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Chemical Equilibrium Equations, balancing and equilibrium.

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Presentation on theme: "Chemical Equilibrium Equations, balancing and equilibrium."— Presentation transcript:

1 Chemical Equilibrium Equations, balancing and equilibrium

2 Equations: A chemical equation uses the chemical symbols and formulas of the chemicals that are used and made by the reaction and other symbolic terms to represent states of matter and reaction conditions.

3 Equations The reactants the chemicals that are put into the reaction. They are placed on the left of the arrow to signify that. The products are the chemicals that are made by the reaction. They are placed on the right of the arrow Reactants  Products

4 Examples A  B PbO + CaS  PbS + CaO NaCl + KNO 3  NaNO 3 + KCl

5 Examples A  B PbO + CaS  PbS + CaO NaCl + KNO 3  NaNO 3 + KCl Reactant Product

6 Equations Conditions required to carry out the reaction may be placed over the arrow. These can include: Heat Acid (H + ) Time for reaction Catalysts

7 Balancing the equations Coefficients are placed in front of formulas to balance the equation and to indicate the number of formula units (atoms, molecules, moles, ions) of each substance reacting or being produced. When no number is shown it is understood that one formula unit of the substance is indicated. This is known as Stoichiometry

8 Balancing Equations The total number of each atom on the left and the right of the periodic table must be the same Therefore when we do reactions the two sides must be balanced We do this by using Stoichiometry (Fancy word for number-in-front-of-chemical)

9 Balancing Equations The easiest way to do these is to balance each atom in the order of ‘least occurring’ and finishing with the atoms that occur in the most chemicals The overall rules can be summarised:

10 Balancing Equations 1. Count and compare atoms on either side to determine which atoms need to be balanced. 2. Select the most complicated formula (the one with the most atoms other than O or H). 3. Treat polyatomic ions as individual units if they occur on either side of the equation. 4. Balance each atom or polyatomic ion by placing coefficients in front of the formula including the unbalanced atom or ion. 5. Check that balancing a formula with a coefficient, has not unbalanced other atoms. Make adjustments as necessary. 6. ALWAYS do a final check making sure each atom or polyatomic ion is balanced and the smallest set of whole numbers has been used.

11 Balancing Equations Examples: S + O 2  SO H 2 + O 2  H 2 O Ca + H 2 O  Ca (OH) 2 + H 2

12 Balancing Equations Examples: 2 S + O 2  2 SO 2 H 2 + O 2  2 H 2 O Ca + 2 H 2 O  Ca (OH) 2 + H 2

13 Balancing Equations Examples: LiCl + Ca(OH) 2  LiOH + CaCl 2 FeBr 3 + H 3 PO 4  FePO 4 + HBr PbCl 2 + H 2 SO 4  PbSO 4 + HCl

14 Balancing Equations Examples: 2 LiCl + Ca(OH) 2  2 LiOH + CaCl 2 FeBr 3 + H 3 PO 4  FePO 4 + 3 HBr PbCl 2 + H 2 SO 4  PbSO 4 + 2 HCl

15 Balancing Equations Examples: C 6 H 12 O 6 + O 2  CO 2 + H 2 O C 7 H 14 O + O 2  CO 2 + H 2 O NaHCO 3 + H 2 SO 4  CO 2 + H 2 O + Na 2 SO 4

16 Balancing Equations Examples: C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O C 7 H 14 O + 10 O 2  7 CO 2 + 7 H 2 O 2NaHCO 3 + H 2 SO 4  2CO 2 + 2H 2 O + Na 2 SO 4

17 States of matter Often you will see letters in brackets after the chemical. It denotes what state the element is in: (s) = solid (l) = liquid (aq) = aqueous i.e. dissolved in water (g) = gas

18 Equilibrium All reactions have a bias towards being products or reactants. Equilibrium is reached when: 1.The rates of opposing reactions are equal. 2. Equilibrium concentrations of reactants and products remain constant. 3. Attainment of equilibrium is not instantaneous and may require an extended period of time.

