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# Properties of Equilibrium Constant

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As the name suggests, equilibrium constants are constant for a certain reaction at a specific temperature. However, if you change the reaction equation or change the temperature, you mess with the value of the equilibrium constant. But luckily for us, we don't necessarily need to work out a new equilibrium constant every time we alter a reaction. Instead, we can use some of the properties of the equilibrium constant to work out its new value.

• We'll consider how Keq is affected by temperature, concentration and pressure.
• We'll then explore what happens to Keq when you reverse a reaction, multiply the reaction by a coefficient or add two reactions together.
• After that, we'll look at using the properties of the equilibrium constant by applying our knowledge to real-life reactions.
• Finally, we'll discuss the importance of the equilibrium constant.

## What is the equilibrium constant?

We explored in "Equilibrium Constant" how if you leave a "reversible reaction" in a closed system, it will eventually reach a state of "Dynamic Equilibrium".

At dynamic equilibrium, the rate of the forward reaction equals the rate of the backward reaction and the relative amounts of products and reactants don't change. We can express the ratio between the relative amounts of products and reactants in such a system using the equilibrium constant, Keq.

The equilibrium constant, Keq, is a value that tells us the relative amounts of reactants and products in a system at equilibrium.

For a certain reaction at a specific temperature, Keq is always the same. It doesn't matter how much of the products or how much of the reactants you start with - provided you keep the reaction equation and the temperature the same, Keq won't change. On the other hand, if you alter either the temperature or the reaction equation, Keq will change. Let's now look at how and why that is.

## Discuss some properties of the equilibrium constant

We'll now explore the properties of the equilibrium constant, Keq, and how it responds to changes in the system's conditions or the reaction equation.

### Changing conditions

First up, let's look at the effect of changing a system's conditions on the equilibrium constant. We mentioned this in the article "Equilibrium Constant", but we'll remind ourselves of it now. This section will focus on pressure, concentration, the presence of a catalyst and temperature.

It is quite simple, really - the only external condition that affects the equilibrium constant, Keq, is temperature. Changing the pressure or concentration of a system at equilibrium has no effect on the equilibrium constant. Adding a catalyst doesn't change its value either:

• Neither increasing nor decreasing the pressure of a system at equilibrium has any effect on the equilibrium constant.
• Likewise, neither increasing nor decreasing the concentration of a system at equilibrium has any effect on the equilibrium constant.
• The presence of a catalyst also doesn't affect the equilibrium constant.
• Changing the temperature of a system at equilibrium does change the equilibrium constant.
• Increasing the temperature favors the endothermic reaction. If the forward reaction is endothermic, then Keq will increase.
• Decreasing the temperature favors the exothermic reaction. If the backward reaction is exothermic, then Keq will decrease.

### Changing the reaction equation

Next, let's look at what happens to the equilibrium constant when you change the reaction equation itself. Remember, the equilibrium constant is only constant for a particular reaction. This means that by changing the reaction equation, we've created a new reaction. This new reaction will have its own unique equilibrium constant. However, the equilibrium constant changes in predictable ways, thanks to certain properties.

#### Reversing the equation

We'll first look at what happens when you reverse the reaction equation.

Take the reaction $\mathrm{A}\left(\mathrm{g}\right)+\mathrm{B}\left(\mathrm{g}\right)⇌\mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right)$. If we were to write an equation for Kc for this reaction (which we'll call Kc1), we'd get the following:

${\mathrm{K}}_{\mathrm{c}1}=\frac{{\left[\mathrm{C}\right]}_{eqm}{\left[\mathrm{D}\right]}_{eqm}}{{\left[\mathrm{A}\right]}_{eqm}{\left[\mathrm{B}\right]}_{eqm}}$

Check out "Equilibrium Constant" to find out how to write the expression for Kc, a particular type of equilibrium constant. There, you'll also learn that although equilibrium constant measurements are always taken at equilibrium, we often don't bother writing out the subscript eqm in the expression- the formula looks a lot more simple if you leave it out. We'll therefore omit eqm for the rest of this article. This turns the expression for Kc1 into the following:

${\mathrm{K}}_{\mathrm{c}1}=\frac{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}$

In addition, you should note that while we've used Kc for these examples, all of the properties that we're about to explore apply to the equilibrium constant Kp too.

Let's consider what would happen if we reversed this reaction. Our old products become our new reactants, and our old reactants become our new products:

$\mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right)⇌\mathrm{A}\left(\mathrm{g}\right)+\mathrm{B}\left(\mathrm{g}\right)$

This gives us the following expression for Kc2:

${\mathrm{K}}_{\mathrm{c}2}=\frac{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}$

Notice something? The expression for Kc2 is the reciprocal of the expression for Kc1. The equilibrium constant of a reaction in one direction is the reciprocal of the equilibrium constant for the same reaction in the reverse direction. Or, simply put: when you reverse a reaction, you take the reciprocal of its equilibrium constant.

