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AP®︎/College Chemistry

Course: AP®︎/College Chemistry > Unit 11

Lesson 1: Equilibrium constant
Science
AP®︎/College Chemistry
Chemical equilibrium
Equilibrium constant
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The equilibrium constant K

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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 start text, a, A, end text, plus, start text, b, B, end text, \rightleftharpoons, start text, c, C, end text, plus, start text, d, D, end text, the equilibrium constant K, start subscript, start text, c, end text, end subscript, also called K or K, start subscript, start text, e, q, end text, end subscript, is defined as follows:
K, start subscript, start text, c, end text, end subscript, equals, start fraction, open bracket, start text, C, close bracket, end text, start superscript, start text, c, end text, end superscript, start text, open bracket, D, close bracket, end text, start superscript, start text, d, end text, end superscript, divided by, open bracket, start text, A, end text, close bracket, start superscript, start text, a, end text, end superscript, open bracket, start text, B, end text, close bracket, start superscript, start text, b, end text, end superscript, end fraction
  • For reactions that are not at equilibrium, we can write a similar expression called the reaction quotient Q, which is equal to K, start subscript, start text, c, end text, end subscript at equilibrium.
  • K, start subscript, start text, c, end text, end subscript and Q can be used to determine if a reaction is at equilibrium, to calculate concentrations at equilibrium, and to estimate whether a reaction favors products or reactants at equilibrium.

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, \rightleftharpoons, is a good visual reminder that these reactions can go either forward to create products, or backward to create reactants. One example of a reversible reaction is the formation of nitrogen dioxide, start text, N, O, end text, start subscript, 2, end subscript, from dinitrogen tetroxide, start text, N, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 4, end subscript:
start text, N, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 4, end subscript, left parenthesis, g, right parenthesis, \rightleftharpoons, 2, start text, N, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis
Imagine we added some colorless start text, N, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 4, end subscript, left parenthesis, g, right parenthesis to an evacuated glass container at room temperature. If we kept our eye on the vial over time, we would observe the gas in the ampoule changing to a yellowish orange color and gradually getting darker until the color stayed constant. We can graph the concentration of start text, N, O, end text, start subscript, 2, end subscript and start text, N, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 4, end subscript over time for this process, as you can see in the graph below.
A graph with concentration on the y axis and time on the x axis. The concentration of nitrogen dioxide starts at zero and increases until it stays constant at the equilibrium concentration. The concentration of dinitrogen tetroxide starts at an arbitrary initial concentration, then decreases until it reaches the equilibrium concentration. At equilibrium, both the concentration of dinitrogen tetroxide and nitrogen dioxide are not changing with time.
Graph of concentration vs. time for the reversible conversion of dinitrogen tetroxide to nitrogen dioxide. At the time indicated by the dotted line, the concentrations of both species are constant, and the reaction is at equilibrium. Image credit: Graph modified from OpenStax Chemistry, CC BY 4.0
Initially, the vial contains only start text, N, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 4, end subscript, and the concentration of start text, N, O, end text, start subscript, 2, end subscript is 0 M. As start text, N, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 4, end subscript gets converted to start text, N, O, end text, start subscript, 2, end subscript, the concentration of start text, N, O, end text, start subscript, 2, end subscript increases up to a certain point, indicated by a dotted line in the graph to the left, and then stays constant. Similarly, the concentration of start text, N, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 4, end subscript decreases from the initial concentration until it reaches the equilibrium concentration. When the concentrations of start text, N, O, end text, start subscript, 2, end subscript and start text, N, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 4, end subscript remain constant, the reaction has reached equilibrium.
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 K, start subscript, start text, c, end text, end subscript, which is also sometimes written as K, start subscript, start text, e, q, end text, end subscript or K. The start text, c, end text in the subscript stands for concentration since the equilibrium constant describes the molar concentrations, in start fraction, start text, m, o, l, end text, divided by, start text, L, end text, end fraction, at equilibrium for a specific temperature. The equilibrium constant can help us understand whether the reaction tends to have a higher concentration of products or reactants at equilibrium. We can also use K, start subscript, start text, c, end text, end subscript to determine if the reaction is already at equilibrium.

How do we calculate K, start subscript, start text, c, end text, end subscript?