19 Equilibrium For a reaction to reach equilibrium two criteria must be met: 1. The temperature must be kept constant 2. The system must be closed. This means that once the reaction has started no substances can be added or removed.

20 Equilibrium Constant (K) The equilibrium constant is a mathematical ratio between the concentration of the reactants and products. It is given as the formula aA + bB  cC + dD would give K = [C] c x [D] d [A] a x [B] b (Products / Reactants)

21 Equilibrium Constant There are things that must be left out of the K- expression as they do not have concentrations These are SOLIDS and LIQUIDS They are simply considered to be ‘1’ and therefore can be left out

22 Examples C 6 H 12 O 6 (s) + 6 O 2 (g)  6 CO 2 (g) + 6 H 2 O (l) PbCl 2 (s)  Pb (s) + Cl 2 (g) H 2 O (l)  H 2 O (g) 2 H 2 (g) + O 2 (g)  2 H 2 O (g) 2 Al (s) + 6 HCl (aq)  2 AlCl 3 (s) + 3 H 2 (g) 4 Al (s) + 3 O 2 (g)  2 Al 2 O 3 (s) Ca 3 (PO 4 ) 2 (s) + 2 H 2 SO 4 (aq)  2 CaSO 4 (s) + Ca(H 2 PO 4 ) 2 (s)

23 Value of K Because K is a ratio, the value of it can indicate if a reaction is likely to lie with the products or the reactants. If K is a large value (greater than 1000) then it means: Products Reactants Which means the reaction must be working, the products are more favoured than the reactants This must be bigger This must be smaller

24 Value of K If K is a small value (less than 0.001) then it means: Products Reactants Which means the reaction does not favour the products, i.e. It doesn’t go… This must be bigger This must be smaller

25 Le Chatelier An equilibrium can be shifted with different changes to a system – – Change in concentration – Change in pressure – Change in volume – Change in temperature

26 Change in concentration Let’s use the equilibrium 2H 2 (g) + O 2 (g)  2 H 2 O (g) Changing the concentration of the various compounds means the equilibrium will shift to try to re-correct back to the correct ratio

27 Change in concentration Let’s use the equilibrium 2H 2 (g) + O 2 (g)  2 H 2 O (g) e.g. Increasing the concentration of H 2

28 Change in concentration Let’s use the equilibrium 2H 2 (g) + O 2 (g)  2 H 2 O (g) e.g. Increasing the concentration of H 2 This will cause too much to be on the left-hand side of the equilibrium, meaning that the reaction will move to the right, using up the H 2 and making more H 2 O

29 Change in concentration Let’s use the equilibrium 2H 2 (g) + O 2 (g)  2 H 2 O (g) e.g. Decreasing the concentration of O 2

30 Change in concentration Let’s use the equilibrium 2H 2 (g) + O 2 (g)  2 H 2 O (g) e.g. Decreasing the concentration of O 2 This would mean there was now too little on the left hand side, therefore the equilibrium will move to the right, H 2 O would decrease to make more O 2

31 Changes in Pressure (by volume change) Volume and pressure are related. Increasing the volume will result in a decreased pressure and decreasing the volume will result in an increased pressure. This ONLY CONCERNS GAS MOLECULES Let’s use the equilibrium 2H 2 (g) + O 2 (g)  2 H 2 O (g)

32 Changes in Pressure (by volume change) Let’s use the equilibrium 2H 2 (g) + O 2 (g)  2 H 2 O (g) E.g. increasing the volume (decreasing the pressure)

33 Changes in Pressure (by volume change) Let’s use the equilibrium 2H 2 (g) + O 2 (g)  2 H 2 O (g) E.g. increasing the volume (decreasing the pressure) On the left of this equilibrium there are THREE gas molecules, and on the right there are TWO. Having more volume means that more molecules can be present, therefore the reaction will move to the LEFT

