# Equilibrium Constant Kp

Equilibria are super important in chemistry, and they come in handy in many real-life situations, especially in industry. Understanding how equilibria work, how they can be altered, and what factors affect them can give us an upper hand.

One way to describe an equilibrium is by using an equilibrium constant, and Kp is a perfect example. Kp is an equilibrium constant that uses partial pressures to show the ratio of products to reactants in a reaction at equilibrium. In this article, we're going to focus on Kp in physical chemistry. We'll start by explaining what Kp is and then delve into the details of partial pressures. Next, we'll show you how to calculate Kp, and then we'll explore its properties. Finally, we'll compare Kc and Kp, so you can see how they differ. By the end of this article, you'll have a solid understanding of Kp and how it works in equilibria.

## Equilibrium constant Kp for the reaction

When you let a reversible reaction sit in a sealed container long enough, it will eventually reach equilibrium. This is when the concentrations of the reactants and products remain the same, and the rate of the forward reaction equals that of the backward reaction. What's interesting is that for each specific reversible reaction, you always end up with the same ratio of products to reactants, as long as the external conditions stay the same. Equilibrium constants are used to express this ratio, with Kp being one such constant.

Kp is similar to Kc, but instead of using the concentrations of species in equilibrium, it uses their partial pressures. Kp always has the same value for a specific reversible reaction under specific conditions, allowing us to predict the proportions of reactants and products accurately. However, before we can make these calculations, we need to understand partial pressure.

In a gas equilibrium, the total pressure of the system is the sum of each individual gas's partial pressure. If we have a system with gases A, B, and C, the total pressure is the partial pressure of A, plus the partial pressure of B, plus the partial pressure of C. We represent partial pressure using , just like how we represent molar concentrations using square brackets. For example, if the partial pressure of oxygen gas in equilibrium is 100 kPa, we write it as: .

**Calculating partial pressure**

Calculating the partial pressure of a gas can be done by finding the mole fraction of the gas in the system and multiplying that by the total pressure. Mole fractions represent the number of moles of a species compared to the total number of moles in a system, as a fraction. However, if finding the partial pressure using the total pressure and mole fraction isn't possible, we can use the equation for Kp to express the ratio of products to reactants in a gaseous reversible reaction at equilibrium.

Kp uses the partial pressures of each gaseous species to find this ratio. To find Kp, we first determine the partial pressures of each gas at equilibrium, which are raised to the power of the number of moles in the equation. We then multiply the values for the products and place them in the numerator and do the same for the reactants, placing them in the denominator. All partial pressures are taken at equilibrium when working with Kp.

For example, let's say we have the reaction A + B ⇌ C + D and the partial pressures of each gas at equilibrium are 2 atm, 3 atm, 4 atm, and 5 atm, respectively. We would calculate Kp as follows:

Kp = (4 atm)^1 x (5 atm)^1 / (2 atm)^1 x (3 atm)^1 = 6.67

The units of Kp depend on the overall equation of the reaction, but they are usually expressed in terms of pressure raised to a power. It's important to remember that all partial pressures used in Kp calculations are taken at equilibrium.

**Units of Kp**

To find the units of Kp, you need to use the equation for Kp that you just worked out. You take each term in the equation and insert its units, and then cancel all of the units down. Sulphur dioxide reacts with oxygen to form sulphur trioxide according to the following equation: The total pressure is 54 kPa. Write an equation for the equilibrium constant Kp for this reaction, and calculate its units.

We know our product is sulphur trioxide, SO3. The equation tells us that 2 moles of sulfur trioxide are produced. Therefore, we have in the numerator. This means that we take the partial pressure of sulphur trioxide and raise it to the power of 2. Likewise, our reactants are sulphur dioxide and oxygen, SO2 and O2. The equation tells us that there are 2 moles and 1 mole respectively. Therefore, we have in the numerator. This means that we take the partial pressure of sulphur dioxide and raise it to the power of 2 and then we multiply it by the partial pressure of oxygen. Overall, we get the following equation for Kp:

We now need to calculate the units of Kp. The question gives us the total pressure in kilopascals, kPa, so we would measure partial pressures in kPa too. If we substitute the units of partial pressure into each term in the equation for Kp, we get the following: Cancelling the units down, we are left with just , which equals . These are the units of Kp for this particular reaction.

