When a chemical reaction takes place, the atoms or ions in the reactants form new bonds and new products with different chemical formulas and different chemical properties.
In this section, we will be revising the basic concepts involved in chemical equilibria (plural of equilibrium). In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time. Usually, this state results when the forward reaction proceeds (continues) at the same rate as the reverse reaction. NB. This does not mean the reaction has stopped; instead it means that it is continuing to react, but the reactants and products are being produced (and re-produced) at a constant rate.
HOMOGENEOUS REACTIONS
are reactions where all the reactants and products are in the same phase,
i.e. all are solids, or all are liquids, or all are gases
SO2(g) + O2(g) ⇌ SO3(g)
HETEROGENEOUS REACTIONS
are reactions where not all species in the reaction vessel are in the same phase.
CaCO3(s) + HCℓ(aq) → CaCℓ2(aq) + CO2(g) + H2O(ℓ)
Most chemical reactions are non-reversible or irreversible – this means that the products cannot change back into the reactants.
These reactions are essentially like baking. The ingredients, acting as the reactants, are mixed and baked together to form a cake, which acts as the product. This cake cannot be converted back to the reactants (the eggs, flour, etc.), just as the products in an irreversible reaction cannot convert back into the reactants (without e.g. human intervention). An example of an irreversible reaction is combustion. When carbon burns in oxygen to form carbon dioxide, the reaction is said to be non-reversible.
C(s) + O2(g) → CO2(g)
Pay special attention
Some chemical reactions are reversible. The products react with each other to form the original reactants under suitable conditions. (E.g. change in pressure, temperature, reactant concentration, etc.).
N2(g) + 3 H2(g) ⇋ 2 NH3(g)
hint
Therefore:
Forward reaction | Reverse reaction | |
N2(g) + 3 H2(g) → 2 NH3(g) | Reaction | 2 NH3(g) → N2(g) + 3 H2(g) |
N2(g) + 3 H2(g) | Reactants | 2 NH3(g) |
2 NH3(g) | Products | N2(g) + 3 H2(g) |
NB: DEFINITIONS
c = n where n = m or C = m
v M Mv
hint: You need to know the formula, unit and symbols used to calculate the concentration of a solution
You need to understand what a chemical equilibrium is and know the conditions needed for a chemical equilibrium.
Chemical equilibrium
Write the name of a reaction in which all reactants and products are in the same phase (1) [1]
Solution
If a closed system is in a state of dynamic chemical equilibrium, and one of the factors that influences the equilibrium (temperature, concentration or pressure) is changed:
NB: Le Châtelier’s Principle states that:
3.4.1 Applying Le Châtelier’s Principle to changes in equilibrium conditions
Hint: If any factor changes, we can apply Le Châtelier’s Principle to determine which changes will happen to the system in equilibrium (what upsets the balance).
Changing the temperature of the reaction mixture (ΔT) | |
Increasing the temperature of the reaction system (i.e. heating it)
| Decreasing the temperature of the reaction system (i.e. cooling it)
|
Changing the concentration of a solution or gas reactant (∆c) | |
Increasing the concentrationof a (g) or (aq) reactant (or product)
| Decreasing the concentration of a (g) or (aq) reactant (or product)
|
Changing the pressure of a gaseous system (∆p) | |
Increasing the pressure (or decreasing the volume) of a gaseous system
| Decreasing the pressure (or increasing the volume) of a gaseous system
|
Steps to follow
Follow these steps to predict the change in a closed chemical system in dynamic equilibrium:
Check that the reaction:
Determine:
Decide:
hint
BX2Y2 is a solid. Solids in equilibrium reactions have very little change in mass, hence they are ignored
Consider the following reversible reaction that is in equilibrium in a closed system (the symbols represent chemicals which are unnamed; if you see symbols like X or Y which aren’t on the Periodic Table, it often means you’ve got to work out what kinds of substances they are).
Step 4/5:
∆H < 0
In each of the following cases:
CASE 1 The reaction system is heated. (3) [3]
Solution
Case 1.
If the reaction system is heated, the temperature of the system increases.
According to Le Châtelier’s Principle, increasing the temperature of the reaction system (i.e. heating it) favours the reaction that will decrease the temperature of the system
(Remember: The opposite is true if the reaction vessel is cooled.) [3]
hint
You must be able to use Le Châtelier’s Principle to identify the factor that will increase the yield of the products and what influence that factor will have on the reaction AT EQUILIBRIUM
CASE 2 Y2(g) is added to the vessel at constant pressure and temperature (3) [3]
Solution
Case 2.
Adding Y2(g) increases the concentration of the Y2(g). ✓
According to Le Châtelier’s Principle, increasing the concentrationof Y2(g) favours the reaction that decreases the concentration of Y2(g)
CASE 3 A2(g) is removed from the vessel at constant pressure and temperature. (3) [3]
Solution
Case 3.
Removing A2(g) decreases the concentration of the A2(g). ✓
According to Le Châtelier’s Principle, decreasing theconcentration of the A2(g) favours the reaction that increases the concentration of the A2(g);
CASE 4 Some of the BX2Y2(s) is removed from the system at constant pressure and temperature. (3) [3]
Solution
Case 4.
