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Chemical Equilibrium A B + A B + C D + A B + C D + Reaction begins. No products yet formed. High rate of collisions between A & B. Rate of forward reaction.

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Presentation on theme: "Chemical Equilibrium A B + A B + C D + A B + C D + Reaction begins. No products yet formed. High rate of collisions between A & B. Rate of forward reaction."— Presentation transcript:

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2 Chemical Equilibrium A B + A B + C D + A B + C D + Reaction begins. No products yet formed. High rate of collisions between A & B. Rate of forward reaction HIGH. Products formed Collisions between reactants decrease. Rate of forward reaction DECREASES Reverse reaction begins. Rate of forward reaction EQUAL to rate of reverse reaction. Dynamic equilibrium established. Concentrations constant & 3 4.

3 EQUILIBRIM REACTIONS Most reactions DO NOT go to completion. Reactions that do not go to completion are REVERSIBLE. Reversible reactions exist in a state of EQUILIBRIUM. Equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction. The reaction proceeding from L R as the equation is written is known as the forward reaction. e.g. N 2(g) + 3H 2(g) 2NH 3(g)

4 A B + A B + C D + A B + C D + Chemical Equilibrium ……………… …………… ………….. …………… ……………… ………………. ……………………………………………

5 A B + A B + C D + A B + C D + Chemical Equilibrium Concentration Reactants Products time Reaction Rate Forward Reverse DYNAMIC CHEMICAL EQUILIBRIUM

6 A B + C D + Concentration time Reactants Products The equilibrium reaction does not mean the amounts of products and reactants are equal. If the products react more easily than the reactants but BOTH RATES are very slow. We say the Equilibrium is shifted to the LEFT. There will be higher concentration of reactants. Ratio of: Products Reactants Would be LESS than one. Equilibrium Position - Left < 1

7 A B + C D + Concentration time Reactants Products Reactants = 1 Equilibrium Position Middle If the reactants and the products both have similar reaction rates. The Equilibrium position will lie in the middle. There will be the same concentration of reactants & Products. Ratio of: Products Reactants Would be EQUAL TO one. = 1

8 A B + C D + Conc time ……………. …………… If the reactants react ……… ………… than the products. The Equilibrium is ………….. to the ………………. There will be higher concentration of …………. Ratio of: Products Reactants Would be ………………. than ONE. Equilibrium Position - Right 1

9 A B + C D + Conc time Reactants Products If the reactants react more easily than the products. The Equilibrium is shifted to the RIGHT. There will be higher concentration of Products. Ratio of: Products Reactants Would be GREATER than one. Equilibrium Position - Right > 1

10 A B + C D + Concentration time ………………. Products Reactants < 1 Equilibrium Position SHIFT Concentration time …………… Products Reactants > 1 Concentration time Reactants Products Reactants = 1 To ……… To …………

11 A B + C D + Concentration time Reactants Products Reactants < 1 Equilibrium Position SHIFT Concentration time Reactants Products Reactants > 1 Concentration time Reactants Products Reactants = 1 To right SHIFTED To Left

12 A B + C D + Conc time reactants products 1.More …………………….added 2.Reactant concentration …………………… 3.Forward reaction rate ………………….. 4.Reactant concentration ………………/Product concentration …………. 5.Reverse reaction rate ………………………. 6.New equilibrium established - ………….. ratio of PRODUCTS/REACTANTS - equilibrium shifted to ………………… Adding Reactants

13 A B + C D + Conc. time reactants products 1.More Reactant added 2.Reactant concentration increases & Forward reaction rate increases 3.Product concentration decreases & Reverse reaction rate decreases 4.Reactant concentration decreases as reactants used up. 5.Product concentration increases as new products formed. 6.New equilibrium established - higher ratio of PRODUCTS/REACTANTS - equilibrium shifted to RIGHT Adding Reactants Reactant added Equilibrium Shifts New equilibrium established

14 A B + C D + Conc time 1.More PRODUCT added 2.PRODUCT concentration increases (Instantly - Reactant decreases ) Adding Products Products Reactants New Equilibrium Eqm. Shifts 3. REVERSE reaction rate increases (fast then slower) 4. PRODUCT concentration decreases, REACTANT concentration increases (fast then slower) 5. Forward reaction rate increases 6. New equilibrium established - lower ratio of PRODUCTS/REACTANTS - equilibrium shifted to LEFT. Product added

15 Conditions of an equilibrium At equilibrium both reactions continue to occur - the system is DYNAMIC. The system is CLOSED – nothing added or removed. The concentrations of reactants and of products remain constant. Rate of the forward reaction equals rate of the back reaction. Equilibrium can be obtained from either side.

16 H 2 (g) + I 2 (g) 2HI(g) ( H = -13kJ/mol) The H value always refers to the forward reaction. 13 kJ of energy is liberated for every mole of HI formed. H 2 (g) + I 2 (g) 2HI(g) ( H = -26kJ) For the whole reaction: Equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction.

