Chemical Reactions and Equilibrium Acid/Bases Redox. Revision What is a chemical reaction? What factors affect the rate of reaction?

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1 Chemistry Fundamental Topics These notes will provide a brief coverage of topics that will be important for your Course in instrumentation. The notes are supplementary to the Instrumentation notes and will provide an understanding of many of the terms and concepts covered. The topics will include: Chemical Reactions and Equilibrium Acid/Bases Redox Revision What is a chemical reaction? What factors affect the rate of reaction? What is collision theory? Chemical Reactions The collision theory explains how and when a chemical reaction will occur. For a chemical reaction to occur the particles must collide in the correct orientation and with the appropriate amount of energy. E a Energy Reactants Products Progress of reaction Fig Reaction path for an exothermic reaction Where E a is the Activation Energy or the energy required for a reaction to go. An exothermic reaction is a reaction that gives out energy to the environment 1

2 ow does the energy compare between the reactants and the products? An endothermic reaction is one where the products have more energy than the reactants Draw a diagram of an endothermic reaction in the space available Where does the energy come from for an endothermic reaction.. Remember energy cannot be created nor destroyed! Reactions in Equilibrium Reactions studied to date have been reactions that go to completion ie they are not reversible aa bb cc dd C 4 2O 2 CO O Many reactions are reversible. This means they can go from left to right and also from right to left. They are indicated with double arrows. aa bb cc dd Many physical changes are easily reversible such as the freezing of water to form ice, which can easily be melted by applying some heat. Write the equation to show the process of freezing and melting as one equation 2

3 A famous reversible chemical reaction is the aber Process for the production of ammonia N 2(g) 3 2(g) 2N 3(g) In chemistry equilibrium refers to the exact balancing of two processes, with one being the opposite of the other. An example of this is a sealed bottle of water. As long as the lid is left on the water level stays fairly constant, but if the lid is removed the equilibrium is disturbed and the water evaporates. Write the equation to represent water in a sealed bottle. Characteristics of a chemical reaction An equilibrium reaction is reversible ie the reactants in the forward reaction are the products in the reverse reaction Reactants Products An equilibrium system is closed ie no substance can be added or removed In an equilibrium system both forward and reverse reactions are occurring at the same rate. An equilibrium system is dynamic on the particle level In an equilibrium system the concentrations of all species remains constant even though both forward an reverse reactions are occurring, since these are happening at the same rate In an equilibrium system the temperature and pressure conditions are constant Draw diagrams to show graphically equilibrium in relation to reaction rate, concentrations etc 3

4 The Equilibrium constant The Law Of Chemical Equilibrium states that, when a reaction reaches equilibrium, the product of the concentration of each substance on the right hand side of the equation raised to the power of their co-efficients, divided by the product of the concentration of each substance on the left hand side raised to the power of their co-efficients is a constant, at a given temperature. This is known as the equilibrium constant, K. Thus, for a general equilibrium reaction, aa bb cc dd K = [C] c [D] d [A] a [B] b The square brackets indicate concentration in moles per litre. Using the change of ozone (O 3 ) an important atmospheric pollutant to oxygen (O 2 ) as an example of how to construct a chemical equilibrium expression. The equation is 2O 3 (g) 3O 2(g) Coefficient Reactant Product To obtain the equilibrium expression, place the concentration of the product on the top, and the concentration of the reactant on the bottom. [O 2 ] Product [O 3 ] Reactant Then use the co-efficients as powers K = [O 2 ] 3 [O 3 ] 2 Co-efficients become powers Write the equilibrium constant expressions for the following: 1. When nitrogen gas (N 2 ) and hydrogen gas ( 2 ) are reacted together, they form an equilibrium with the product, ammonia gas (N 3 ). 2. The reaction of sulfur dioxide (SO 2 ) with oxygen to produce sulfur trioxide (SO 3 ). 4

