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Equilibrium Concentrations

- Chemical Reactions
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- Weak Acid and Base Equilibria
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In **Calculating Equilibrium Constant**, we learned how you can use** **initial, equilibrium, and change in concentration to **calculate the equilibrium constant** for a particular reaction. But this also works the other way. If we know both initial concentrations and the equilibrium constant, we can **calculate the equilibrium concentration** of each species involved in the reaction.

- This article is about
**calculating equilibrium concentrations**in chemistry. - We'll start by introducing you to the
**formulae**and**units**you'll use when calculating equilibrium concentrations. - We'll then look at how you can use
**initial concentrations**and the**equilibrium constant**to work out equilibrium concentrations of species in a reversible reaction. - This will involve
**forming and solving a quadratic equation**. You'll be able to practice your skills with the help of our

**worked examples**.

To calculate equilibrium concentrations, we need to know two things:

The

**value of the****equilibrium constant****K**_{c }for this particular reaction.The

**formula for the equilibrium constant K**_{c}.The

**initial concentration**of each of the species involved in the expression for K_{c}. In a homogeneous reaction, this means all of the species in the reaction.It might be helpful to know the

**quadratic formula**too.

The **formula for the equilibrium constant K**_{c} can be derived from the reaction equation. For example, take the reaction \(aA(g)+bB(g) \rightleftharpoons cC(g)+dD(g)\) . To find K_{c}, we'd use the following expression:

$$K_c=\frac{{[C]_{eqm}}^c\space {[D]_{eqm}}^d}{{[A]_{eqm}}^a\space {[B]_{eqm}}^b}\qquad K_p=\frac{{{(P_C)}_{eqm}}^c\space {{(P_D)}_{eqm}}^d}{{{(P_A)}_{eqm}}^a\space {{(P_B)}_{eqm}}^b}$$

Check out **Equilibrium**** Constant** for a reminder on the exact method of writing K_{c}. You'll also learn about how we write formulae for K_{c} for heterogeneous reactions.

Calculating equilibrium concentrations also involves solving a quadratic equation. Most scientific calculators can solve these for you, but you might instead want to use the **quadratic formula**. For the equation \(ax^2+bx+c=0\) , the quadratic formula looks a little something like this:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

It is typical for concentration to be expressed in M, which means moles per liter. This is equivalent to moles per decimeter cubed, or mol dm^{-3}. Don't be surprised if you see it written in this alternative form!

Now that we've discussed the formulae and units used in calculating equilibrium concentrations, we're ready to move on to the method. But before we go any further, let's remind ourselves how you calculate the equilibrium constant from initial, equilibrium, and change in concentration. This will help us understand how we can adapt the process to work out equilibrium concentration from the initial concentration and the equilibrium constant:

- Use the
**initial**and**equilibrium concentration**of a species that you do know to calculate its**change in concentration**. - Use the
**balanced chemical equation**to calculate the**change in concentration**and thus the**equilibrium concentration**of the remaining species involved in the reaction. - Substitute the equilibrium concentration values into the expression for the equilibrium constant to get your final answer.

So, how does this differ from calculating equilibrium concentrations? In the case of the latter, we don't know the equilibrium concentration or the change in concentration of any of the species- we only know their *initial* concentrations. However, we can represent the change in concentration of each of the species involved using multiples of an arbitrary letter, such as x. We then show the equilibrium concentration using the initial concentration and the change in x.

When we substitute the equilibrium concentrations into the formula for the equilibrium constant K_{c}, we end up with a quadratic equation. We can solve this to find x. Remember that x represents the change in concentration - and so we can therefore calculate the equilibrium concentration of each species using their initial concentrations and our value for x.

The process can seem a little tricky, but once you've got your head around it, it isn't too complicated. We'll tell you the steps and then walk through them with two worked examples:

- Create a table with rows for the
**initial**,**change in**, and**equilibrium concentrations**of the reactants and products. Fill in the initial concentrations of all of the species involved. - Use the
**balanced chemical equation**to represent the**change in concentration**of each species in terms of x. - Write the
**equilibrium concentration**of each species using its initial concentration and its change in x. - Substitute the equilibrium concentrations into the expression for K
_{c}. You should end up with a fraction containing factors of x on one side, and a constant on the other. - Multiply the denominator of the fraction through both sides of the equation and rearrange to find a
**quadratic equation**. - Solve the equation using your calculator or the
**quadratic formula**to find two values for x. Discard the inappropriate value. - Go back to your table of initial, change in, and equilibrium concentrations of the reactants and products, and use your value of x to calculate the
**equilibrium concentration**of each species.

