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Equilibrium DP Chemistry R. Slider. Characteristics of Equilibrium Closed system – nothing in, nothing out Forward and reverse reactions are occurring.

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Presentation on theme: "Equilibrium DP Chemistry R. Slider. Characteristics of Equilibrium Closed system – nothing in, nothing out Forward and reverse reactions are occurring."— Presentation transcript:

1 Equilibrium DP Chemistry R. Slider

2 Characteristics of Equilibrium Closed system – nothing in, nothing out Forward and reverse reactions are occurring at the same rate – known as ‘dynamic equilibrium’ Macroscopic properties (e.g. Colour, concentration, pH) remain constant Changes in temp, pressure, volume and concentration can change the equilibrium position. Catalysts do not.

3 Equilibrium Graph - Concentration This reaction starts with A only, with the [A] (concentration of A) at a maximum at Time = 0 s. As the reaction proceeds, A gets used up and [B] steadily increases until equilibrium is reached which can be seen as no change in either concentration. In this graph, [B] > [A] which means the forward reaction is more favoured than the reverse. What would the graph look like if the reverse was favoured? A B

4 Equilibrium Graph – Rate of Reaction A B This reaction starts with A only, so the rate of the forward reaction is at a maximum and slows down as [A] decreases As the reaction proceeds and [B] steadily increases, the rate of the reverse reaction increases until equilibrium is reached when the rates of the forward and reverse reactions are equal.

5 Equilibrium constant - K c products reactants Indices are from the coefficients in the balanced chemical equation

6 Equilibrium constant - K c Units for K c will vary depending upon the reaction. In fact, there may be no units. The value of K c for a particular reaction is only affected by changes in temperature homogeneousIn a homogeneous reaction, all of the states of matter are the same heterogeneousIn a heterogeneous reaction, there are different states of matter. Solids do not take part in the equilibrium constant

7 Equilibrium constant - K c When K c >> 1 When K c << 1 Equil goes far right (forward rxn almost to completion) Equil goes far left (forward rxn hardly proceeds) What happens when K c = 1? K c = 0?

8 Le Chatelier’s Principle “An equilibrium system that is exposed to a stress will shift the equilibrium position to oppose that stress” Note: a stress can be a change in temperature, pressure, volume or concentration. Catalysts do not affect equilibrium; they simply affect the rate of the forward and reverse reactions.

9 Le Chatelier’s Principle – Concentration Consider the following reaction Consider the following reaction: A + B C + D Adding a reactant: This will stress the system To relieve the stress, the system can produce more products Equilibrium shifts right Adding a reactant: This will stress the system To relieve the stress, the system can produce more products Equilibrium shifts right Adding a product: This will stress the system To relieve the stress, the system can produce more reactants Equilibrium shifts left Adding a product: This will stress the system To relieve the stress, the system can produce more reactants Equilibrium shifts left What happens if you remove some reactant or product?

10 Le Chatelier’s Principle – Temperature (exothermic) Consider the following exothermic reaction Consider the following exothermic reaction: A + B C + D + heat Increasing the temperature: This is like adding a product To relieve the stress, the system can reduce the heat by producing more reactants Equilibrium shifts left Increasing the temperature: This is like adding a product To relieve the stress, the system can reduce the heat by producing more reactants Equilibrium shifts left Decreasing the temperature: This is like removing a product To relieve the stress, the system can produce more products Equilibrium shifts right Decreasing the temperature: This is like removing a product To relieve the stress, the system can produce more products Equilibrium shifts right

11 Le Chatelier’s Principle – Temperature (endothermic) Consider the following endothermic reaction Consider the following endothermic reaction: heat + A + B C + D Increasing the temperature: This is like adding a reactant To relieve the stress, the system can reduce the heat by producing more products Equilibrium shifts right Increasing the temperature: This is like adding a reactant To relieve the stress, the system can reduce the heat by producing more products Equilibrium shifts right Decreasing the temperature: This is like removing a reactant To relieve the stress, the system can produce more reactants Equilibrium shifts left Decreasing the temperature: This is like removing a reactant To relieve the stress, the system can produce more reactants Equilibrium shifts left

12 Le Chatelier’s Principle – Pressure Consider the following reaction Consider the following reaction: 2A + B C + D Increasing the pressure: This means there is less room for the particles To relieve the stress, the system can reduce the pressure by producing less moles of gas Equilibrium shifts right Increasing the pressure: This means there is less room for the particles To relieve the stress, the system can reduce the pressure by producing less moles of gas Equilibrium shifts right Decreasing the pressure: This means there is more space for particles To relieve the stress, the system can increase the pressure by producing more moles of gas Equilibrium shifts left Decreasing the pressure: This means there is more space for particles To relieve the stress, the system can increase the pressure by producing more moles of gas Equilibrium shifts left Notice that there are 3 moles of gas on the left and 2 moles on the right Assume all species are gases What happens if you increase or decrease the volume? How does this relate to pressure?