## How do we calculate Kp?

To calculate Kp for the given reaction of sulphur dioxide and oxygen to form sulphur trioxide, we first found the mole fractions of each gas by dividing the number of moles of each gas by the total number of moles in the system. We then multiplied each mole fraction by the total pressure of the system to find the partial pressure of each gas. Substituting these partial pressures into the equation for Kp, we calculated a value of 81 kPa^-1.

Kp is affected by changes in temperature, which can shift the position of the equilibrium and change the value of Kp. Increasing the temperature favors the endothermic reaction and increases the partial pressure of the products, leading to an increase in Kp. Decreasing the temperature favors the exothermic reaction and decreases the partial pressure of the products, lowering the value of Kp. Le Chatelier's Principle helps to account for the shifted position of the equilibrium.

**Kp and pressure**

Changing the pressure of a gaseous system does not change the value of Kp. However, it does shift the position of the equilibrium in order to counteract the change. Increasing the pressure shifts the position of the equilibrium to favour the side that produces less moles of gas. This decreases the pressure in the system, counteracting the change in conditions.Decreasing the pressure shifts the position of the equilibrium to favour the side that produces the greater number of moles of gas. This increases the pressure in the system, counteracting the change in conditions.

**Kp and concentration**

Changing the concentration of a gaseous system at equilibrium does not change the value of Kp, but it does shift the position of the equilibrium. According to Le Chatelier's Principle, increasing the concentration of the reactants shifts the position of the equilibrium to favor the forward reaction, while increasing the concentration of the products shifts the position of the equilibrium to favor the backward reaction. This helps to counteract the change in conditions and maintain the value of Kp.

**Kp and catalysts**

You are absolutely right! Adding a catalyst to a gaseous system at equilibrium does not affect the value of Kp or shift the position of the equilibrium. Catalysts increase the rate of reaction by lowering the activation energy of both the forward and backward reactions equally, so the equilibrium position remains the same.

When it comes to changing the pressure or concentration of a gaseous system at equilibrium, Le Chatelier's Principle helps to explain how the position of the equilibrium shifts without changing the value of Kp. For example, if the concentration of products is decreased, the equilibrium shifts to favor the forward reaction, producing more products to increase the numerator in the Kp equation and maintain a constant value of Kp. Similarly, if the total pressure of the system is increased, the equilibrium shifts to favor the reaction that produces the fewest moles of gas, increasing the numerator and maintaining a constant value of Kp.

Kc and Kp are both equilibrium constants, but Kp is more useful when working with gas equilibria, as it is calculated using the partial pressures of gaseous species in an equilibrium, while Kc is calculated using the concentrations of aqueous, liquid, or gaseous species in an equilibrium.

You are absolutely correct! Kp is an equilibrium constant based on partial pressures and tells you the ratio of products to reactants in a reaction at equilibrium. Partial pressure is the pressure that a single constituent gas exerts on a closed system and can be determined using the mole fraction of the gas and the total pressure of the system. The expression for Kp depends on the number of moles of reactants and products involved in the reaction.

Changing the temperature of a gaseous equilibrium changes the value of Kp and shifts the position of the equilibrium. On the other hand, changing the concentration or pressure has no effect on the value of Kp but does shift the position of the equilibrium. Additionally, adding a catalyst neither affects the value of Kp nor shifts the position of the equilibrium. Kp is very similar to Kc, except that Kc is calculated using molar concentrations while Kp uses partial pressures.

## Equilibrium Constant Kp

**What is the equilibrium constant Kp of a reaction?**

The equilibrium constant, Kp, is found using the partial pressures of the reactants and products of the reaction.

**How do you find the equilibrium constant Kp?**

To find the equilibrium constant Kp you need to know the partial pressure that each reactant and each product exerts on the reaction mixture. Once you know each partial pressure, you input them into the expression for Kp, which is different for different gas equilibria, to find the value of the constant.

**What are equilibrium constants Kp and Kc?**

The equilibrium constants Kc and Kp are both values that tell you about the rates of the forward and backward reactions involved in a dynamic equilibrium. The difference between Kp and Kc is that Kp is only applicable to gas equilibria.

**Write an expression for the equilibrium constant Kp?**

For the gas equilibrium:aA + bB -> dD + eEWhere all the reactants and products are in their gaseous state, the expression for Kp is:(p(D)d • p(E)e) ÷ (p(A)a • p(B)b)

**What are the properties of Kp?**

The properties of Kp are based on how the value of Kp changes depending on reaction conditions. Kp is only constant at fixed temperatures, but is unaffected by pressure, or the addition of a catalyst.