#### Multiplying by a coefficient

Now let's consider what happens if you multiply the reaction equation by a coefficient. We've seen above that for the reaction $\mathrm{A}\left(\mathrm{g}\right)+\mathrm{B}\left(\mathrm{g}\right)⇌\mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right)$, we get the following expression for Kc1:

${\mathrm{K}}_{\mathrm{c}1}=\frac{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}$

What if we multiplied the entire equation by three? We'd get the following:

$3\mathrm{A}\left(\mathrm{g}\right)+3\mathrm{B}\left(\mathrm{g}\right)⇌3\mathrm{C}\left(\mathrm{g}\right)+3\mathrm{D}\left(\mathrm{g}\right)$

Note that this equation is still balanced - it is simply three times larger in magnitude than the original. But it means that the expression for Kc changes too:

${\mathrm{K}}_{\mathrm{c}2}=\frac{{\left[\mathrm{C}\right]}^{3}{\left[\mathrm{D}\right]}^{3}}{{\left[\mathrm{A}\right]}^{3}{\left[\mathrm{B}\right]}^{3}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{c}2}={\left(\frac{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}\right)}^{3}={\left({\mathrm{K}}_{\mathrm{c}1}\right)}^{3}$

This is the same as our original expression for Kc, but cubed. Multiplying a balanced chemical equation by a coefficient raises the equilibrium constant to the power of that coefficient. If you times an equation by two, you square Keq. If you times an equation by four, you raise Keq to the power of four.

Last of all, let's explore the effect of adding multiple reactions together. Suppose that the products of the reaction $\mathrm{A}\left(\mathrm{g}\right)+\mathrm{B}\left(\mathrm{g}\right)⇌\mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right)$ then react to form two new products, E(g) and F(g). Here are the two reactions and their expressions for Kc:

$\mathrm{A}\left(\mathrm{g}\right)+\mathrm{B}\left(\mathrm{g}\right)⇌\mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right){\mathrm{K}}_{\mathrm{c}1}=\frac{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{C}\left(\mathrm{g}\right) +\mathrm{D}\left(\mathrm{g}\right)⇌\mathrm{E}\left(\mathrm{g}\right)+ \mathrm{F}\left(\mathrm{g}\right){\mathrm{K}}_{\mathrm{c}2}=\frac{\left[\mathrm{E}\right]\left[\mathrm{F}\right]}{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}$

We can write this as one overall equation, with its own respective expression for Kc:

$\mathrm{A}\left(\mathrm{g}\right)+\mathrm{B}\left(\mathrm{g}\right)⇌\mathrm{E}\left(\mathrm{g}\right)+\mathrm{F}\left(\mathrm{g}\right){\mathrm{K}}_{\mathrm{c}3}=\frac{\left[\mathrm{E}\right]\left[\mathrm{F}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}$

What can you see? The expression for Kc3 is simply the product of the expressions for Kc1 and Kc2:

$\frac{\left[\mathrm{E}\right]\left[\mathrm{F}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}=\frac{\overline{)\left[\mathrm{C}\right]\left[\mathrm{D}\right]}}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}×\frac{\left[\mathrm{E}\right]\left[\mathrm{F}\right]}{\overline{)\left[\mathrm{C}\right]\left[\mathrm{D}\right]}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{c}3}={\mathrm{K}}_{\mathrm{c}1}×{\mathrm{K}}_{\mathrm{c}2}$

Therefore, we can deduce that the equilibrium constant for the overall reaction made up of two or more reactions is equal to the product of their individual equilibrium constants. In other words, when you add up individual reactions, you multiply their equilibrium constants together.

## Summary of properties of the equilibrium constant

To help consolidate your learning, we've created a handy table summarizing the properties of the equilibrium constant:

The properties of the equilibrium constant. StudySmarter Originals

## Application of properties of the equilibrium constant

Let's now have a go at calculating the equilibrium constant using what we've just learned about its properties.

Use the following information to work out Kc for the reaction $2{\mathrm{CO}}_{2}+8{\mathrm{H}}_{2}⇌2{\mathrm{CH}}_{4}+4{\mathrm{H}}_{2}\mathrm{O}$:

$1\right){\mathrm{CH}}_{4}+{\mathrm{H}}_{2}\mathrm{O}⇌\mathrm{CO}+3{\mathrm{H}}_{2}{\mathrm{K}}_{\mathrm{c}1}=6.5\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2\right)\mathrm{CO}+{\mathrm{H}}_{2}\mathrm{O}⇌{\mathrm{CO}}_{2}+{\mathrm{H}}_{2}{\mathrm{K}}_{\mathrm{c}2}=0.12$

Well, we have been given two equations. With a bit of manipulation, they can be turned into the desired reaction. First of all, notice that whilst we can see CO in both reaction 1 and reaction 2, it isn't present in the overall reaction. We need to add reactions 1 and 2 together to eliminate CO. Remember that when we add two reactions to each other, we multiply their equilibrium constants together. Therefore, this new reaction's equilibrium constant, Kc3, equals the product of Kc1 and Kc2:

