Download Equilibrium 4 Noteform - IndustrialProcesses

Chemistry 12 (HL)
Unit 4 / IB Topic 7.2
Chemical Equilibrium 4
Industrial Processes – Haber Process and Contact Process
Reference: Higher Level Chemistry, p. 248-251
The industrial production of chemical substances requires careful planning. Chemical engineers need to
consider 2 criteria when designing a process to produce any chemical:
1. How can the yield be maximized?
(Think about how you would use Le Chatelier’s Principle to do this.)
2. How can the rate be maximized?
(Think about the factors that affect the rate of a chemical reaction.)
In many industrial processes, these two criteria are at odds with each other, and compromises must be
made.
e.g. 90% yield and a reaction time of one year vs 50% yield and a reaction time of several days
THE HABER PROCESS (aka The Haber Bosch Process)
The Haber process was developed by the German chemist Fritz Haber in 1913.
This process is used to manufacture
used to make:
, which is
• F
• E
• P
The chemical equation for this reaction is:
This reaction does not readily occur in nature because ….
p. 1
Chemistry 12 (HL)
Unit 4 / IB Topic 7.2
How can Le Chatelier’s Principle and kinetics be applied in order to maximize the yield of ammonia?
1. concentration
reactants
products
2. pressure
equilibrium considerations:
kinetics (rate) considerations:
economic factors:
3. temperature
equilibrium considerations:
kinetics (rate) considerations:
the compromise:
4. catalyst
p. 2
Chemistry 12 (HL)
Unit 4 / IB Topic 7.2
Interpret this graph showing the effects of reaction conditions on the yield of ammonia:
Source: http://www.ausetute.com.au/haberpro.html
A summary of the Haber Process:
Source: http://www.chemguide.co.uk/physical/equilibria/haber.html
Typical yield of the Haber process:
p. 3
Chemistry 12 (HL)
Unit 4 / IB Topic 7.2
THE CONTACT PROCESS
The Contact Process is used to manufacture
which is used for:
,
• Fe
• Po
• De
• Pa
• Me
• Ba
There are 3 reactions involved in this process:
1 combustion of sulfur
2 oxidation of sulfur dioxide
3 combination of sulfur trioxide with water
The overall rate of the process depends on the STEP 2. We can apply Le Chatelier’s Principle to this
reaction to find the conditions that most favour the formation of the product.
2 SO2(g) + O2(g) ⇔ 2 SO3(g)
∆H = - 196 kJ mol-1
1 pressure
2 temperature
3 catalyst
p. 4
Chemistry 12 (HL)
Unit 4 / IB Topic 7.2
Here is a summary diagram of this step in the Contact Process:
Source: http://www.chemguide.co.uk/physical/equilibria/contact.html
QUESTIONS
1.
N2(g) + 3H2(g) ⇔ 2NH3(g)
In the reaction
∆H = –92 kJ
which of the following changes will increase the amount of ammonia at equilibrium?
I.
Increasing the pressure
II.
Increasing the temperature
III.
Adding a catalyst
A.
2.
I only
B.
II only
2 SO2(g) + O2(g) ⇔ 2 SO3(g)
C.
I and II only
D.
II and III only
∆H = -200 kJ
According to the equation above, what temperature and pressure conditions produce the greatest
amount of SO3?
temperature
pressure
A
low
low
B
low
high
C
high
high
D
high
low
p. 5
Chemistry 12 (HL)
Unit 4 / IB Topic 7.2
3.
The diagrams below represent equilibrium mixtures for the reaction Y + X2 XY + X at 350 K and
550 K respectively. Deduce and explain whether the reaction is exothermic or endothermic.
3.
Consider the following reaction in the Contact process for the production of sulfuric acid for parts (a)
to (d) in this question.
2SO2 + O2 ⇔ 2SO3
(a)
Write the equilibrium constant expression for the reaction. (1)
(b)
(i)
State the catalyst used in this reaction of the Contact process.(1)
ii)
State and explain the effect of the catalyst on the value of the equilibrium
constant and on the rate of the reaction. (4)
(c)
Use the collision theory to explain why increasing the temperature increases the rate
of the reaction between sulfur dioxide and oxygen. (2)
(d)
Using Le Chatelier’s principle state and explain the effect on the position of equilibrium of
(i)
increasing the pressure at constant temperature. (2)
(ii)
removing of sulfur trioxide. (2)
(iii)
using a catalyst. (2)
p. 6