Calculating equilibrium concentrations doesn't necessarily just involve solving a quadratic equation. There could be higher powers of x in your equation - it all depends on the number of species involved in the reaction and their coefficients in the balanced chemical equation. However, for your AP exam, you only need to know how to calculate equilibrium concentrations using equations featuring multiples of x and x^{2}.

Ready to give it a go? Here are two problems for you to try your hand at.

**2.00 M CO _{2} and 2.00 M H_{2} are mixed in a sealed container and the system is left to reach equilibrium. Use the following information to calculate the equilibrium concentrations of each species involved in the reaction:**

$$CO_2(g)+H_2(g)\rightleftharpoons CO(g)+H_2O(g)\qquad K_c=3.00$$

Right - let's start by making a table showing the initial, change in, and equilibrium concentration of each species:

Species | CO_{2} + H_{2} ⇌ CO + H_{2}O | ||||

Concentration (M) | Initial | 2.00 | 2.00 | 0.00 | 0.00 |

Change | |||||

Equilibrium |

We've started with just CO_{2} and H_{2}. We know that they will react to form CO and H_{2}O. Looking at the balanced chemical equation, we can see that the molar ratio of CO_{2}:H_{2}:CO:H_{2}O is 1:1:1:1. For every mole of CO_{2} that reacts, one mole of H_{2} also reacts, forming one mole each of both CO and H_{2}O. This also means that their change in concentration is the same. But whilst the concentrations of CO_{2} and H_{2} decrease, the concentrations of CO and H_{2}O increase.

Let's represent the change in concentration of CO_{2} using the letter x. Because the concentration of CO_{2} decreases, we'll use -x. Thanks to the molar ratio, we also know that the change in concentration of H_{2} is -x and the change in concentration of both CO and H_{2}O is +x. Here are these values, added to our table:

Species | CO_{2} + H_{2} ⇌ CO + H_{2}O | ||||

Concentration (M) | Initial | 2.00 | 2.00 | 0.00 | 0.00 |

Change | -x | -x | +x | +x | |

Equilibrium |

We can now find the equilibrium concentration of each species by adding the change in concentration to the initial concentration:

Species | CO_{2} + H_{2} ⇌ CO + H_{2}O | ||||

Concentration (M) | Initial | 2.00 | 2.00 | 0.00 | 0.00 |

Change | -x | -x | +x | +x | |

Equilibrium | 2.00 - x | 2.00 - x | x | x |

We're now ready to substitute these values into the expression for the equilibrium constant K_{c}:

$$K_c=\frac{[CO]_{eqm}\space [H_2O]_{eqm}}{[CO_2]_{eqm}\space [H_2]_{eqm}}$$

The question gives us a value for K_{c}. We can set our expression for K_{c} equal to that value, giving us a fraction on one side and a constant on the other. If we multiply the denominator through both sides of the equation and bring all of the terms over to one side, we end up with a quadratic equation equalling 0:

$$3.00=\frac{(x)(x)}{(2.00-x)(2.00-x)}$$ $$3.00(2.00-x)(2.00-x)=(x)(x)$$ $$12.00-12.00x+3.00x^2=x^2$$ $$12.00-12.00x+2.00x^2=0$$

If you've got a scientific calculator, now is the time to pull it out and solve the equation. You should end up with two possible values for x:

$$x=3+\sqrt{3}\qquad \qquad x=3-\sqrt{3}$$

How do we know which of these values is correct?

- 3 + √3 = 4.73 M, whilst 3 - √3 = 1.27 M.
- Remember that x represents a change in concentration, and that we know that the equilibrium concentration of CO
_{2}is (2.00 - x). - If x equals 4.73 M, then the equilibrium concentration of CO
_{2}turns out to be (2.00 - 4.73) = -2.73 M. - This can't be the case - we can't have a negative concentration!
- Therefore, the only feasible value of x is 1.27 M.

The final step is to go back to our starting table and work out the equilibrium concentrations of each species and replace our expressions involving x with actual numbers. Here are our final answers:

Species | CO_{2} + H_{2} ⇌ CO + H_{2}O | ||||

Concentration (M) | Initial | 2.00 | 2.00 | 0.00 | 0.00 |

Change | -x | -x | +x | +x | |

Equilibrium | 2.00 - x= 0.73 | 2.00 - x= 0.73 | x = 1.27 | x = 1.27 |

Hopefully, the process makes a little more sense now. Here's another example for you to try.