$3\right){\mathrm{CH}}_{4}+2{\mathrm{H}}_{2}\mathrm{O}+\overline{)\mathrm{CO}}⇌\overline{)\mathrm{CO}}+{\mathrm{CO}}_{2}+ 4{\mathrm{H}}_{2}{\mathrm{K}}_{\mathrm{c}3}=6.5×0.12\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Overall}:\phantom{\rule{0ex}{0ex}}3\right){\mathrm{CH}}_{4}+2{\mathrm{H}}_{2}\mathrm{O}⇌{\mathrm{CO}}_{2}+ 4{\mathrm{H}}_{2}{\mathrm{K}}_{\mathrm{c}3}=0.78$

Reaction 3 looks a little closer to our desired reaction. However, the reactants and products are on the wrong sides. We, therefore, need to reverse reaction 3. Remember that when we do this, we take the reciprocal of the equilibrium constant:

$4\right){\mathrm{CO}}_{2}+4{\mathrm{H}}_{2}⇌{\mathrm{CH}}_{4}+2{\mathrm{H}}_{2}\mathrm{O}{\mathrm{K}}_{\mathrm{c}4}=\frac{1}{0.78}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{c}4}=1.28$

We're almost there. The last step is to multiply reaction 4 by two. Remember that this means we need to raise the equilibrium constant to the power of two:

$5\right)2{\mathrm{CO}}_{2}+8{\mathrm{H}}_{2}⇌2{\mathrm{CH}}_{4}+4{\mathrm{H}}_{2}\mathrm{O}{\mathrm{K}}_{\mathrm{c}5}=1.{28}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{c}5}=1.64$

## Importance of the equilibrium constant

The equilibrium constant has a few uses:

• We can compare it to the reaction quotient to find out the direction a reaction is traveling in.
• We can use its magnitude to estimate how far the reaction will go to completion.
• We can use it to calculate the relative amounts of species in a system at equilibrium.

You can learn more about the reaction quotient in the article with the same name, "Reaction Quotient", and practice working with it in "Using the Reaction Quotient". In "Magnitude of Equilibrium Constant", you'll see how the value of Keq relates to the extent of the reaction and the position of the equilibrium. And in "Calculating Equilibrium Concentrations", you'll be able to find equilibrium concentrations using the equilibrium constant.

You've reached the end of this article. By now, you should be able to list the properties of the equilibrium constant, including the effect of changing the system's conditions and altering the reaction equation. You should also be able to apply this knowledge to real-life reactions.

## Properties of Equilibrium Constant - Key takeaways

• The equilibrium constant, Keq, is constant for a certain reaction at a specific temperature. Changing the temperature or the reaction equation changes the value of Keq.
• Changing the external conditions of an equilibrium system has different effects on Keq:
• Changing the pressure has no effect on Keq.
• Changing the concentration also has no effect on Keq.
• Adding a catalyst has no effect on Keq.
• Changing the temperaturedoes affect Keq.
• Keq is only constant for a particular reaction. This means that changing the reaction equation changes the value of Keq:
• Reversing the direction of a reaction takes the reciprocal of Keq.
• Multiplying a reaction by a coefficient raises Keq to the power of the coefficient.
• Adding two reactions to each other multiples their respective values of Keq together.

The equilibrium constant is constant for a certain reaction at a specific temperature. It isn't affected by changes in pressure or concentration, or the presence of a catalyst. However, it is affected by temperature. If you change the reaction equation, you also change the value of the equilibrium constant - check out this article to find out more.

The equilibrium constant is a value that tells us the relative amounts of reactants and products in a system at equilibrium.

At equilibrium, the rate of the forward reaction equals the rate of the backward reaction and the relative amounts of products and reactants don't change.

The equilibrium constant is unaffected by changes in pressure or concentration, or the presence of a catalyst. However, it is affected by temperature. Changing the reaction equation also changes the equilibrium constant. Reversing the equation takes the reciprocal of Keq, whilst multiplying the reaction by a coefficient raises Keq to the power of that coefficient. On the other hand, adding two reactions to each other multiplies their respective values of Keq together.

The equilibrium constant is unaffected by changes in temperature, pressure or the presence of a catalyst, but is affected by temperature. The equilibrium constant also changes when you change the reaction equation, and you can find out how exactly it responds in this article. We can use the equilibrium constant to find out the direction a reaction is travelling, estimate how far a reaction will go to completion, and calculate the relative amounts of species in a system at equilibrium.

## Final Properties of Equilibrium Constant Quiz

Question

What is the equilibrium constant?

The equilibrium constant, Keq, is a value that tells us the relative amounts of reactants and products in a system at equilibrium.

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Question

Which of the following affect the equilibrium constant?

Just temperature

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Question

True or false? Increasing the temperature of a system at equilibrium always increases the value of the equilibrium constant.

False

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Question

True or false? Increasing the concentration of a system at equilibrium has no effect on the equilibrium constant.

True

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Question

Complete the following sentence: Reversing a reaction ____.

Takes the reciprocal of Keq.

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Question

Complete the following sentence: Multiplying a reaction by a coefficient ____.

Multiplies Keq by that coefficient.

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Question

Complete the following sentence: Adding two reactions to each other ____.

Multiplies their respective values of Keq together.

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Question

True or false? Reversing a reaction always makes Keq smaller.

False

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