Equilibrium Syllabus > equilibrium (sl) > 8.2

# 8.2 - Position of equilibrium

8.1.1: State the equilibrium constant expression (Kc) for a homogeneous reaction. Consider equilibria involving one phase, gases or species in aqueous solution. The equilibrium constant is specific to a given system and varies with temperature. No calculations are required.

The equilibrium constant

This is given the symbol Kc (not to be confused with the rate constant K) and gives us a measure of the positions of an equilibrium.

For the equilibrium:

x moles reactants y moles products

It is given by the formula:

Kc = [products]x/[reactants]y

where x and y are the stoichiometries of the products and the reactants respectively (i.e. the balancing numbers in the equation)

If the value of Kc is large then this means that the products concentration is much larger than the reactants concentration at equilibrium and so the equilibrium lies to the right hand side (i.e. there is a much greater concentation of the products than the reactants)

If the value of Kc is small then the equilibrium lies to the left hand side - i.e. there is a much greater concentration of reactants than products.

Example:

For the Haber Process equilibrium

N2 + 3H2 2NH3

The equilibrium law gives the expression:

 Kc = [NH3]2 [N2] x [H2]3

In one experiment a mixture of H2, N2 and NH3 was allowed to reach equilibrium at 472°C. The concentration of gases at equilibrium was analysed and found to contain the following:

 component concentration hydrogen 0.1207M nitrogen 0.0402M ammonia 0.00272M

What value did Haber come up with for the equilibrium constant, Kc?

 Kc = [0.00272]2 [0.0402] x [(0.1207]3

Kc = 0.105

We can see that under the conditions of this reaction the equilibrium lies to the left, not much ammonia will be present at equilibrium.

8.1.2 Deduce the extent of a reaction from the magnitude of the equilibrium constant. When Kc >> 1, the reaction goes almost to completion, When Kc << 1, the reaction hardly proceeds.

The extent of equilibrium

This section is fairly straightforward and just requires the student to appreciate the relationship between the actual values of the equilibrium constant Kc and the extent of the reaction. When the syllabus statement says 'goes to completion' it means that the reaction goes all the way to the products and that effectively no reactants are left, or that the quantity of reactants remaining at equilibrium is so small that we can forget about it.

Likewise, to say that 'the reaction hardly proceeds' means that to all intents and pruposes there is no reaction. At equilibrium virtually all of the reactants remain and the quantity of products formed is so tiny that we can say that there is effectively no reaction.

8.1.3 Describe and predict the qualitative effects of changes of temperature, pressure and concentration on the position of equilibrium and the value of the equilibrium constant. Use Le Chatelier's principle to predict the effects of these changes on the position of equilibrium. The value of the equilibrium constant (Kc) is only affected by temperature. The position of equilibrium may change without the value of Kc changing.

The effect of changing conditions on an equilibrium

Reactions are subject to the effects of changing conditions (i.e. initial concentrations, pressure and temperature) careful consideration of the dependence of the forward and backward reactions involved in the equilibrium on these conditions leads to an understanding of the effect of changing them.

The effect of Concentration changes

For a system at equilibrium the concentrations of both reactants and products are constant (but not the same). The value of Kc depends on these concentrations and Kc is constant unless the temperature is changed.

If a further quantity of reactant is added to the mixture already at equilibrium then the value of [products]/[reactants] no longer equals the value of Kc and the equilibrium must make adjustments to reestablish the equilibrium concentrations.

In this example the value of the ratio [products]/[reactants] is too small as more reactants have been added. The only way to adjust this is by making more product (and at the same time using up some of the reactants)

The equilibrium responds by making more product until the equilibrium ratio is reestablished at the Kc value.

In summary the effect of adding reactants has been to make more product.

This gives a general rule.

• If we add to the left hand side of an equilibrium we make more of the right hand side
• If we add to the right hand side of an equilibrium we make more of the left hand side
• If we remove from the left hand side the equilibrium makes more of the left hand side component
• If we remove from the right hand side the equilibrium makes more of the left hand side component

Explanation

All of this can be easily explained by considering the rates of the forwards and back reactions.

At equilibrium the rates of the forward and back reactions are equal

rate forwards = rate backwards

When we study rates of reaction one of the first conclusions drawn is that the rate of a chemical reaction depends on the concentrations of the reactants. The greater the concentration the more collisions occur and the faster the rate of the reaction.

