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Chapter 16.  Ammonia is prepared from nitrogen and hydrogen gas. N 2 (g) + 3H 2 (g) → 2NH 3 (g)  According to the equation 1 mol of nitrogen and 3 mol.

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Presentation on theme: "Chapter 16.  Ammonia is prepared from nitrogen and hydrogen gas. N 2 (g) + 3H 2 (g) → 2NH 3 (g)  According to the equation 1 mol of nitrogen and 3 mol."— Presentation transcript:

1 Chapter 16

2  Ammonia is prepared from nitrogen and hydrogen gas. N 2 (g) + 3H 2 (g) → 2NH 3 (g)  According to the equation 1 mol of nitrogen and 3 mol of hydrogen gas should produce 2 mol of ammonia.  In fact no matter how long you wait, the reaction seems to ‘stop’ when much less than 2 mol of ammonia is present.  The stage when the quantities of reactants and products in the reaction remain unchanged is called chemical equilibrium.

3  The fact that this reaction does not proceed to completion has serious consequences for manufacturing companies.  The presence of large amounts of unreacted starting materials in reactions mixtures is wasteful and costly.  The profitability of the industry is dependent upon the reaction yield.  The reaction yield is the extent of conversion of reactants into products.

4  Ammonia is not the only reaction that reaches equilibrium before the theoretical amount of product is reached.  To maximise the yield for such reactions, the following questions are vital: Why do some reactions reach equilibrium? How can the amount of product from a reaction that reaches equilibrium be increased?

5  Some physical and chemical changes can be reversed.  For example: Ice to water and back again H 2 O(s) H 2 O(l) What is the forward reaction? What is the reverse reaction?  The double arrow is used to indicate that this is a reversible reaction.

6  Nickel-metal hydride batteries that are rechargeable. NiOOH(s) + MH(s) Ni(OH) 2 (s) + M(s)  Hydrated copper(II) sulfate CuSO 4.5H 2 O(s) CuSO 4 (s) + 5H 2 O(l)  Ammonia production N 2 (g) + 3H 2 (g) 2NH 3 (g) discharging recharging

7  We use the idea that process can be reversed to understand why some reactions reach an equilibrium  In these reactions the forward and reverse reactions occur simultaneously.  Suppose nitrogen gas and hydrogen gas were added to a sealed container at a constant temperature. The nitrogen and hydrogen start to react immediately forming ammonia. The following sequence of events then occur:

8  As the forward reaction proceeds, the concentration of nitrogen and hydrogen decrease, so the rate of the ammonia production decreases.  At the same time as ammonia is being formed, some ammonia molecules react to re-form nitrogen and hydrogen. The rate of this reverse reaction increases as the concentration of ammonia increases.

9  Eventually the forward and back reactions proceed at the same rate. When this situation is reached ammonia is formed at exactly the same rate as it is breaking down. The concentration so ammonia, nitrogen and hydrogen will then remain constant.  At the equilibrium position no further change will take place in the rate of either the forward or reverse reaction.


11  Equilibrium is a dynamic state, since the forward and back reactions have not ceased. Instead they occur simultaneously at the same rate.  During equilibrium: The amounts and concentrations of chemical substances remains constant The total gas pressure is constant The temperature is constant The reaction is ‘incomplete’

12  Do all reactions proceed to the same extent?  Read the experiment performed on page 266 to find out.

13  So many reactions of importance involve equilibria that chemists have tried to find mathematical relationships for.  Look at table 16.1 on page 267.  The fraction seems to give a constant value for each mixture.  This concentration fraction is known as the reaction quotient for the reaction. [NH 3 ] 2 [N 2 ][H 2 ] 2

14  While the concentration fraction can be calculated for any mixture of reactants and products, it is only at equilibrium that it has a constant value.  We give this the value K   K=  Where K is known as the equilibrium constant. The equilibrium constant for the reaction at 400°C is 0.052 M -2 [NH 3 ] 2 [N 2 ][H 2 ] 2

