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Properties of Equilibrium Constant

- Chemical Reactions
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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.

- This article is about the
**properties of the equilibrium constant**,**K**._{eq} - We'll consider how K
_{eq}is affected by**temperature**,**concentration**and**pressure.** - We'll then explore what happens to K
_{eq}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**.

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, K**_{eq}.

The **equilibrium constant**,** K _{eq}**, is a value that tells us the

**For a** **certain reaction at a specific temperature****, K _{eq} 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, K

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

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, K _{eq}, is temperature**. Changing the pressure or concentration of a system at equilibrium has

**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 K_{eq}will increase.**Decreasing the temperature favors the exothermic reaction**. If the backward reaction is exothermic, then K_{eq}will decrease.

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.

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

Take the reaction $\mathrm{A}\left(\mathrm{g}\right)+\mathrm{B}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right)$. If we were to write an equation for K_{c} for this reaction (which we'll call K_{c1}), 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 K_{c}, 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 K_{c1} 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 K_{c} for these examples, all of the properties that we're about to explore apply to the equilibrium constant K_{p} 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:

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

This gives us the following expression for K_{c2}:

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

Notice something? The expression for K_{c2} is the reciprocal of the expression for K_{c1}. **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.

Now let's consider what happens if you **multiply the reaction equation by a coefficient**. We've seen above that for the reaction $\mathrm{A}\left(\mathrm{g}\right)+\mathrm{B}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right)$, we get the following expression for K_{c1}:

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

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

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

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 K_{c} 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 K_{c}, 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 K_{eq}. If you times an equation by four, you raise K_{eq} 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 $\mathrm{A}\left(\mathrm{g}\right)+\mathrm{B}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right)$ then react to form two new products, E(g) and F(g). Here are the two reactions and their expressions for K_{c}:

$\mathrm{A}\left(\mathrm{g}\right)+\mathrm{B}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{C}\left(\mathrm{g}\right)+\mathrm{D}\left(\mathrm{g}\right){\mathrm{K}}_{\mathrm{c}1}=\frac{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}{\left[\mathrm{A}\right]\left[\mathrm{B}\right]}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{C}\left(\mathrm{g}\right)\hspace{0.17em}+\mathrm{D}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{E}\left(\mathrm{g}\right)+\hspace{0.17em}\mathrm{F}\left(\mathrm{g}\right){\mathrm{K}}_{\mathrm{c}2}=\frac{\left[\mathrm{E}\right]\left[\mathrm{F}\right]}{\left[\mathrm{C}\right]\left[\mathrm{D}\right]}$

We can write this as one overall equation, with its own respective expression for K_{c}:

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

What can you see? The expression for K_{c3} is simply the product of the expressions for K_{c1} and K_{c2}:

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

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.

To help consolidate your learning, we've created a handy table summarizing the 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 K_{c} for the reaction $2{\mathrm{CO}}_{2}+8{\mathrm{H}}_{2}\rightleftharpoons 2{\mathrm{CH}}_{4}+4{\mathrm{H}}_{2}\mathrm{O}$:

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

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, K_{c3}, equals the product of K_{c1} and K_{c2}:

$3){\mathrm{CH}}_{4}+2{\mathrm{H}}_{2}\mathrm{O}+\overline{)\mathrm{CO}}\rightleftharpoons \overline{)\mathrm{CO}}+{\mathrm{CO}}_{2}+\hspace{0.17em}4{\mathrm{H}}_{2}{\mathrm{K}}_{\mathrm{c}3}=6.5\times 0.12\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Overall}:\phantom{\rule{0ex}{0ex}}3){\mathrm{CH}}_{4}+2{\mathrm{H}}_{2}\mathrm{O}\rightleftharpoons {\mathrm{CO}}_{2}+\hspace{0.17em}4{\mathrm{H}}_{2}{\mathrm{K}}_{\mathrm{c}3}=0.78$

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:

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

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:

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

This is our final answer.

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 K_{eq} 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**.

- The
**equilibrium constant, K**_{eq}, is constant for a**certain reaction at a specific temperature**.**Changing the temperature or the reaction equation changes the value of K**_{eq}. **Changing the external conditions**of an equilibrium system has different effects on K_{eq}:**Changing the pressure**has no effect on K_{eq}.**Changing the concentration**also has no effect on K_{eq}.**Adding a catalyst**has no effect on K_{eq}.**Changing the temperature***does*affect K_{eq}.

- K
_{eq }is only constant*for a particular reaction*This means that**.****changing the reaction equation**changes the value of K_{eq}:**Reversing the direction of a reaction**takes the reciprocal of K_{eq}.**Multiplying a reaction by a coefficient**raises K_{eq}to the power of the coefficient.**Adding two reactions to each other**multiples their respective values of K_{eq}together.

_{eq}, whilst multiplying the reaction by a coefficient raises K_{eq} to the power of that coefficient. On the other hand, adding two reactions to each other multiplies their respective values of K_{eq} together.

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