5 The meaning of K What does the numerical value of K mean? Because it is a ratio of concentrations, we can tell something about the equilibrium position, and whether it favours the products or the reactants. If K is large (> 1), then the top of the ratio is larger than the bottom, so the concentration of the products must be greater than the reactants. The opposite case applies whether K is small (< 1). The value of K is fixed for a given reaction at a given temperature. Its value is not affected by concentration or pressure. Le Chatelier s Principle When a system at equilibrium is subjected to a change in temperature, pressure, concentration or volume, the equilibrium position may undergo a spontaneous change (shift) in such a direction so as to minimise the imposed change. Ie it reacts to change what was done to it! The shifts can take many hours to complete until a new equilibrium position is reached and the rates of the forward and reverse reactions are once again equal Examples a) If the temperature of a system is increased, the equilibrium position will automatically shift in the direction which uses up heat (absorbs heat). If the temperature of a system is decreased, the equilibrium position will automatically shift in the direction that Example Fe 3 (aq) SCN (aq) FeSCN 2 (aq) energy (pale yellow) (colourless) (brown-red). i) ow does heating affect the colour of the system? Adding heat favours the reverse reaction which uses up heat (energy). The equilibrium position shifts to the left until a new equilibrium position is reached at the higher constant temperature. The concentration of FeSCN 2 will decrease as the concentrations of Fe 3 and SCN - will increase in molar proportion. The colour of the system will lighten. ii) ow does cooling affect the colour of the system? Explain 5

6 b) If the pressure of a system is increased (more molecules of a gas from the system ae added or the volume is decreased), the equilibrium will shift in the direction that lowers the pressure (smaller number of molecules) If the pressure of a system is decreased, the equilibrium will shift in the direction that. Example: N 2 O 4(g) 2NO 2(g) (colourless) (brown) What happens to the colour when the pressure is increased? (Explain) Note: Changes in pressure will only affect the.state! c) If the concentration of a substance is increased, the equilibrium will shift in the direction that lowers the concentration of that substance If the concentration of a substance is decreased,.. Example: Fe 3 (aq) SCN (aq) FeSCN 2 (aq) What happens to the equilibrium position if: (i) more Fe 3 is added. (ii) More SCN is added (iii) A Fe 3 scavenger is added (ie it removes Fe 3 ). d) If the volume is increased, the pressure of any gaseous components will decrease, and the equilibrium will automatically shift in the direction that. Example: 2 O (l) 2 O (g) Use Le Chatelier s Principle to explain why wet carpets will dry out better with the car doors open than when they are shut! 6

7 Exercise Use Le Chatelier s Principle to predict the effects of changes in the following equilibrium systems. The aber process is used to manufacture ammonia, by the following equilibrium reaction, N N 3 heat Which reaction (forward or reverse) will be favoured by the following changes? (a) adding more N 2 (b) removing N 3 as it forms (c) increasing the temperature (d) decreasing the temperature (e) increasing the container volume (f) decreasing the container volume 7

8 ACIDS AND BASES Acids and bases are among the oldest recognised group of chemicals. Long before atoms and molecules, elements and compounds were understood, the properties and behaviour of certain substances led them to be grouped under these two headings. The table below lists some of the significant properties of acids and bases. Comparison of properties of acids and bases Property Acid Base Taste sour bitter Feel nothing characteristic soapy Reaction metals with Reaction with each other React with many metals, forming salt and hydrogen Formation of a solution lacking in characteristic properties - neutralisation Reaction with litmus turns blue litmus red React with only a few metals such as Al Formation of a solution lacking in characteristic properties - neutralisation turns red litmus blue Early theories such as the Arrhenius theory explained acids as those solutions that contained hydrogen ions. ydrochloric acid Cl (aq) (aq) Cl (aq) Sulfuric Acid 2 SO 4(aq) 2 (aq) 2- SO 4 (aq) Acetic (ethanoic) Acid C 3 COO (aq) (aq) C 3 COO (aq) Bases contained hydroxide ion. Soluble bases were called alkalis Sodium hydroxide (caustic soda) NaO (aq) Na (aq) O - (aq) Calcium hydroxide (limewater) Ca(O) 2(aq) Ca 2 (aq) 2O - (aq) Some common names for acids and bases Formula Systematic Name Common Name Cl hydrochloric acid muriatic acid 2 SO 4 sulfuric acid oil of vitriol NO 3 nitric acid NaO sodium hydroxide caustic soda Na 2 CO 3 sodium carbonate CaO calcium oxide lime KO potassium hydroxide caustic potash K 2 CO 3 potassium carbonate 8