Consider the balanced reversible reaction below:
start text, a, A, end text, plus, start text, b, B, end text, \rightleftharpoons, start text, c, C, end text, plus, start text, d, D, end text
If we know the molar concentrations for each reaction species, we can find the value for K, start subscript, start text, c, end text, end subscript using the relationship
K, start subscript, start text, c, end text, end subscript, equals, start fraction, open bracket, start text, C, close bracket, end text, start superscript, start text, c, end text, end superscript, start text, open bracket, D, close bracket, end text, start superscript, start text, d, end text, end superscript, divided by, open bracket, start text, A, end text, close bracket, start superscript, start text, a, end text, end superscript, open bracket, start text, B, end text, close bracket, start superscript, start text, b, end text, end superscript, end fraction
where open bracket, start text, C, close bracket, end text and start text, open bracket, D, close bracket, end text are equilibrium product concentrations; open bracket, start text, A, end text, close bracket and open bracket, start text, B, end text, close bracket are equilibrium reactant concentrations; and start text, a, end text, start text, b, end text, start text, c, end text, and start text, d, end text are the stoichiometric coefficients from the balanced reaction. The concentrations are usually expressed in molarity, which has units of start fraction, start text, m, o, l, end text, divided by, start text, L, end text, end fraction.
Five glass ampules. The colors vary, with the leftmost vial frosted over and colorless and the second vial to the left containing a dark yellow liquid and gas. The liquid and gas inside the third, fourth, and fifth vials from the left are increasingly darker orange-brown in color.
Dinitrogen tetroxide, a colorless liquid and gas, is in equilibrium with nitrogen dioxide, an orange-brown gas. The equilibrium constant and the equilibrium concentrations of both species depend on the temperature! Temperature of ampoules from left to right: -196 degreesC, 0 degreesC, 23 degreesC, 35 degreesC, and 50 degreesC. Image credit: Eframgoldberg on Wikimedia Commons, CC BY-SA 3.0
There are some important things to remember when calculating K, start subscript, start text, c, end text, end subscript:
  • K, start subscript, start text, c, end text, end subscript is a constant for a specific reaction at a specific temperature. If you change the temperature of a reaction, then K, start subscript, start text, c, end text, end subscript also changes.
  • Pure solids and pure liquids, including solvents, are not included in the equilibrium expression.
  • K, start subscript, start text, c, end text, end subscript is often written without units, depending on the textbook.
  • The reaction must be balanced with the coefficients written as the lowest possible integer values in order to get the correct value for K, start subscript, start text, c, end text, end subscript.
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 K, start subscript, start text, p, end text, end subscript to tell it apart from the equilibrium constant using concentrations in molarity, K, start subscript, start text, c, end text, end subscript. In this article, however, we will be focusing on K, start subscript, start text, c, end text, end subscript.

What does the magnitude of K, start subscript, start text, c, end text, end subscript tell us about the reaction at equilibrium?

The magnitude of K, start subscript, start text, c, end text, end subscript can give us some information about the reactant and product concentrations at equilibrium:
  • If K, start subscript, start text, c, end text, end subscript is very large, ~1000 or more, we will have mostly product species present at equilibrium.
  • If K, start subscript, start text, c, end text, end subscript is very small, ~0.001 or less, we will have mostly reactant species present at equilibrium.
  • If K, start subscript, start text, c, end text, end subscript is in between 0.001 and 1000, we will have a significant concentration of both reactant and product species present at equilibrium.
By using these guidelines, we can quickly estimate whether a reaction will strongly favor the forward direction to make products—very large K, start subscript, start text, c, end text, end subscript—strongly favor the backward direction to make reactants—very small K, start subscript, start text, c, end text, end subscript—or somewhere in between.

Example

Part 1: Calculating K, start subscript, start text, c, end text, end subscript from equilibrium concentrations

Let's take a look at the equilibrium reaction that takes place between sulfur dioxide and oxygen to produce sulfur trioxide:
2, start text, S, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, plus, start text, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, \rightleftharpoons, 2, start text, S, O, end text, start subscript, 3, end subscript, left parenthesis, g, right parenthesis
The reaction is at equilibrium at some temperature, start text, T, end text, and the following equilibrium concentrations are measured:
[SO2]=0.90 M[O2]=0.35 M[SO3]=1.1 M\begin{aligned} {[}\text{SO}_2{]}&= 0.90 \,\text {M}\\ \\ [\text O_2] &= 0.35 \,\text M\\ \\ [\text{SO}_3] &= 1.1 \,\text M\end{aligned}
We can calculate K, start subscript, start text, c, end text, end subscript for the reaction at temperature start text, T, end text by solving following expression:
K, start subscript, start text, c, end text, end subscript, equals, start fraction, open bracket, start text, S, O, end text, start subscript, 3, end subscript, close bracket, squared, divided by, open bracket, start text, S, O, end text, start subscript, 2, end subscript, close bracket, squared, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket, end fraction
If we plug our known equilibrium concentrations into the above equation, we get:
Kc=[SO3]2[SO2]2[O2]=[1.1]2[0.90]2[0.35]=4.3\begin{aligned} K_\text c &= \dfrac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text O_2]} \\ \\ &= \dfrac{[1.1]^2}{[0.90]^2[0.35]} \\ \\ &= 4.3 \end{aligned}
Note that since the calculated K, start subscript, start text, c, end text, end subscript value is between 0.001 and 1000, we would expect this reaction to have significant concentrations of both reactants and products at equilibrium, as opposed to having mostly reactants or mostly products.

