Google Classroom
Reversible reactions, equilibrium, and the equilibrium constant K. How to calculate K, and how to use K to determine if a reaction strongly favors products or reactants at equilibrium.
Key points
A reversible reaction can proceed in both the forward and backward directions.
Equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction. All reactant and product concentrations are constant at equilibrium.
Given a reaction
For reactions that are not at equilibrium, we can write a similar expression called the reaction quotient
Introduction: reversible reactions and equilibrium
A reversible reaction can proceed in both the forward and backward directions. Most reactions are theoretically reversible in a closed system, though some can be considered to be irreversible if they heavily favor the formation of reactants or products. The double half-arrow sign we use when writing reversible reaction equations,
Imagine we added some colorless
Initially, the vial contains only
All reactions tend towards a state of chemical equilibrium, the point at which both the forward process and the reverse process are taking place at the same rate. Since the forward and reverse rates are equal, the concentrations of the reactants and products are constant at equilibrium. It is important to remember that even though the concentrations are constant at equilibrium, the reaction is still happening! That is why this state is also sometimes referred to as dynamic equilibrium.
Based on the concentrations of all the different reaction species at equilibrium, we can define a quantity called the equilibrium constant
How do we calculate
Consider the balanced reversible reaction below:
If we know the molar concentrations for each reaction species, we can find the value for
where
There are some important things to remember when calculating
Pure solids and pure liquids, including solvents, are not included in the equilibrium expression.
The reaction must be balanced with the coefficients written as the lowest possible integer values in order to get the correct value for
Note: If any of the reactants or products are gases, we can also write the equilibrium constant in terms of the partial pressure of the gases. We typically refer to that value as
What does the magnitude of
The magnitude of
If
If
If
By using these guidelines, we can quickly estimate whether a reaction will strongly favor the forward direction to make products—very large
Example
Part 1: Calculating
Let's take a look at the equilibrium reaction that takes place between sulfur dioxide and oxygen to produce sulfur trioxide:
The reaction is at equilibrium at some temperature,
We can calculate
If we plug our known equilibrium concentrations into the above equation, we get:
Note that since the calculated
Part 2: Using the reaction quotient
Now we know the equilibrium constant for this temperature:
We would like to know if this reaction is at equilibrium, but how can we figure that out? When we aren't sure if our reaction is at equilibrium, we can calculate the reaction quotient,
At this point, you might be wondering why this equation looks so familiar and how
If we calculate
Because our value for
The short answer
When
The longer answer
Since all reactions at a constant temperature will adjust concentrations until they reach equilibrium, we can use
- If
- If
For more details, check out this article on the reaction quotient Q.
Example 2: Using
Let's consider an equilibrium mixture of
We can write the equilibrium constant expression as follows:
We know the equilibrium constant is
What is the concentration of
Since
If we know that the equilibrium concentrations for
If we plug in our equilibrium concentrations and value for
As predicted, the concentration of
Summary
A reversible reaction can proceed in both the forward and backward directions.
Equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction. All reactant and product concentrations are constant at equilibrium.
Given an equation
For reactions that are not at equilibrium, we can write a similar expression called the reaction quotient
Attributions
This article was adapted from the following articles:
“Equilibrium constants” from OpenStax Chemistry, CC BY 4.0
"Writing equilibrium constant expressions involving liquids and solids" from UC Davis ChemWiki, CC BY-NC-SA 3.0 US
The modified article is licensed under a CC-BY-NC-SA 4.0 license.
Log in S Chung 8 years agoPosted 8 years ago. Direct link to S Chung's post “This article mentions tha...” This article mentions that if Kc is very large, i.e. 1000 or more, then the equilibrium will favour the products. I thought that if Kc is larger than one (1), then that's when the equilibrium will favour the products. Conversely, if Kc is less than one (1), the equilibrium will favour the reactants. • (57 votes) abhishekppatil99 8 years agoPosted 8 years ago. Direct link to abhishekppatil99's post “If Kc is larger than 1 it...” If Kc is larger than 1 it would mean that the equilibrium is starting to favour the products however it doesnt necessarily mean that that the molar concentration of reactants is negligible. (78 votes) Lily Martin 7 years agoPosted 7 years ago. Direct link to Lily Martin's post “why aren't pure liquids a...” why aren't pure liquids and pure solids included in the equilibrium expression? • (13 votes) awemond 7 years agoPosted 7 years ago. Direct link to awemond's post “Equilibrium constant are ...” Equilibrium constant are actually defined using activities, not concentrations. The activity of pure liquids and solids is 1 and the activity of a solution can be estimated using its concentration. So, pure liquids and solids actually are involved, but since their activities are equal to 1, they don't change the equilibrium constant and so are often left out. (18 votes) doctor_luvtub 8 years agoPosted 8 years ago. Direct link to doctor_luvtub's post “"Kc is often written with...” "Kc is often written without units, depending on the textbook." When Kc is given units, what is the unit? • (11 votes) Azmith.10k 8 years agoPosted 8 years ago. Direct link to Azmith.10k's post “Depends on the question....” Depends on the question. (16 votes) Afigueroa313 8 years agoPosted 8 years ago. Direct link to Afigueroa313's