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Equilibrium & Reversible Reactions

Reversible Reaction

A reaction that either can go backwards or forwards

A reversible reaction can also be when the backward or forward reaction are occuring simultaneously but, this is not a must

A good example is the hydration of copper sulfate. This is done when you add water to the Anhydrous copper sulfate :

CuSO4 + 5H2O → CuSO4.5H2O

This releases a lot of heat and turns the added water to steam. There is also a color change from white to blue

The reaction could go back also:

CuSO4.5H2O → CuSO4 + 5H2O

This is done when the hydrated salt is heated. The color changes from blue to white

This could be combined to one reversible reaction

CuSO4 + H2O ⇋ CuSO4.5H2O

Notice the forward and the backward reactions don't occur simultaneously and also this reaction is used as a chemical test for water

The ⇋ sign indicates it is a reversible reaction

Dynamic Equilibirium

This is a special type of reversible reaction:

When the rate of the forward and backward reaction are the same and so the concentration of products and reactants remains constant

There are more conditions for an equilibirium to occur:

  • Must be in a Closed system
  • A system in which no matter or energy enters or leaves the system

    The Google definition for a closed system is not suitable as it says energy can leave out of the system but, the marksheme definition is the one above

    A closed system could be either a solution in which the products and reactants are in aqueous forms or it could be gases trapped in a container

  • It is Dynamic
  • When both the reactants and products are continuously reacting simultaneously

    This shows that the rate of forward and backward reaction are not Zero but happening continously. In other words the net rate is Zero as it cancels out

In fact, when a question ask what is a "Dynamic Equilibirium"? It is best if you include all of these points:

When the rate of the forward and backward reaction are the same and is continously reacting simultaneously and so the concentration of products and reactants remains constant. It also occurs in a closed system

Equilibirium Position

An equilibrium position is not a real thing but an idea of relativity

For example is a equilibrium, if there are more products formed than the reactants, then we say the equilibrium position is to the right or shifted to the right

If there are more reactants formed than the products, then we say the equilibrium position is to the left or shifted to the left

Usually the concentration of products and reactants are not equal or the same. They remain constant and it is usually rare for them to be the same

Factors Affecting the Equilibrium

Before we go through the factors that affect the equilibrium, we must go through a very important principle which is used to explain the effects of each factor

Le Chatelier's principle

When two or more factors affecting the equilibrium are changed, the equilibrium position shifts to the direction that opposes the change made to it

The system does not like to change and any change made to it will be opposed by the system to reduce the change made to it. You will understand more as we give the factors:

Below are the factors which affect the equilibrium position of the system. These are the model answers so follow these:

N2 + 3H2 ⇋ 2NH3
This an example used to explain the below factors

  • Concentration
  • Refer the above example

    Increasing the concentration of N2 will cause the equilibrium position to be disturbed. The system will try to oppose it by decreasing the concentration of N2 by increasing the forward reaction rate. The equilibrium position shifts to the right and more NH3 is formed

  • Pressure
  • When you increase the pressure, the equilibrium is disturbed. The system will try to reduce the pressure by favouring/ increasing the rate towards fewer molecules. So this increase the rate of the forward reaction and so equilibrium position shifts to the right and produce more NH3

  • Temperature
  • The above forward reaction is exothermic

    When the temperature is increased the equilibrium is disturb and the system will try to decrease the temperature by increase the rate of the endothermic reaction which is the backward reaction. So the equilibrium position shifts to the left and produce more N2 and H2

    Further Explanation

    If you don't understand, do not worry we will explain!

    The 1st Factor is pretty straight forward and it applies to different situation also

    For example if we reduce the concentration of N2 the rate of backward reaction will increase to make more N2

    If we increase the concentration of NH3, then the backward reaction rate increases. That is why we collect and remove Ammonia as soon as it is made

    The 2nd factor is a bit more complicating. We did talk about this in ideal gas that P ∝ n. So more gases moles or particles in the system, the higher the pressure. So when we increase the pressure, the system has to decrease it by having less molecules in the system. As we can see the left side has 4 Molecules of gas where as the right has only 2!

    The same concept is applied when we reduce the pressure, the system will try to increase it by having more particles in the system which is the left!

    The 3rd factor is also easy to understand - endothermic is a cooling effect which absorbs energy and exothermic is an heating effect which releases energy. So using Le Chatelier Principle we can explain these

    What about a Catalyst?