Part 2: Using the reaction quotient Q to check if a reaction is at equilibrium

Now we know the equilibrium constant for this temperature: K, start subscript, start text, c, end text, end subscript, equals, 4, point, 3. Imagine we have the same reaction at the same temperature start text, T, end text, but this time we measure the following concentrations in a different reaction vessel:
[SO2]=3.6 M[O2]=0.087 M[SO3]=2.2 M\begin{aligned}{[}\text{SO}_2] &= 3.6 \,\text {M}\\ \\ [\text O_2] &= 0.087 \,\text M\\ \\ [\text{SO}_3] &= 2.2 \,\text M\end{aligned}
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, Q:
Q, equals, start fraction, open bracket, start text, S, O, end text, start subscript, 3, end subscript, close bracket, squared, divided by, open bracket, start text, S, O, end text, start subscript, 2, end subscript, close bracket, squared, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket, end fraction
At this point, you might be wondering why this equation looks so familiar and how Q is different from K, start subscript, start text, c, end text, end subscript. The main difference is that we can calculate Q for a reaction at any point whether the reaction is at equilibrium or not, but we can only calculate K, start subscript, start text, c, end text, end subscript at equilibrium. By comparing Q to K, start subscript, start text, c, end text, end subscript, we can tell if the reaction is at equilibrium because Q, equals, K, start subscript, start text, c, end text, end subscript at equilibrium.
If we calculate Q using the concentrations above, we get:
Q=[SO3]2[SO2]2[O2]=[2.2]2[3.6]2[0.087]=4.3\begin{aligned} Q &= \dfrac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text O_2]} \\ \\ &= \dfrac{[2.2]^2}{[3.6]^2[0.087]} \\ \\ &= 4.3 \end{aligned}
Because our value for Q is equal to K, start subscript, start text, c, end text, end subscript, we know the new reaction is also at equilibrium. Hooray!

Example 2: Using K, start subscript, start text, c, end text, end subscript to find equilibrium compositions

Let's consider an equilibrium mixture of start text, N, end text, start subscript, 2, end subscript, start text, O, end text, start subscript, 2, end subscript and start text, N, O, end text:
start text, N, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, plus, start text, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, \rightleftharpoons, 2, start text, N, O, end text, left parenthesis, g, right parenthesis
We can write the equilibrium constant expression as follows:
K, start subscript, start text, c, end text, end subscript, equals, start fraction, start text, open bracket, N, O, end text, close bracket, squared, divided by, open bracket, start text, N, end text, start subscript, 2, end subscript, close bracket, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket, end fraction
We know the equilibrium constant is 3, point, 4, times, 10, start superscript, minus, 21, end superscript at a particular temperature, and we also know the following equilibrium concentrations:
open bracket, start text, N, end text, start subscript, 2, end subscript, close bracket, equals, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket, equals, 0, point, 1, start text, M, end text
What is the concentration of start text, N, O, end text, left parenthesis, g, right parenthesis at equilibrium?
Since K, start subscript, start text, c, end text, end subscript is less than 0.001, we would predict that the reactants start text, N, end text, start subscript, 2, end subscript and start text, O, end text, start subscript, 2, end subscript are going to be present in much greater concentrations than the product, start text, N, O, end text, at equilibrium. Thus, we would expect our calculated start text, N, O, end text concentration to be very low compared to the reactant concentrations.
If we know that the equilibrium concentrations for start text, N, end text, start subscript, 2, end subscript and start text, O, end text, start subscript, 2, end subscript are 0.1 M, we can rearrange the equation for K, start subscript, start text, c, end text, end subscript to calculate the concentration of start text, N, O, end text:
K, start subscript, start text, c, end text, end subscript, equals, start fraction, start text, open bracket, N, O, end text, close bracket, squared, divided by, open bracket, start text, N, end text, start subscript, 2, end subscript, close bracket, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket, end fraction, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, space, start text, G, e, t, space, t, h, e, space, N, O, space, t, e, r, m, space, b, y, space, i, t, s, e, l, f, space, o, n, space, o, n, e, space, s, i, d, e, point, end text
open bracket, start text, N, O, end text, close bracket, squared, equals, K, open bracket, start text, N, end text, start subscript, 2, end subscript, close bracket, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket, space, space, space, space, space, space, space, start text, T, a, k, e, space, t, h, e, space, s, q, u, a, r, e, space, r, o, o, t, space, o, f, space, b, o, t, h, space, s, i, d, e, s, space, t, o, space, s, o, l, v, e, space, f, o, r, space, open bracket, N, O, close bracket, point, end text
open bracket, start text, N, O, end text, close bracket, equals, square root of, K, open bracket, start text, N, end text, start subscript, 2, end subscript, close bracket, open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket, end square root
If we plug in our equilibrium concentrations and value for K, start subscript, start text, c, end text, end subscript, we get:
[NO]=K[N2][O2]=K[N2][O2]=(3.4×10−21)(0.1)(0.1)=5.8×10−12 M\begin{aligned}[\text{NO}]&=\sqrt{K [\text N_2] [\text O_2]}\\ \\ &=\sqrt{K [\text N_2] [\text O_2]}\\ \\ &=\sqrt{(3.4 \times 10^{-21})(0.1)(0.1)}\\ \\ &=5.8 \times 10^{-12}\,\text M\end{aligned}
As predicted, the concentration of start text, N, O, end text, 5, point, 8, times, 10, start superscript, minus, 12, end superscript, start text, M, end text, is much smaller than the reactant concentrations open bracket, start text, N, end text, start subscript, 2, end subscript, close bracket and open bracket, start text, O, end text, start subscript, 2, end subscript, close bracket.