When we add more reactant the forward rate will now be greater than the back rate. The reaction is now not at equilibrium and the forward reaction proceeds faster than the back reaction until the equilibrium conditions are reestablished.

Similiarly removal of a component from one side will reduce the rate of its reaction and case the equilibrium to make more of it.

In all of these changes the value of Kc remains unchanged. The equilibrium is temporarily disturbed and then the equilibrium concentrations are re-established.

The effect of temperature change

Reactions are either exothermic or endothermic i.e. when a reaction proceeds it does so accompanied by either a release of energy or an absorption of energy. This is normally shown on an energy coordinate graph.

If the following represents the reactants and products in an equilibrium:

x moles reactants y moles products

 Endothermic change Exothermic change

You can see that the activation energy needed for an endothermic change is much greater than for the corresponding reverse exothermic change.

If a reaction is exothermic in the forward direction in an equilibrium it will be endothermic in the reverse direction and vice-versa (law of conservation of energy)

Increasing the temperature of an equilibrium mixture will always favour the endothermic direction over the exothermic process. The rate of the endothermic process will increase more than the exothermic direction of change and the equilibrium will be re-established with new concentrations based on more of the endothermic product (i.e. the product of the endothermic direction of change)

To express this in non-scientific terms it can be considered that the endothermic direction is absorbing heat therefore giving it more heat by increasing the temperature will favour it.

The reverse argument is true for the direction of exothermic change. The exothermic reaction is giving out heat and therefore applying more will hinder this process. Raising the temperature retards the reaction in the exothermic direction.

This will become clearer with an example.

 Example: For the Haber Process equilibrium N2 + 3H2 2NH3 enthalpy change (forward direction) = -92 kJ (negative sign = exothermic) The reaction is exothermic in the direction of ammonia formation. Increasing the temperature will therefore hinder the forward reaction and favour the reverse reaction. Increasing the temperature will produce less ammonia at equilibrium.

The effect of pressure change

Pressure change only affects gaseous equilibria and only then when there are an unequal number of moles of gas on either side of the equilibrium.

When the pressure is increased on a gaseous mixture at equilibrium the equilibrium will respond so as to release the applied pressure. In other words the equilibrium will move towards the side of the fewer number of moles.

It should be remembered that the pressure of a gas is directly proportional to the number of moles of gas and that this does not depend on the nature of the gas in question - all gases behave in the same way in terms of their physical properties. see section 5 states of matter.

Again this can be made clearer by considering an example.

 Example: For the Haber Process equilibrium N2 + 3H2 2NH3 four moles gas two moles of gas There are four moles of gas on the left hand side and only two moles of gas on the right hand side. Increasing the pressure will move the equilibrium to the right hand side and have the effect of releasing the pressure.

This is rather a difficult concept to provide a satisfactory simple answer. However it is enough to know that the equilibrium responds to the change in conditions by copunteracting the change in conditions (i.e. increase the pressure and it responds releasing the pressure by making a fewer number of moles)

The actual explanantion involves considering the consequences of changing the pressure of a gas.

If the number of moles of gas is constant and the temperature doesn't change then the only way to change the pressure is to change the volume of the container.

If the volume of the container changes then the concentration (moles/volume) must also change. Hence the effect of changing the pressure is a corresponding change in the concentration - double the pressure and the concentration doubles.

If the equilibrium expression has a different number of moles on the top and the bottom then doubling the concentration of all the components will change the ratio of the equilibrium expression. The equilibrium expression no longer equals Kc and so the system responds to re-establish the ratio to be equal to Kc again.

Example

Consider the equilibrium:

2NO2 N2O4

The equilibrium law:

 Kc = [N2O4] [NO2]2

If we represent the equilibrium concentrations as.... [N2O4] = x ; and [NO2] = y

then at equilibrium:

Kc =x/y2

If we double the pressure then the concentration of both [N2O4] and [NO2] will also double i.e. the new concentrations are 2x and 2y

The equilibrium expression ratio is now:

 K'c = 2x 4y2

8.1.4 State and explain the effect of a catalyst on an equilibrium reaction.

Catalysts

A catalyst offers an alternative mechanism for the reaction with a lower activation energy. As the energy required for reaction is lower it is easier to achieve and therefore gives a faster reaction.

8.1.5 Describe and explain the application of equilibrium and kinetics concepts to the Haber process and the Contact process.

The Haber process

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