17 Reaction Rate Time H 2 + I 2 2 HI H 2 + I 2 2 HI

18 Reaction Rate Time H 2 + I 2 2 HI H 2 + I 2 2 HI

19 At equilibrium the concentration of all sub- stances are constant at a fixed temperature. Each equilibrium has a constant e.g. for H 2 (g) + I 2 (g)2HI(g) K c = K c is only determined by the concentrations of solutions and gases. Pure liquids & solids are not included in the equation – their concentrations are constant. N.B [H 2 ] means conc. of...

20 In General: aA + bB cC + dD When K c has a high value, there will be more PRODUCTS – (on the RIGHT) - we say the equilibrium lies to The RIGHT (vice versa for a low value). K c = [C] c [D] d [A] a [B] b

21 Calculations 1.The balanced equation must be known. 2. The concentration of a solid or liquid remains constant - these are not included in the equation. 3.The concentration of the solvent is constant and not included in the equation. 4.The value of K c is given without units.

22 Kc Calculation Examples Write expressions for Kc for each of the following reactions: A.C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) B.Ca(s) + 2H 2 O() Ca(OH) 2 (aq) + H 2 (g) C.AgNO 3 (aq) + NaC(s) AgC(s)+ NaNO 3 (aq) D. Na 2 CO 3 (s) + 2HC (aq) 2NaC(aq) + H 2 O() + CO 2 (g)

23 Kc Calculation Examples Write expressions for Kc for each of the following reactions: A.C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) B.Ca(s) + 2H 2 O() Ca(OH) 2 (aq) + H 2 (g) C.AgNO 3 (aq) + NaC(s) AgC(s)+ NaNO 3 (aq) D. Na 2 CO 3 (s) + 2HC (aq) 2NaC(aq) + H 2 O() + CO 2 (g) K c = [CO 2 ] 3 [H 2 O] 4 [C 3 H 8 ] [O 2 ] 5 K c = [Ca(OH) 2 ][H 2 ] K c = [NaNO 3 ] [AgNO 3 ] K c = [NaCl] 2 [CO 2 ] [HCl] 2

24 Kc Calculation Examples Kc = [NH 3 ] 2 [H 2 ] 3 [N 2 ] 3H 2 + N 2 2NH 3 Initial: 7mol8mol0mol Reacts: 6mol2mol Eqm:1mol6mol4mol [ ]: 1/26/24/2 7 = (4/2) 2 (1/2) 3 (6/2) = 10.66

25 Calculating K

26 Methane gas reacts with water vapour to produce carbon monoxide gas and hydrogen gas according to this equation: CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) At equilibrium: [CO] = M [H2] = M [CH4] = M. If K is 5.67, calculate the concentration of water vapour Homework: complete the worksheet: Equilibrium Exercise 1 Calculating Concentrations at Equilibrium

27 Carbon monoxide is a primary starting material in the synthesis of many organic compounds, including methanol, CH 3 OH(l). At C, K is 6.4 x 107 for the decomposition of carbon dioxide into carbon monoxide and oxygen. Calculate the concentrations of all entities at equilibrium if mol of CO 2 (g) is placed in a closed container heated at C.

28 If 0.50 mol N 2 O 4 (g) is placed in a 1.0 L closed container at C, what will be the concentrations of N 2 O 4 (g) and NO 2 (g) at equilibrium? (K=4.50) When hydrogen and iodine are placed in a closed container at C, they react to form hydrogen iodide. At this temperature, the equilibrium constant, K is Determine the concentrations of all entities at equilibrium if 4.00 mol of hydrogen and 2.00 mol of iodine are placed in a 2.00 –L reaction vessel. H 2 (g) + I 2 (g) 2HI(g)

29 The following reaction has an equilibrium constant of K. H 2 (g) + I 2 (g) 2HI(g) If 2.00 mol of hydrogen gas, H 2 (g) and 3.00 mol of iodine gas, I 2 (g) are placed in a 1.00 L reaction vessel at 1100K, what is the equilibrium concentration of each gas?

30 Predicting the effect Le Chateliers Principle If the conditions of an equilibrium system are changed, a process takes place which tends to oppose or cancel the effect of the change.

31 2.The equilibrium constant K c = N 2 (g) + 3H 2 (g) 2NH 3 (g) ( H < 0) When the concentration of the N 2 is increased, the ratio (Kc) will be smaller. To restore the value of K c and the equilibrium,more N 2 will have to react with H 2 This will diminish the concentrations of reactants.

32 K c = N 2 (g) + 3H 2 (g) 2NH 3 (g) ( H < 0) When the concentration of the N 2 is increased: the concentration of reactants increases instantaneously The ratio K c (momentarily) decreases Forward reaction rate increases and the concentrations of REACTANTS is LOWERED whilst the concentration of NH 3 INCREASES. The reverse reaction rate INCREASES as more NH 3 forms. A new equilibrium is established which has shifted to the RIGHT. The value of the ratio K c is restored to its original value KcKc time Overall Kc UNCHANGED!

