 atmospheres of pressure. At equilibrium, 4.00 moles of N2 gas had been converted into ammonia according to the following equation: N2 gas plus 3H2 gas reacting reversibly to give 2NH3 gas. Calculate πΎ π for this equilibrium, giving your answer to two decimal places in scientific notation.

We are asked to calculate πΎ π. πΎ π is a special equilibrium constant, which we can use when the substances in a reaction are gases, when the reaction is reversible and at equilibrium, and this can only take place in a closed system. πΎ π is the ratio of the concentrations of the products and reactants but expressed as partial pressures.

In this problem, πΎ π is equal to the partial pressure of the product, ammonia, NH3, raised to the power of two, since ammoniaβs stoichiometric coefficient is two. The denominator will be the partial pressure of the reactant, N2, nitrogen gas, raised to the power of one, since nitrogenβs stoichiometric coefficient is one, multiplied by the partial pressure of the other reactant, H2, which is hydrogen gas, raised to the power of three, since this substance has a stoichiometric coefficient of three.

We have now formulated the πΎ π expression for this reversible reaction. It is not necessary to show powers of one, and so to simplify, letβs remove it. If we choose, we can simplify further by removing these parentheses here. To calculate πΎ π, all we need are the partial pressures of ammonia, nitrogen, and hydrogen. But we are not given these values. So we have to use data that we have been given to first calculate these three partial pressures and then calculate πΎ p.

Letβs clear some space to do the calculation. We can draw a handy table to see what data we have and what is missing. We are told that six moles of nitrogen and 20 moles of hydrogen react. We can assume that initially there is no ammonia. We are also told that at equilibrium four moles of nitrogen have been converted or changed into ammonia.

Because we know the starting number of moles of the reactants and the moles of one of the reactants that has been converted or changed, we can use an ICE table, where I is the initial or starting number of moles, C the changed or converted moles, and E the equilibrium number of moles for each species. We can fill in these initial values. Changed or converted moles are the number of moles of each substance that reacted or were produced. We know that 4.00 moles of nitrogen are converted to ammonia. But how many moles of hydrogen reacted and how many moles of ammonia were produced overall? We need to use the mole ratio one as to three as to two to calculate these values.

Since the ratio of nitrogen to ammonia is one as to two, then if 4.00 moles of nitrogen reacted or converted or changed, then 8.00 moles of ammonia must have been produced. In a similar way, if the moles of nitrogen to hydrogen is one as to three, then the moles of hydrogen that changed must have been three times that of the moles of nitrogen which were changed, which gives 12.00 moles of hydrogen that reacted.

The third line on the ICE table, equilibrium moles, is the number of moles of each species present at equilibrium. Initially, there were 6.00 moles of nitrogen. 4.00 moles of this were changed or reacted. And so the difference, 2.00 moles, must be the moles of nitrogen present at equilibrium. And this is the number of moles of nitrogen that did not react. Doing the same calculation for the other reactant, we get 8.00 moles of hydrogen present at equilibrium or unreacted. Initially, there was no ammonia in the system. At the end of the reaction, 8.00 moles of ammonia had been produced. This means at equilibrium there were 8.00 moles of ammonia present. Can you see that for the product we are doing an addition and not a subtraction?

We are now going to use the equilibrium number of moles of each species to calculate the partial pressure for each gas. But this is a two-step process. First, we need to calculate the mole fraction of each substance and, from this, the partial pressure of each substance. Then, we can calculate πΎ π. To calculate the mole fraction of a substance, letβs call it π΄, we need to know the number of moles of that substance at equilibrium divided by the total number of moles of all the substances.

If we take the sum of these three values, we will calculate the total number of moles of particles in the system at equilibrium. Performing this sum, we get a total of 18.00 moles. Therefore, the denominator value for the mole fraction calculation for each substance is 18.00 moles. For nitrogen, the numerator value is 2.00 moles, and for hydrogen and ammonia, 8.00 moles. Solving for each, we get 0.111 recurring for nitrogen, 0.444 recurring for hydrogen, and 0.444 recurring for ammonia. Letβs keep too three significant figures for simplicity.

Note that mole units will cancel, and mole fraction is a unitless or dimensionless quantity. These values are the fraction of the total number of moles present at equilibrium. There is about 11.1 percent of nitrogen present at equilibrium and about 44.4 percent hydrogen and 44.4 percent ammonia.

Now we can use the equation the partial pressure of a substance, letβs call it π΄, is equal to the mole fraction of that substance multiplied by the total pressure. We know the mole fraction of each substance at equilibrium. And we are told the total pressure is 50 atmospheres. So nitrogenβs partial pressure is equal to 0.111, the mole fraction, multiplied by 50 atmospheres, giving us 5.55 atmospheres, the partial pressure of nitrogen. This value means that of the total pressure of 50 atmospheres, 5.55 atmospheres comes from nitrogen. We can do a similar calculation for hydrogen and ammonia. And we get a partial pressure for each of 22.2 atmospheres.

Finally, we are ready to calculate πΎ π. We can substitute the partial pressures for each substance into the πΎ π expression. And we get a πΎ π value for this reaction in scientific notation to two decimal places, which is what we were asked, of 8.12 times 10 to the negative three. We have not included units because πΎ π is normally a dimensionless or unitless quantity.

Letβs wrap up what weβve learnt about πΎ π. For a gas equilibrium reaction in a closed system, the equilibrium constant for partial pressures is given by this expression. πΎ π is equal to the partial pressures of the products C and D multiplied by each other divided by the partial pressures of the reactants A and B multiplied by each other and each value raised to the power of the stoichiometric coefficient of that species from the chemical equation.

We learnt that πΎ π is usually dimensionless and is temperature dependent. We learnt that the mole fraction of a substance at equilibrium is equal to the number of moles of that substance divided by the total number of moles and that mole fraction can be used with total pressure to work out the partial pressure of a particular substance.

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