    Catalyst does speed the reaction but, it increases both the backward and forward reaction rate by the same amount. SO there is no net change. However, the equilibrium position or equilibrium is reached faster.

    In other words, this does not change the yield but affects the time taken to get this yield!

    No change in Equilibirium position

    Sometimes changing a factor may not affect its equilibrium position

    For example:

    H2(g) + I2(g) ⇋ 2HI(g)

    In this case both sides have 2 moles or molecules so an increase / decrease in pressure has no affect on the equilibrium position but it will affect the rate or the rate at which the equilibrium is reached

    Also keep in mind that, we need to count the number of GASES molecules on both side of the equaution and not other ones because, gas is the state which causes pressure (theoretically)

    Equilibrium Constants

    An equilibrium constant is the ratio between the products and the reactants. It has a constant value for a specific temperature

    There are 2 types of Equilibirium constants in the AS syllabus:

    1. Kc

    1. Kp

    We will see how we can calculate and the situations we use each constant for.

    Kc Constant

    This is how we calculate it:

     Kc=   ____________

    For example:

    H2(g) + I2(g) ⇋ 2HI(g)
     Kc=   ____________
    This is known as the equilibrium expression

    The brackets are used to show that we take the concentration of each gas in the equilibrium and the powers is the mole ratio of that substance

    We use this only when the number of moles or concentration of each product and reactant is given

    The above equation is the equilibrium expression and the value is the Kc value

    But as these are gases, we know gases exert pressure so we can introduce another constant

    Kp Constant

    This is only used when the PARTIAL Pressures of the gases in the system are given

    What are partial pressures?

    It is the pressure exert by a single type of gas in a mixture of gas

    For example in a container of gas, there is HI gas, I2 and H2 gas in the mixture. Each of them exert a specific amount of pressure. For example, if the pressure of the container is 10Pa and, the partial pressure of I2 and H2 is 2pa each. So the partial pressure of HI should be 6Pa

    As we know, the pressure is directly proportional to the number of moles so we can calculate the partial pressures if the number of moles of each substance is given

    Partial Pressure = ( Number of Moles/Total Moles in the System ) * Total pressure

    For example in the container, there is 2 moles of I2 and 2 moles of H2 and 4 moles of HI. The total pressure is 10Pa. Find the partial pressure of HI:

    Partial Pressure of HI = ( 4/8 ) * 10 = 5 Pa

    After finding these partial pressures for each substance, we need to substitue them to a similar expression

    H2(g) + I2(g) ⇋ 2HI(g)
     Kp=   ____________
             P1H P1I
    This is known as the equilibrium expression
    The P2HI means partial pressure of HI to the power 2

    So we only use this for gases.

    Units for Kc and Kp

    The units can change from equation to equation

    The units on both side of the equilibrium expression must be the same

    For example:

     Kc=   ____________
    In the form of units:
     Kc=   ____________
           moldm-3 * moldm-3		  

    So after cancelling we get 1/moldm-3 which is the same as mol-1dm3

    The same could be applied for other reactions also but always remember that this can be different and sometimes there could be no units also

    The same can be applied for Kp

    H2(g) + I2(g) ⇋ 2HI(g)
     Kp=   ____________
             P1H P1I
     Kp=   ____________
             Pa * Pa
    so the final units will be 1/Pa or Pa-1

    Exam Examples & Questions

    We will see now some common questions in exams

    In a container of volume 2dm3, 0.1 moles of I2 and 0.1 moles of H2 are left to reach equilibrium. When equilibrium is reached there is 0.1 moles of HI. Find the Kc

    We will put this in a proper format:

                         H2(g)              I2(g)                      2HI(g)
    Before               0.1                0.1                        -
    At equilibrium        X                  X                        0.1

    So we don't know the final moles of the reactant. To do this we first need to identify the mole ratio.

    It is 1 : 1 : 2

    So any moles reacted or lost in H2 and I2 must be doubled to form HI

    So HI gains by 0.1 Moles that means H2 and I2 must decrease 0.05 moles to form 0.1 moles of HI (Basic mole ratio).


                         H2(g)              I2(g)                      2HI(g)
    Before               0.1                0.1                        -
    At equilibrium      0.05                0.05                       0.1

    As 0.05 moles is reacted from the H2 and I2

    So after you find the moles for each substance, then find the concentration - remember the volume remains the same for all concentration as it is common for each

    Then after finding the concentration. Find Kc

    Kp questions are much simpler but, we recommend you practice more questions

    End of Chapter Questions
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