9 This model of acids and bases is very limiting as: It only applies to aqueous solutions Does not account for substances like ammonia (N 3 ) which produce solution s with the characteristics of a basic solution yet N 3 contains no hydroxide ions Does not explain some substances which can act as both acids and bases. They are called amphiprotic. Substances which react with both acids and bases are amphoteric. A better theory was proposed in 1923 by two chemists, Lowry and Brönsted, who were working independently of each other in the same way that Mendeleev and Mayer produced almost identical periodic tables at the same time. owever, in this case, both chemists have gained recognition for their work: the Lowry-Brönsted theory acids and bases. A Lowry-Brönsted acid is any substance that donates a proton to another in a chemical reaction. A Lowry-Brönsted base is any substance that accepts a proton from another in a chemical reaction. Example 1 Ammonia hydrogen chloride ammonium chloride N 3 Cl (aq) N 4 Cl - ydrogen ion has been donated (ie lost by the Cl) ydrogen ion has been accepted In the above the Cl is donating a hydrogen ion to the N 3 ie Cl is an acid The N 3 is acting as a base as it accepts a proton and becomes N 4. Class exercise For the reversible (equilibrium) reactions identify the acid and base combination in both the forward and reverse directions. Or N 4 (aq) O - (aq) N 3(aq) 2 O (l) N O N O _ Ammonia (base) accepting a proton from water (acid) 9

10 2 O (l) C 3 COO (aq) 3 O (aq) C 3 COO - (aq) Water is an example of an amphiprotic substance ie it can accept and donate protons Class exercise In which of the following reactions is water acting as an acid? a) 2 O (l) CO 3 - (aq) O - (aq) 2 CO 3(aq) b) 3 O (aq) PO 4 - (aq) 2 O (l) 2 PO 4 - (aq) c) N 4 (aq) 2 O (l) N 3(aq) 3 O (aq) d) 3 O (aq) CN - (aq) CN (aq) 2 O (l) In each of the reactions above there are particles which are closely related- they only differ by one hydrogen ion ie N 4 / N 3 and 2 O / O - These particles are called conjugate acid/base pairs. Are there any other pairs in the reactions above? Acid and Base Strength Strong acids and bases are those that dissociate totally (break-up into ions) in solution. This means that when the strong acid or base is in solution, there is none of the original acid or base form left. The terminology is that the acid or base is totally ionised Examples Cl (aq) (aq) Cl - (aq) Or for the dissolution process Cl (aq) 2 O (l) 3 O (aq) Cl - (aq) This enables us to determine the concentration of the ions formed ie if the original acid concentration is 0.1M (for the Cl above) then the concentration of ions when in solution will be 0.1M for both the and also for the Cl - Other strong acids include nitric (NO 3 ) and sulfuric ( 2 SO 4 ) What does this mean for the concentration of the concentration in sulfuric acid 10

