Classwork Series and Exercises {Chemistry – SS2}: Chemical Equilibrium

Chemistry SS 2 Week 4

Topic: Chemical Equilibrium

Introduction

Chemical equilibrium has been reached in a reaction when the rate of the forward reaction is equal to the rate of the reverse reaction. When a chemical reaction has reached equilibrium, collisions are still occurring: the reaction is now happening in each direction at the same rate. This means that reactants are being formed at the same rate as products are being formed, and this is indicated by double arrows. At equilibrium, the reaction can lie far to the right, meaning that there are more products in existence at equilibrium, or far to the left, meaning that at equilibrium there are more reactants. The concentration of the reactants and products in a reaction at equilibrium can be expressed by equilibrium constant, symbolized K or K_eq

For the general reaction

 

In the above expression, the brackets, as always, symbolize the concentration of the reactants and products in molarity. However, while in the above expression we used the plain symbol K to symbolize the equilibrium constant, there are several types of equilibrium constants. For example, K_c symbolizes the equilibrium constant in an aqueous solution, K_p symbolizes the partial pressures of gases in equilibrium, and K_sp symbolizes the solubility product of solids classified as insoluble. K values have no units, and a K > 1 means that the reaction favours the products at equilibrium, while a K < 1 means that the reaction favours the reactants at equilibrium.

Here are a couple of rules to follow when using equilibrium constant expressions:

1. Pure solids do not appear in the equilibrium expression.
2. Pure liquids do not appear in the equilibrium expression.
3. Water, either as a liquid or solid, does not appear in the equilibrium expression.
4. When a reactant or product is preceded by a coefficient, its concentration is raised to the power of that coefficient in the K_eq expression.
5. When the K_eq of a reaction has been multiplied by a number, the K is raised to the power of the multiplication factor (Kⁿ), so if it has been multiplied by 2, K is squared, if it has been multiplied by 3, K is cubed, and so on.

Reversible Reactions

A reversible reaction is one which can be made to go in either direction depending on the conditions.

If you pass steam over hot iron the steam reacts with the iron to produce a black, magnetic oxide of iron called triiron tetroxide, Fe₃O₄.

These reactions are reversible, but under the conditions normally used, they become one-way reactions. The products aren’t left in contact with each other, so the reverse reaction can’t happen.

Reversible reactions happening in a closed system

A closed system is one in which no substances are either added to the system or lost from it. Energy can, however, be transferred in or out at will.

In the example, iron being heated in steam in a closed container. Heat is being added to the system, but none of the substances in the reaction can escape. The system is closed.

As the tri-iron tetroxide and hydrogen start to be formed, they will also react again to give the original iron and steam. This has established what is known as a dynamic equilibrium.

During a dynamic equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This means that both reactants and products will be present at any given point in time.

Le-Chatelier’s Principle

Once an equilibrium is established, no further change is apparent as long as the external conditions remain unchanged.

On changing the external conditions, whether the equilibrium will shift towards the reactants or the products can be predicted by the application of Le-Chatelier’s principle which states that when a stress (change in concentration, temperature or total pressure) is applied to a system at equilibrium, the system readjusts so as to relieve or offset the stress. In other words, it basically states that if stress is applied to a system at equilibrium, the position of the equilibrium will shift in the direction that reduces the stress to reinstate equilibrium.

Application of Le-Chatelier’s Principle

In the synthesis of Ammonia (Haber’s process)

In the example above, if more reactants are added to the system, the reaction will shift in the forward direction, and if more products are added, the reaction will shift in the reverse direction. If the reaction is endothermic and heat is added, the reaction will shift to the right. If heat is added to the system and the reaction is exothermic, heat should be thought of as a product and the reaction will shift to the left The addition of pressure will cause a shift in the direction that results in the fewer number of moles of a gas, while if pressure is relieved, the reaction will shift in the direction that produces more moles of a gas.

There are numerous types of equilibrium problems one may encounter, both qualitative and quantitative.

Qualitative Problems

2 N₂ (g) + 3 H₂ (g) 2 NH₃ (g)

Suppose additional nitrogen (N₂) was added to the system. First, identify the stress. The stress is too much nitrogen. The equilibrium will shift to the right, towards products, in order to remove the excess nitrogen.

Suppose some ammonia (NH₃) was removed from the system. The stress is not enough ammonia, so the equilibrium will shift to the right to replenish the ammonia removed.

Suppose some hydrogen (H₂) was removed from the system. The stress is not enough hydrogen, so the equilibrium will shift to the left to replenish the hydrogen removed.

Suppose a catalyst is added to the system. Since the catalyst is not a part of the equilibrium, the equilibrium will not shift. The only affect that a catalyst has on an equilibrium is to allow the system to reach equilibrium faster.

Suppose an inert gas, such as helium, as added to the system. Since helium is not a part of this equilibrium, it will not affect the equilibrium.

Quantitative Problem

By examining the amounts of reactants and products at equilibrium, one may numerically see how an equilibrium favours either reactants or products.

For example: Suppose we are given the following equilibrium at 500 K:

CO(g) + 2 H₂ (g) CH₃OH(g)

The equilibrium concentrations are: [fusion_builder_container hundred_percent=”yes” overflow=”visible”][fusion_builder_row][fusion_builder_column type=”1_1″ background_position=”left top” background_color=”” border_size=”” border_color=”” border_style=”solid” spacing=”yes” background_image=”” background_repeat=”no-repeat” padding=”” margin_top=”0px” margin_bottom=”0px” class=”” id=”” animation_type=”” animation_speed=”0.3″ animation_direction=”left” hide_on_mobile=”no” center_content=”no” min_height=”none”][CO] = 0.0911 M, [H₂] = 0.0822 M, [CH₃OH] = 0.00892 M, what is the value of the equilibrium constant? Does the equilibrium favour reactants or products?

Solution:

First, we need to write the mass action expression:

Since the value of the equilibrium constant is greater than one, (K_eq >1), the equilibrium favours the products.

Another type of equilibrium problem deals with finding the equilibrium concentrations given the initial concentrations and the value of the equilibrium constant. Suppose you are given the following equilibrium:

Example 2: CO(g) + H₂O(g) CO₂ (g) + H₂ (g) K_eq = 23.2 at 600 K

If the initial amounts of CO and H₂O were both 0.100 M, what will be the amounts of each reactant and product at equilibrium?

Solution:

For this type of problem, it is convenient to set a table showing the initial conditions, the change that has to take place to establish equilibrium and the final equilibrium conditions.

The amounts of reactants and products present at equilibrium will be the combination of the initial amounts and the change. Just add the quantities together:

Multiply both sides by the denominator, 0.100 – x:

Factors Affecting Chemical Equilibrium

1.   Effect of change of concentration

By increasing or decreasing the concentration of one of the reactants or products, the reaction can be manipulated in the desired direction.

To increase the yield of NH3 the concentration of N₂ or preferably H₂ should be increased in the synthesis of ammonia.

2.  Effect of change of pressure

Reactions which proceed with decrease in volume are favoured by high pressures.

4.  Effect of catalyst

The presence of a catalyst in a reversible reaction does not affect the state of equilibrium but only causes the reaction system to attain equilibrium in a shorter period of time.

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