33 N 2 (g) + 3H 2 (g) 2NH 3 (g) ( H < 0) Increasing the temperature will favour the ………………………. reaction. In an EXOTHERMIC REACTION The ………………………reaction will be favoured More ……………………..produced the reaction shift to the ………….. K c = The ratio (K c ) will therefore ………………………….. KcKc time

34 N 2 (g) + 3H 2 (g) 2NH 3 (g) ( H < 0) Increasing the temperature will favour the ENDOTHERMIC reaction. The REVERSE reaction will be favoured More REACTANTS produced the reaction shift to the LEFT. K c = The ratio (K c ) will therefore DECREASE. KcKc time

35 N 2 (g) + 3H 2 (g) 2NH 3 (g) ( H < 0) Increasing the PRESSURE Equilibrium will shift to …………………..pressure. Reaction will favour the side with the ……………………….number of MOLES of GAS – …………… More ………….. produced. K c goes go ……………….(this eg)! K c = The ratio (K c ) will therefore INCREASE. KcKc time

36 N 2 (g) + 3H 2 (g) 2NH 3 (g) ( H < 0) Increasing the PRESSURE Equilibrium will shift to REDUCE pressure. Reaction will favour the side with the LEAST number of MOLES of GAS – RHS NH 3. More NH 3 produced. K c goes go up (this eg)! K c = The ratio (K c ) will therefore INCREASE. KcKc time

37 Effect of a Catalyst Potential Energy E prod E react E prod E react A CATALYST lowers the ACTIVATION ENERGY of the reaction by providing a different reaction pathway. Activation Energy is lowered for ………………… FORWARD AND REVERSE REACTIONS. EQUILIBRIUM DOES ………………….. SHIFT!!!!BOTH rates are therefore increased by the ……………………and so the EQUILIBRIUM DOES ………………….. SHIFT!!!! Ea

38 Effect of a Catalyst Potential Energy E prod E react E prod E react A CATALYST lowers the ACTIVATION ENERGY of the reaction by providing a different reaction pathway. Activation Energy is lowered for BOTH FORWARD AND REVERSE REACTIONS. EQUILIBRIUM DOES NOT SHIFT!!!!BOTH rates are therefore increased by the same amount and so the EQUILIBRIUM DOES NOT SHIFT!!!! Ea

39 N 2 (g) + 3H 2 (g) 2NH 3 (g) ( H < 0) Favorable conditions: High LOW In practice a ……………………. Temperature is used. Too ………………… will slow the reaction down. K c = KcKc time

40 N 2 (g) + 3H 2 (g) 2NH 3 (g) ( H < 0) Favourable conditions: High [N 2 ] & [H 2 ] High Pressure LOW Temperature LOW [NH 3 ] In practice a compromise temperature is used. Too low will slow the reaction down. K c = KcKc time

41 ChangeShift

42 Equilibrium Disturbance a)Conc of O 2 is increased b)Temperature is increased c)Pressure is increased 2SO 2(g) + O 2(g) --> 2SO 3(g)

43 Depending on circumstances, a solution may be: Under saturated – forward rate greater than reverse (salt dissolving) Saturated – forward & reverse the same rate (equilibrium) Over saturated – reverse greater than forward – salt precipitates Heterogeneous equilibrium

44 Depending on circumstances, a solution may be: Under saturated – forward rate greater than reverse (salt dissolving) Saturated – forward & reverse the same rate (equilibrium) Over saturated – reverse greater than forward – salt precipitates Heterogeneous equilibrium

45 1.All nitrates are soluble 2.All alkali metal & ammonium salts are soluble 3.Chlorides, bromides & iodides are soluble - except Ag, Hg, Cu & Pb 4.Sulphates are soluble – except Pb, Ca, Ag &Hg 5.Carbonates, phosphates & sulphates of alkali metals & ammonium are soluble 6.Hydroxides of alkali metals, ammonium & barium are soluble 7.Sulphides of alkali metals, alkaline earth metals and ammonium are soluble

46 The equation for the equilibrium reaction of a saturated salt solution can be represented as follows: AB (s) A + (aq) + B - (aq)

47 The equilibrium constant is called the solubility product and is calculated as follows: Ca(OH) 2 (s) Ca 2+ (aq) + 2 OH - (aq) K sp = [Ca 2+ ][OH - ] 2

48 Temperature Change: Solubility curves show us that solubility's of most salts increase with increase in temperature.

49 Change in concentration : NaCl (s) Na + (aq) + Cl - (aq) Adding HCl to the above equilibrium, causes the equilibrium to shift to the left. NaCl is therefore precipitated until the equilibrium is restored. Disturbance of the equilibrium by increas- ing the concentration of one kind of ion is called the common ion effect.

50 Take note that the common ion effect is not restricted to solubility equilibria only. NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) Adding a few drops of indicator and NH 4 Cl will show that the equilibrium will shift to the left - Explain.


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