post “Any suggestions for where...” Any suggestions for where I can do equilibrium practice problems? • (7 votes) Rippy 8 years agoPosted 8 years ago. Direct link to Rippy's post “Try googling "equilibrium...” Try googling "equilibrium practise problems" and I'm sure there's a bunch. Some will be PDF formats that you can download and print out to do more. Sorry for the British/Australian spelling of practise. (15 votes) Alejandro Puerta-Alvarado 6 years agoPosted 6 years ago. Direct link to Alejandro Puerta-Alvarado's post “I get that the equilibr...” I get that the equilibrium constant changes with temperature. I don't get how it changes with temperature. The formula for calculating Kc or K or Keq doesn't seem to incorporate the temperature of the environment anywhere in it, nor does this article seem to specify exactly how it changes the equilibrium constant, or whether it's a predicable change. With this in mind, can anyone help me in understanding the relationship between the equilibrium constant and temperature? • (8 votes) Bhagyashree U Rao 6 years agoPosted 6 years ago. Direct link to Bhagyashree U Rao's post “You forgot *main* thing. ...” You forgot main thing. Kc depends on Molarity and Molarity depends on volume of the soln, which in turn depends on 'temperature'. Hope you get it! (10 votes) Cynthia Shi 8 years agoPosted 8 years ago. Direct link to Cynthia Shi's post “If the equilibrium favors...” If the equilibrium favors the products, does this mean that equation moves in a forward motion? Or would it be backward in order to balance the equation back to an equilibrium state? • (6 votes) Matt B 8 years agoPosted 8 years ago. Direct link to Matt B's post “If it favors the products...” If it favors the products then it will favourite the forward direction to create for products (and fewer reactants). As the reaction proceeds, the reaction will approach the equilibrium, and this will cause the forward reaction to decrease and the backward reaction to increase until they are equal to each other. (10 votes) Becky Anton 8 years agoPosted 8 years ago. Direct link to Becky Anton's post “Any videos or areas using...” Any videos or areas using this information with the ICE theory? • (4 votes) S Chung 8 years agoPosted 8 years ago. Direct link to S Chung's post “Check out 'Buffers, Titra...” Check out 'Buffers, Titrations, and Solubility Equilibria'. (7 votes) Carissa Myung 8 years agoPosted 8 years ago. Direct link to Carissa Myung's post “Say if I had H2O (g) as e...” Say if I had H2O (g) as either the product or reactant. Would I still include water vapor (H2O (g)) in writing the Kc formula? • (4 votes) Emily 8 years agoPosted 8 years ago. Direct link to Emily's post “YES! Very important to kn...” YES! Very important to know that with equilibrium calculations we leave out any solids or liquids and keep gases. So with saying that if your reaction had had H2O (l) instead, you would leave it out! (5 votes) Isaac Nketia 8 years agoPosted 8 years ago. Direct link to Isaac Nketia's post “What happens if Q isn't e...” What happens if Q isn't equal to Kc? • (2 votes) Srk 8 years agoPosted 8 years ago. Direct link to Srk's post “If Q is not equal to Kc, ...” If Q is not equal to Kc, then the reaction is not occurring at the Standard Conditions of the reaction (7 votes) Osama Shammout 6 years agoPosted 6 years ago. Direct link to Osama Shammout's post “Excuse my very basic voca...” Excuse my very basic vocabulary. • (3 votes) RogerP 6 years agoPosted 6 years ago. Direct link to RogerP's post “That's a good question! H...” That's a good question! However, the position of the equilibrium is temperature dependent and lower temperatures favour dinitrogen tetroxide. In fact, dinitrogen tetroxide is stable as a solid (melting point -11.2 °C) and even in the liquid state is almost entirely dinitrogen tetroxide. Only in the gaseous state (boiling point 21.7 °C) does the position of equilibrium move towards nitrogen dioxide, with the reaction moving further right as the temperature increases. There are really no experimental details given in the text above. Therefore, the experiment could be done by adding liquid dinitrogen tetroxide and allowing it to warm up and become a gas whereupon an equilibrium will be established. (At 100 °C, only 10% of the mixture is dinitrogen tetroxide.) (3 votes)Want to join the conversation?
for example - is the value of Kc is 2, it would mean that the molar concentration of reactants is 1/2 the concentration of products. In this case though the value of Kc is greater than 1, the reactants are still present in considerable amount.
Khan academy was trying to show us all the extreme cases, so the case in which Kc is 1000 the molar concentration of reactants is so less that practically the equilibrium has shifted almost completely to the product side and vice versa in case of Kc being 0.001.
And if you read carefully, they dont say that when Kc is very large products are favoured but they are saying that when Kc if very large mostly products are present and vice versa.
Hope this helps :-)
For example, in Haber's process: N2 +3H2<---->2NH3
Kc=[NH3]^2/[N2][H2]^3
Using molarity(M) as unit for concentration:
Kc=M^2/M*M^3=M^-2
i.e Kc will have the unit M^-2 or Molarity raised to the power -2.
Hope you can understand my vague explanation!!
but the reaction will take place.There can be two cases :
1) If Q>Kc - The reaction will proceed in the direction of reactants.
2) If Q<Kc - The reaction will proceed in the direction of products.
So basically we are saying that N2O4 (Dinitrogen tetroxide) is put in a vial or a container, it reacts to become 2NO2 overtime until they are constant (forward and reverse). What I keep wondering about is: Why isn't it already at a constant? I mean, so while we are taking the dinitrogen tetroxide why isn't it turning? so that it disappears? I don't know if my vague terms get the idea explained but why aren't things if they have the same conditions change so that they always are in equilibrium.
Why we can observe it only when put in a container? Why until the time we put it, it starts changing why not since it formulated, it changes, and if it does, then how come hasn't the reactants finish (becomes all used)?
