chapter 16. ammonia is prepared from nitrogen and hydrogen gas. n 2 (g) + 3h 2 (g) → 2nh 3 (g) ...

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Page 1: 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

Chapter 16

Page 2: 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

Ammonia is prepared from nitrogen and hydrogen gas. N2(g) + 3H2(g) → 2NH3(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.

Page 3: 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

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.

Page 4: 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

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?

Page 5: 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

Some physical and chemical changes can be reversed.

For example:• Ice to water and back again• H2O(s) H2O(l)• What is the forward reaction?• What is the reverse reaction?

The double arrow is used to indicate that this is a reversible reaction.

Page 6: 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

Nickel-metal hydride batteries that are rechargeable.

NiOOH(s) + MH(s) Ni(OH)2(s) + M(s)

Hydrated copper(II) sulfateCuSO4.5H2O(s) CuSO4(s) + 5H2O(l)

Ammonia productionN2(g) + 3H2(g) 2NH3(g)

discharging

recharging

Page 7: 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

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:

Page 8: 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

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.

Page 9: 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

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.

Page 10: 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
Page 11: 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

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’

Page 12: 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

Do all reactions proceed to the same extent?

Read the experiment performed on page 266 to find out.

Page 13: 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

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.

[NH3]2

[N2][H2]2

Page 14: 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

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

[NH3]2

[N2][H2]2

Page 15: 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

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]bK =

Page 16: 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

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

104, there will be significant amounts of reactants and products present at equilibrium

• For values of K that are very large (>104) 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.

Page 17: 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

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.

Page 18: 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

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.

Page 19: 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

Calculate the value of the equilibrium constant for the reaction represented by the equation:H2(g) + I2(g) 2HI(g)at 460°C, if a 2.00 L vessel contains an equilibrium mixture of 0.0860 mol of H2, 0.124 mol of I2 and 0.716 mol of HI.

Where might we start??

Page 20: 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

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

Page 21: 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

Page 271 Questions 2 - 8

Page 22: 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

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.

Page 23: 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

For an equilibrium mixture ofN2(g) + 3H2(g) 2NH3(g)

Without changing the volume or temperature

Page 24: 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

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 N2 and H2 increases

• Ultimately the rates of the forward and back reaction become equal again and a new equilibrium is established

Page 25: 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

If we added more NH3 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.

Page 26: 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

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).

Page 27: 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

“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.

Page 28: 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

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

Page 29: 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

“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.”

Page 30: 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

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.

Page 31: 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

For exampleFe3+(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.

Page 32: 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

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.

Page 33: 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

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.