Chemical Equilibrium

Equilibrium – A state of balance Static Equilibrium – When a meter stick is balanced in the middle neither end is moving this is an example of static equilibrium

Dynamic Equilibrium – When a person is running on a treadmill they are moving forward while the belt is moving forward but over all neither is progressing in either direction as the two movements balance each other this is known as Dynamic equilibrium

In chemical reactions a state of dynamic equilibrium exists When magnesium ribbon is burned in air, a bright light is given off and a white substance called magnesium oxide is formed 2Mg + O2 2MgO It is very difficult to turn the magnesium oxide back into magnesium and oxygen chemists often say the reaction has “gone to completion” meaning the reactants are almost completely turned to products.

Many reactions do not go to completion German chemist Fritz Haber discovered that nitrogen gas and hydrogen gas react to form ammonia Ammonia is used in the manufacture of fertilizers, explosive sand nylon

Ammonia production Haber + THE NITROGEN AND HYDROGEN ARE CONTINUALLY FORMING AMMONIA WHILST THE AMMONIA IS CONTINUALLY DECOMPOSING TO FOR NITROGEN AND HYDROGEN

The above reaction is reversible ie The above reaction is reversible ie. It takes place in the forward and backward direction Such reactions are said to be equilibrium reactions and are indicated using the equilibrium arrow sign

When a reaction is in dynamic equilibrium the rate of forward reaction is the same as the rate of backwards reaction This means that for the above reaction there is no change in the overall amounts of either nitrogen or hydrogen or ammonia present in the reaction system

Definition Chemical equilibrium is a state of dynamic balance where the rate of the forward reaction equals the rate of the reverse reaction NB examine figure 17.2 p233

Le Chatelier’s Principle Many industrial chemical processes involve equilibrium reactions It is important for chemists to understand these so they can try to shift the balance to the right ( ie. To maximise the amount of product formed) French Chemist Henri Le Chetalier studied equilibrium reactions and put forward a rule known as Le Chetalier’s Principle

Changes in Temperature Increase in temp an endothermic reaction is favoured Decrease in temp an exothermic reaction is favoured H2 + I2 → 2HI ΔH = -12.6 kJ/mol 2HI → H2 + I2 ΔH = +12.6kJ/mol

Changes in concentration H2 or I2 (increased) SHIFTS TO THE RIGHT to make more 2HI HI (increased) SHIFTS TO THE LEFT to make more H2 or I2 A decrease in the concentration of HI has the opposite effect , hydrogen and iodine react forming more HI SHIFTS TO THE RIGHT A decrease in H2 or I 2 SHIFTS TO THE LEFT

Adding a catalyst It has no effect on the on the equilibrium A catalyst increases the rate of the reaction by lowering the activation energy and the activation energy for the reverse reaction is lowered to the same extent.

Changes in pressure Pressure changes only effect gases. Increase in pressure the reaction that produces less molecules is favoured N2 + 3H2 produces 2NH3 A decrease in pressure the reaction that produces more molecules is favoured

Changes in the vol of the container An increase in volume decreases the pressure. A decrease in volume increases the pressure

Le Chatelier’s Principle To demonstrate the effect of temp and concentration changes CoCl42- + 6H2O Co(H2O)62+ 4Cl- Blue Red Cr2O72- + H2O 2CrO42- + 2H+ Orange Yellow FeCl3 + CNS- Fe(CNS)2+ + 3Cl- Yellow Red Adding acid shifts to left and adding base shifts to right

Industrial applications of Le Chatelier’s Principle Manufacture of Ammonia by the Haber Process - Ammonia is one of the most important chemicals made by the chemical industry 80% of ammonia is used to make fertilisers It is made by reacting nitrogen gas (from air) with hydrogen gas (from natural gas) in the presence of an iron catalyst

Reaction iron catalyst Nitrogen + hydrogen ammonia N2 + 3H2 2NH3 H = -92.4kj/mol The amount of ammonia made depends on the temperature and pressure inside the vessel

Using Le Chatelier’s principle Fritz Haber selected the right conditions for the manufacture of ammonia There are 4 moles on the left hand side but only 2 moles on the right hand side using le Chatelier’s principle this tells us a good yield of ammonia will be obtained at high pressure NB: A high pressure plant is expensive to build and maintain a compromise pressure of 200 atmospheres is used in practice

The reaction is exothermic going from left to right, therefore decreasing the temperature will push the reaction in this direction However if temperature is too low the rate of reaction becomes too slow and even though more ammonia will be formed it will be formed much more slowly A compromise temperature of 500⁰C is used in practice

Le Chatelier’s principle tells us that manufacture of ammonia by the Haber process is best achieved at high pressure and low temperature The iron catalyst does not change the equilibrium but is very important as it helps the system reach equilibrium much more quickly

Manufacture of Sulfuric Acid by the Contact Process Sulfuric acid is used in the manufacture of many materials such as paints, detergents, fertilizers, plastics, fibres, car batteries etc. It is manufactured in the Contact process which involves passing sulfur dioxide gas and oxygen gas over a catalyst There needs to be very close contact between the catalyst and the two gases for the reaction to proceed satisfactorily

Reaction vanadium pentoxide Sulfur dioxide + Oxygen sulfur trioxide 2SO2 + O2 2SO3 H = -196 kj/mol The sulfur trioxide formed is reacted with water to make sulfuric acid NB: Platinum could also be used as a catalyst but is found to be easily poisoned by impurities in the reactants

Le Chatelier’s Principle indicates high pressure would favour the formation of product This is true but due to high costs a pressure just above atmospheric pressure is used in practice Another reason is that high pressure tend to liquefy sulfur dioxide

As the reaction is exothermic from left to right low temperature will drive the reaction in this reaction However too low a temperature will greatly slow the rate of the reaction so a compromise temperature of 450⁰C is used as it is found the catalyst works best at this temperature

The equilibrium constant (Kc) When a system reaches a state of equilibrium chemists and chemical engineers need to know the concentrations of reactants and products that exist in the equilibrium mixture This can be found either by experiment or by performing calculations

Law of Chemical Equilibrium Provided the temperature remains constant there is a simple mathematical relationship between the concentrations of reactants and products in an equilibrium mixture Consider the reaction where a moles of reactant A react with b moles of reactant B to form c moles of reactant C and d moles of reactant D

The equilibrium constant

For the reaction between nitrogen and hydrogen to form ammonia the Equilibrium constant expression is N2 + 3H2 2NH3

When writing Equilibrium constant expressions ............. Products go above the line and reactants below the line The concentrations of the products are multiplied together, the concentration of the reactants are multiplied together (multiplication sign is left out) The concentration of all species is raised to the power of their coefficients in the balanced equation

Every Equilibrium reaction has its own value for Kc at the particular temperature at which the reaction takes place At 458⁰C the value of Kc for the reaction H2 + I2 2HI is 46 This constant value is always obtained at this temperature no matter what concentrations of hydrogen + iodine are reacted

NB: In the above example the value of Kc at 46 is quite large it tells us that the numerator is much greater than the denominator (ie the reactants “went into” the product 46 times!) This indicates how far the reaction is pushed to the right, the larger the value of Kc the more the reaction is pushed towards products A low value of Kc indicates only a small fraction of reactants have formed products

2 SO2 + O2 2SO3 At equilibrium the conc of SO2 , O2 and SO3 were 0.07 mol/l 0.035mol/l and 0.03mol/l respectively. Calculate the value of the equilibrium constant KC for this reaction Kc = [SO3] (0.03)2 [SO2]2[O2] (0.07)2(0.035) KC = 5.25mol/l