Summary

A photograph of an oceanside beach. The yellowish sand is covered with people on beach towels, and there are also some swimmers in the blue-green ocean. The beach is also surrounded by houses from a small town.
If the rate of people entering the water is equal to the rate of people getting out of the water, then the system is at equilibrium! The total number of people on the beach and the number of people in the water will stay constant even though beachgoers are still moving between the sand and the water. Image credit: penreyes on flickr, CC BY 2.0
  • 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 start text, a, A, end text, plus, start text, b, B, end text, \rightleftharpoons, start text, c, C, end text, plus, start text, d, D, end text, the equilibrium constant K, start subscript, start text, c, end text, end subscript, also called K or K, start subscript, start text, e, q, end text, end subscript, is defined using molar concentration as follows:
K, start subscript, start text, c, end text, end subscript, equals, start fraction, open bracket, start text, C, close bracket, end text, start superscript, start text, c, end text, end superscript, start text, open bracket, D, close bracket, end text, start superscript, start text, d, end text, end superscript, divided by, open bracket, start text, A, end text, close bracket, start superscript, start text, a, end text, end superscript, open bracket, start text, B, end text, close bracket, start superscript, start text, b, end text, end superscript, end fraction
  • For reactions that are not at equilibrium, we can write a similar expression called the reaction quotient Q, which is equal to K, start subscript, start text, c, end text, end subscript at equilibrium.
  • K, start subscript, start text, c, end text, end subscript can be used to determine if a reaction is at equilibrium, to calculate concentrations at equilibrium, and to estimate whether a reaction favors products or reactants at equilibrium.

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  • marcimus orange style avatar for user S Chung
    Posted 7 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.
    (55 votes)
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    • starky tree style avatar for user abhishekppatil99
      Posted 7 years ago. Direct link to abhishekppatil99's post “If Kc is larger than 1 it...”
      Incredible Answer
      Great Answer
      Good Answer
      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.
      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 :-)
      (74 votes)
  • duskpin ultimate style avatar for user Lily Martin
    Posted 6 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)
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    • orange juice squid orange style avatar for user awemond
      Posted 6 years ago. Direct link to awemond's post “Equilibrium constant are ...”
      Good Answer
      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)
  • mr pink red style avatar for user doctor_luvtub
    Posted 7 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)
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    • aqualine ultimate style avatar for user Azmith.10k
      Posted 7 years ago. Direct link to Azmith.10k's post “Depends on the question. ...”
      Good Answer
      Depends on the question.
      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!!
      (15 votes)
  • blobby green style avatar for user Afigueroa313
    Posted 7 years ago. Direct link to Afigueroa313's post “Any suggestions for where...”
    Any suggestions for where I can do equilibrium practice problems?
    (7 votes)
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  • duskpin tree style avatar for user Alejandro Puerta-Alvarado
    Posted 5 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)
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  • mr pants teal style avatar for user Cynthia Shi
    Posted 7 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)
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    • old spice man green style avatar for user Matt B
      Posted 7 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.
      (9 votes)
  • piceratops sapling style avatar for user Becky Anton
    Posted 7 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)
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  • duskpin ultimate style avatar for user Carissa Myung
    Posted 7 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)
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  • blobby green style avatar for user Isaac Nketia
    Posted 7 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)
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    • piceratops tree style avatar for user Srk
      Posted 7 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
      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.
      (6 votes)
  • female robot grace style avatar for user Osama Shammout
    Posted 5 years ago. Direct link to Osama Shammout's post “Excuse my very basic voca...”
    Excuse my very basic vocabulary.
    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)?
    (3 votes)
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    • piceratops seed style avatar for user RogerP
      Posted 5 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)