11 Calcium hydroxide (Ca(O) 2 )is an example of a strong base. Write the equation for the dissociation of calcium hydroxide If the concentration of the calcium hydroxide solution is 0.10M then the concentration of hydroxide ions is 0.20M (or 1:2 from the equation) Other examples of strong bases are most metal hydroxides such as sodium hydroxide (NaO) and potassium hydroxide (KO). Weak acids and bases only partially dissociate into ions in aqueous solutions. An equilibrium forms between the original acid and base particles and the ions in solution. Concentrations of the ions formed are therefore less than the original concentrations of the acid or base would suggest. An example of a weak acid is ethanoic acid (acetic). All organic acids are weak and will only partially dissociate. Or for the aqueous dissociation C 3 COO (aq) (aq) C 3 COO - (aq) C 3 COO (aq) 2 O (l) 3 O (aq) C 3 COO - (aq) Analysis of the concentrations of the species in solution indicates that only 10% of the ethanoic acid dissociates. ie a 0.1M ethanoic acid solution contains only about 0.01M hydrogen ions Strong Acids Strong Bases Weak Acids Weak Bases Cl Br 2 SO 4 NO 3 ClO 4 all soluble metal hydroxide salts, e.g. NaO, KO, Ca(O) 2 Representation of strong and weak F all alkanoic acids, e.g. ethanoic acid 2 CO 3 carbonic acid N 3 metal salts of carbonate and hydrogen carbonate ions STRONG WEAK molecular form dissociated form 11

12 A common example of a weak base is ammonia N 3. N 3(aq) 2 O (l) N 4 (aq) O - (aq) Other weak bases are sodium carbonate (washing soda) and sodium hydrogen carbonate (baling soda). It is the carbonate and hydrogen carbonate ions that act as bases CO 3 - (aq) 2 O (l) 2 CO 3(aq) O - (aq) Other terms associated with solutions of acids an bases are: Concentrated Dilute These relate to the amount of solute that is in the solution and should not be confused with the terms strong and weak A 0.001M Cl solution is a dilute solution of a strong acid A 10M solution of Na 2 CO 3 is a concentrated solution of a weak base. p the Chemists Shorthand Water dissociates to a very small extent at 25 o C 2 O (l) (aq) O - (aq) An equilibrium constant can be written for this reaction, (with the omission of 2 O (l) as pure solvents are not included, nor are solids in an equilibrium expression). K w = [ (aq)] [O - (aq)] = 1 x at 25 o C In pure water the concentration of hydrogen ions and the concentration of hydroxide ions are equal ie [ (aq)] = [O - (aq)] = 1.0 x 10-7 M When an acid is added to the water the concentration of hydrogen ions increases as the concentration of hydroxide ions falls, however both ions are present [ (aq)] > [O - (aq)] but, the product of the concentrations of each ion still equals 1.0 x When a base is added to water, the concentration of hydroxide ions rises as the concentration of hydrogen ions falls [ (aq)] < [O - (aq)] At temperatures other than 25 o C the equilibrium constant K w value will change. As the concentrations of and O - ions in water solutions are mostly very small, chemists have developed a system using base 10 logarithms to give more manageable numbers 12

13 This system uses p values where: px = - log 10 X p = - log 10 [ ] or conversely [ ] = 10 -p M po = - log 10 [O - ] or conversely [O - ] = 10 -po M Example If the concentration of hydrogen ions in a solution is 0.01 M the p = -log 10 [0.01] =2 For pure water at 25 o C [ ] = [O - ] = 1 x 10-7 Therefore p = - log 10 [10-7 ] = 7 and po = - log 10 [10-7 = 7 thus p po = = 14 In the laboratory most solutions have values between 0 and 14 p units Exercises Calculate the p for the following: [ ] = 0.10 M [ ] = 10-4 M [ ] = 2.4 x 10-3 M [ ] = 5.8 x 10-5 M [ ] = 3.0 x M Calculate the [ ] for the following p = 3 p = 9 p = 1.5 p =12.2 p = 0 Write the dissociation equation for each of the following strong acids and then calculate the p of the solution M Cl 0.10 M 2 SO M NO 3 Find the p of the following solutions 0.10 M NaO M Ba(O) M KO M Ca(O) 2 13