13 Le Chatelier’s Principle – Catalysts Consider the following reaction Consider the following reaction: A + B C + D Adding a catalyst This will not affect the equilibrium position or K c Catalysts reduce the activation energy This speeds up the forward and reverse reactions equally Equilibrium is reached faster Adding a catalyst This will not affect the equilibrium position or K c Catalysts reduce the activation energy This speeds up the forward and reverse reactions equally Equilibrium is reached faster

14 Production of Ammonia – The Haber process In 1912, German scientist, Fritz Haber developed a process for manufacturing ammonia from nitrogen and hydrogen. N 2 (g) + 3H 2 (g)  2NH 3 (g) + 92kJ Source: Notice that this reaction is reversible which can establish equilibrium. This means that Le Chatelier’s Principle applies to the chemistry of this process. Also, note that the forward reaction is exothermic

15 Ammonia and Le Chatelier N 2 (g) + 3H 2 (g)  2NH 3 (g) + 92kJ Using your knowledge of Le Chatelier’s Principle, describe what the optimum conditions (in terms of yield and rate) for this reaction will be in relation to the following: Temperature Pressure Use of a catalyst

16 Optimum ammonia production Rate Yield Cost Source: Chemistry Contexts 2, 2006

17 Conditions for Haber Process Pressure - high(250 atm) – to shift equilibrium right and increase rate Temperature – moderate (450 0 C) – low favours equilibrium, high favours rate. This temperature is a trade-off Catalyst – use of an iron catalyst helps to increase the rate and overcome the relatively low temperature required. Removal of ammonia – shifts the equilibrium towards the products. Pressure - high(250 atm) – to shift equilibrium right and increase rate Temperature – moderate (450 0 C) – low favours equilibrium, high favours rate. This temperature is a trade-off Catalyst – use of an iron catalyst helps to increase the rate and overcome the relatively low temperature required. Removal of ammonia – shifts the equilibrium towards the products.

18 Production of Sulfuric Acid – The Contact Process Sulfuric acid is made in 3 steps: Steps to make sulfuric acid 1.Sulfur dioxide is made 2.Sulfur trioxide is made from sulfur dioxide 3.Sulfuric acid is made from sulfur trioxide

19 Contact Process Chemistry Step 1: Sulfur is roasted in oxygen to produce sulfur dioxide S (s) + O 2(g)  SO 2(g) Step 2: Sulfur dioxide is reacted with oxygen using a Vanadium catalyst to produce sulfur trioxide in a reversible exothermic reaction SO 2(g) + O 2(g)  SO 3(g) + heat Step 3: Sulfur trioxide is converted to sulfuric acid through a series of reactions

20 Contact Process conditions Using Le Chatelier's principle, the equilibrium yield of sulfur trioxide (step 2) should increase when: - temperatures are low, since the reaction is exothermic; - pressure is high; - excess reactants are present. However the rate of the reaction is high when: - temperature is high, hence an obvious conflict exists with the equilibrium yield; - the pressure is high; - a catalyst is used. Using Le Chatelier's principle, the equilibrium yield of sulfur trioxide (step 2) should increase when: - temperatures are low, since the reaction is exothermic; - pressure is high; - excess reactants are present. However the rate of the reaction is high when: - temperature is high, hence an obvious conflict exists with the equilibrium yield; - the pressure is high; - a catalyst is used. Predict the conditions that would be used in this process. Justify your answer

21 Contact Process conditions Pressure - low(1-2 atm) – the process already favours the foward reaction. The extra pressure is not cost effective Temperature – moderate (450 0 C) – low favours equilibrium, high favours rate. This temperature is a trade-off same as Haber Catalyst – use of an vanadium catalyst helps to increase the rate and overcome the relatively low temperature required. Excess oxygen– shifts the equilibrium towards the products. Pressure - low(1-2 atm) – the process already favours the foward reaction. The extra pressure is not cost effective Temperature – moderate (450 0 C) – low favours equilibrium, high favours rate. This temperature is a trade-off same as Haber Catalyst – use of an vanadium catalyst helps to increase the rate and overcome the relatively low temperature required. Excess oxygen– shifts the equilibrium towards the products.


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