34 Changes in Pressure (by volume change) Let’s use the equilibrium 2H 2 (g) + O 2 (g)  2 H 2 O (g) E.g. decreasing the volume (increasing the pressure)

35 Changes in Pressure (by volume change) Let’s use the equilibrium 2H 2 (g) + O 2 (g)  2 H 2 O (g) E.g. decreasing the volume (increasing the pressure) On the left of this equilibrium there are THREE gas molecules, and on the right there are TWO. Having less volume means that less molecules can be present, therefore the reaction will move to the RIGHT

36 Changes in Pressure (by volume change) Let’s use the equilibrium C 6 H 12 O 6 (s) + 6 O 2 (g)  6 CO 2 (g) + 6 H 2 O (l) E.g. decreasing the volume (increasing the pressure)

37 Changes in Pressure (by volume change) Let’s use the equilibrium C 6 H 12 O 6 (s) + 6 O 2 (g)  6 CO 2 (g) + 6 H 2 O (l) E.g. decreasing the volume (increasing the pressure). On the left there are SIX gas molecules, and on right there are SIX gas molecules. Because they are the same, changing the volume will not affect the equilibrium

38 Changes in pressure (Using inert gases) By adding another gas into the same volume, the pressure can also be altered. This has NO EFFECT on the equilibrium. Only if the gas is involved in the equilibrium or the volume is changed will pressure be an issue.

39 Change in Temperature Endothermic means the reaction requires heat Exothermic means the reaction releases heat

40 Change in Temperature Endothermic means the reaction requires heat A + heat  B Exothermic means the reaction releases heat A  B + heat Therefore heat can be though of as a product or a reactant when considering the equilibrium

41 Changes in Temperature What happens to an exothermic reaction if the temperature is increased? What happens to an endothermic reaction if the temperature is increased? What happens to an exothermic reaction if the temperature is decreased? What happens to an endothermic reaction if the temperature is decreased?

42 Changes in Temperature What happens to an exothermic reaction if the temperature is increased? LEFT What happens to an endothermic reaction if the temperature is increased? RIGHT What happens to an exothermic reaction if the temperature is decreased? RIGHT What happens to an endothermic reaction if the temperature is decreased? LEFT

43 Use of a Catalyst A catalyst is a substance which increases the rate of a chemical reaction without being used up in the reaction. A reaction at equilibrium is occurring at equal rates in opposite directions. A catalyst increases the rates of the opposing reactions equally and has no effect on the position of equilibrium, but it allows the reaction to reach equilibrium faster.

44 Rates of reaction This is how long it takes to get to equilibrium, not which way the equilibrium will shift. Some reactions are naturally very quick, other reactions are known to take millennia! Rates are affected by two main factors: – Kinetic energy of particles – The number of collisions taking place

45 Rate can be affected by: – Concentrations – Surface area – Temperature – Catalysts / Inhibitors Rates of reaction

46 Rate measurement Rates are measured by the change in the concentration of a product or reactant, over a known amount of time. Rate = Change in concentration Change in time For products they are reported as positive rates (being made) For reactants they are reported as negative rates (being used up)

47 Rates: Concentration Concentration affects rates by changing the number of collisions taking place. More particles = more collisions = faster reactions Therefore higher concentrations make for faster reactions. Lower concentrations have slower reactions

48 Rates: Surface Area Surface area affects rates by changing the number of collisions taking place. More surface area = more collisions = faster reactions Therefore higher surface areas make for faster reactions. Less surface areas have slower reactions

49 Rates: Temperature Temperature affects rates by changing the kinetic energy in the reaction Higher temperature = more kinetic energy = faster reactions Therefore higher temperature make for faster reactions. Colder reactions are slower

50 Rates: Catalysts / Inhibitors Catalysts increase the rate of a reaction without being used up They alter the mechanism of the reaction to make it faster Inhibitors have the opposite action to catalysts. They slow the reaction down without being used up


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