15  Chemists have found that: Different chemical reactions have different values of K For a particular reaction, K is constant for all equilibrium mixtures at a fixed temperature.  For the equation aW + bX cY + dZ at a particular temperature  The equation must be specified when an equilibrium constant is quoted. [Y] c [Z] d [W] a [X] b K =

16  It gives us an indication of the extent of reaction; How far the forward reaction proceeds before equilibrium was established. For values of K that are between about 10 -4 and 10 4, there will be significant amounts of reactants and products present at equilibrium For values of K that are very large (>10 4 ) the equilibrium mixture contains mostly products, with relatively small amounts of reactants. For values of K that are very small (<10 -4 ) the equilibrium mixture contains mostly reactants with very small amounts of products.

17  The value of K for a particular reaction depends only upon temperature. It is not affected by pressure, concentration or catalysts.  The effect of a change in temperature on an equilibrium depends on whether the reaction is exothermic or endothermic.

18  As temperature increases: For exothermic reactions, the amount of product decreases and so the value of K decreases For endothermic reactions, the amount of products increases and so the value of K increases.  Since the value of K depends on temperature, it is important to specify the temperature at which an equilibrium constant has been measured.

19  Calculate the value of the equilibrium constant for the reaction represented by the equation: H 2 (g) + I 2 (g) 2HI(g) at 460°C, if a 2.00 L vessel contains an equilibrium mixture of 0.0860 mol of H 2, 0.124 mol of I 2 and 0.716 mol of HI.  Where might we start??

20  Check out pages 269 – 270 to find some more examples of calculations.

21  Page 271  Questions 2 - 8

22  The equilibrium position (the relative amounts of reactants and products) of a reaction may be changed by: Adding or removing a reactant or product Changing the pressure by changing the volume (for equilibria involving gases) Dilution (for equilibria involving solutions) Changing the temperature.

23  For an equilibrium mixture of N 2 (g) + 3H 2 (g) 2NH 3 (g) Without changing the volume or temperature

24  The following events occur as the composition of the mixture adjust to return to equilibrium: The increased concentration of nitrogen gas causes the rate of the forward reaction to increase and more ammonia is formed As the concentration of ammonia increases, the rate of the back reaction to re-form N 2 and H 2 increases Ultimately the rates of the forward and back reaction become equal again and a new equilibrium is established

25  If we added more NH 3 we would increase the rate of the reverse reaction.  When equilibrium is re-established we would have an increase in the concentration of the reactants.  It is also possible to remove a reactant or product from the equilibrium mixture.  This has the reverse effect to adding a substance.

26  Addition of a reactant leads to the formation of more products (a new forward reaction)  Addition of a product leads to the formation of more reactants (a new reverse reaction).

27  “If an equilibrium system is subjected to a change, the system will adjust itself to partially oppose the effect of the change.”  This principal can be used to predict the consequences of adding or removing substances.  It can also be used to predict the effect of changes of pressure and temperature.

28  The pressure of gases in an equilibrium mixture can be changed by increasing or decreasing the volume of the container while keeping the temperature constant.  Look at the examples on page 273 and 274

29  “In general the effect of a change of pressure, by changing the container volume, depends on the relative number of gas particles on both sides of the equation.”

30  The effect of diluting the solution by adding water is similar to changing the volume in gaseous equilibria.  Where possible, a net reaction occurs in the direction that produces the greater number of particles.

31  For example Fe 3+ (aq) + SCN - (aq) Fe(SCN) 2+ (aq) 2 particles in solution 1 particle in solution  Dilution results in an increase in the amounts of reactants.  In terms of Le Chaterlier’s Principal, a net back reaction increases the total concentration of particles in solution, offsetting the effects of dilution.

32  The effect temperature has on equilibrium depends on whether the reaction is exothermic or endothermic.  An increase in temperature in an equilibrium mixture results in: A net backwards reaction for exothermic reactions A net forward reaction for endothermic reactions.

33  Read page 276 – 277  Both 16.4 and the Extension information.  Look at the summary on page 279.  Your Turn  Page 279  Questions 9 – 12  Chapter Review Questions are now due.

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