**6.4 M H _{2}, 4.5 M Cl_{2}, and 1.0 M HCl are combined in a closed system. Use the following information to find the equilibrium concentration of each species.**

$$H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g)\qquad K_c=2.8$$

Once again, we'll begin by making a table of initial, equilibrium, and change in concentration:

Species | H_{2} + Cl_{2} ⇌ 2HCl | |||

Concentration (M) | Initial | 6.4 | 4.5 | 1.0 |

Change | ||||

Equilibrium |

Let's call the change in concentration of H_{2} x. We don't actually know which direction the reaction will move in; H_{2} and Cl_{2} could react to produce HCl, or HCl could react to produce H_{2} and Cl_{2}. However, we'll assume that the reaction moves from left to right and H_{2} and Cl_{2} are used up. This means that the change in concentration of H_{2} is -x. Looking at the balanced chemical equation, we can see that the molar ratio of H_{2}:Cl_{2}:HCl is 1:1:2. Therefore, the change in concentration of Cl_{2} is also -x, whilst the change in concentration of HCl is +2x. We can use these values to write equilibrium concentrations for each of the three species:

Species | H_{2} + Cl_{2} ⇌ 2HCl | |||

Concentration (M) | Initial | 6.4 | 4.5 | 1.0 |

Change | -x | -x | +2x | |

Equilibrium | 6.4 - x | 4.5 - x | 1.0 + 2x |

Let's now substitute our equilibrium concentrations into the expression for K_{c}. Once again, the value of K_{c} was given to us in the question. We can set our K_{c} expression equal to this value and rearrange in order to find a quadratic equation equalling 0.

$$K_c=\frac{{[HCl]_{eqm}}^2}{[H_2]_{eqm}\space [Cl_2]_{eqm}}$$ $$2.8=\frac{{(1.0+2x)}^2}{(6.4-x)\space (4.5-x)}$$ $$2.8(6.4-x)(4.5-x)={(1.0+2x)}^2$$ $$1.2x^2+34.52x-79.64=0$$

We solve the equation and arrive at two possible values for x:

$$x=-30.914\qquad \qquad x=2.147$$

Look at the table of concentrations again to work out the correct value of x.

- If x equals -30.914 M, then that makes the equilibrium concentration of HCl equal (1.0 + 2 (-30.914)) = -60.828 M.
- We can't have a negative concentration, and so we discard this value of x.
- Therefore, x equals 2.147 M.

The last step is to substitute the correct value of x into the equilibrium concentration expressions for each species:

Species | H_{2} + Cl_{2} ⇌ 2HCl | |||

Concentration (M) | Initial | 6.4 | 4.5 | 1.0 |

Change | -x | -x | +2x | |

Equilibrium | 6.4 - x= 4.253 | 4.5 - x= 2.353 | 1.0 + 2x= 5.294 |

These are our final answers.

The same method can also be used to calculate equilibrium partial pressures. Instead of representing a change in *concentration*, the letter x simply represents a change in *partial pressure* and is measured in atm.

That's it for this article. By now, you should understand how to** calculate equilibrium concentrations** using **initial concentrations** and the **equilibrium constant K**_{c}. This involves **representing the change in concentration in terms of x**, and** forming** and** solving a quadratic equation**.

**Equilibrium concentrations**can be calculated using**initial concentrations**and the**equilibrium constant K**_{c}.**Calculating equilibrium concentrations**uses the**formula for K**_{c}and involves**solving a quadratic equation**. You use the units**M**, or**mol dm**^{-3}.- To calculate equilibrium concentrations, take the following steps:
- Use the balanced chemical equation to represent the change in concentration of each species in terms of x.
- Write the equilibrium concentration of each species using its initial concentration and its change in x.
- Substitute the equilibrium concentrations into the expression for K
_{c}. - Rearrange to form a quadratic equation.
- Solve the quadratic equation to find two values for x. Discard the inappropriate value.
- Use your value of x to calculate the equilibrium concentration of each species.

_{c}. This process involves representing the change in concentration as x, and finding and solving a quadratic equation. Check out this article for some worked examples.

_{c}. This process involves representing the change in concentration as x, and finding and solving a quadratic equation. Although this may seem a little tricky, we have a couple of great worked examples in this article to help you out.

_{c} with equilibrium concentrations, you first create an expression for K_{c} using the balanced chemical equation. You then substitute the equilibrium concentrations into this expression to get your final answer.

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