Removing BX2Y2(s) has no effect on the equilibrium as it is a solid. ✓
Removing a solid from an equilibrium system does not disturb the concentration. ✓ (Solids in equilibrium reactions have very little change in mass, hence they are ignored)
The concentration of A2(g) remain the same. ✓ [3]
CASE 5 The pressure on the system is increased (or the volume is decreased) (3) [3]
Solution
Case 5.
According to Le Châtelier’s Principle, increasing the pressure (or decreasing the volume) of a gaseous system favours the reaction that decreases the pressure of the system by decreasing the total number of gaseous moles in the system. ✓
hint
There are 4 gaseous moles on the left, and 3 gaseous moles on the right.
Say whether the statement is TRUE or FALSE: If the equilibrium constant for the reaction A2(g) + B2(g) ⇌ 2AB(g) is equal to K, then the equilibrium constant for the reverse reaction
2AB(g) ⇌ A2(g) + B2(g) is also equal to K. (3) [3]
Solution
FALSE✓✓ . ... for the reverse reaction 2AB(g) ⇌ A2(g) + B2(g) is equal to 1/K ✓✓ OR ... less (smaller) than K [3]
The industrial preparation of hydrogen gas is represented by the equation below:
CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g) ∆H > 0
The reaction reaches equilibrium at 1 000 °C in a closed container.
hint
There are various ways to state Le Châtelier’s Principle correctly.
Solution
hint
In physical chemistry, saturation is the point at which a solution of a substance can dissolve no more of that substance. Additional amounts of the solute will appear as a separate phase (usually as a precipitate solid). This point of maximum concentration, the saturation point, depends on the temperature and pressure of the solution as well as the chemical nature of the substances involved.
3.4.2 Dynamic chemical equilibrium in solutions
A dynamic chemical equilibrium in a solution:
The common ion effect
Steps to follow
When a solution is added to a reaction system:
Solution added to the equilibrium mixture: | Effect on equilibrium mixture: H+(aq) + OH−(aq) ⇌ H2O(ℓ) |
Hydrochloric acid, HCℓ(aq) |
|
Sulphuric acid, H2SO4(aq) |
because: |
Concentrated sulphuric acid, H2SO4(ℓ) |
|
A base like NaOH(aq) or KOH(aq) |
|
A solution containing Cℓ(aq) ions e.g. NaCℓ(aq) |
|
e.g. Worked example 1
Consider the following equilibrium:
CoCℓ42–(aq) + 6H2O(ℓ) ⇋ Co(H2O)62+(aq) + 4Cℓ–(aq) ΔH < 0
blue purple-pink pink
When cobalt chloride crystals (CoCℓ2⋅6H2O) are dissolved in ethanol (the solvent) in a flask, a blue solution is formed with CoCℓ42– ions. By carefully adding water to the blue solution until it just turns purple-pink because of the formation of the pink Co(H2O)62+ ions, the above equilibrium is reached.
Five test tubes each contain a small amount of the above purple-pink equilibrium solution. Describe and explain the colour change observed in each of the following instances:
1. The following hypothetical reaction reaches equilibrium in a closed container at a certain temperature:
X2(g) + Y2(g) ⇋ 2XY(g) ∆H < 0
Which ONE of the following changes will increase the AMOUNT of XY(g)?
2. The equation below represents a chemical reaction at equilibrium in a closed container.
H2(g) + I2(g) ⇌ 2HI(g) ∆H < 0
Which ONE of the following changes will increase the yield of HI(g) in the above reaction?
3. A chemical reaction reaches equilibrium. Which ONE of the following statements regarding this equilibrium is TRUE?
4. The reaction represented by the equation below reaches equilibrium.
2CrO24−(aq) + 2H+(aq) ⇌ Cr2O27−(aq) + H2O(ℓ)
yellow orange
Which ONE of the following changes to the reaction mixture will change its colour from yellow to orange?
5. Give one word for the following phrase:
The stage reached in a reversible chemical reaction when the rate of the forward reaction is equal to the rate of the reverse reaction. (2) [10]
Solution
1. A ✓✓ (2)
2. B✓✓ (2)
3. A ✓✓(2)
4. D ✓✓ (2)
5. (Dynamic/Chemical) equilibrium ✓✓ (2) [10]
NB
Solids and pure liquids are omitted from the Kc expression as their concentration is [1], as multiplying by 1 has no effect.
3.5 The Equilibrium Constant (Kc) (The Law of Mass Action)
The concentrations of all the compounds (solutions and gases) in a closed system in dynamic chemical equilibrium are related by a mathematical equation. The numerical value of this equation is called the equilibrium constant (Kc).
In the hypothetical equation below the equilibrium expression for this reaction is:
2P+3R⇋2S+4T
Kc = [S]2 × [T]4
[P]2 × [R]3
hint
Kc: equilibrium constant (no unit)
[substance]: concentration of reactant or product (in mol·dm–3) mol: number of moles of each compound in the balanced reaction equation.
NB
Write Kc expressions for each of the following reactions:
Solution
3.5.1 Factors that influence the equilibrium constant
For a closed system in equilibrium:
The value of Kc DOES NOT change if:
hintYou need to know the:
that play a role in HOW/WHEN Kc changes/does NOT change
The effect of a change in temperature on Kc
Kc for a specific reaction is constant at a specific temperature.
If the change in the temperature of a closed equilibrium system: | |
favours the forward reaction: | favours the reverse reaction: |
|
|
3.5.2 The meaning of Kc-values
The significance of the Kc value is important in industrial processes. The economic viability of an industrial process in the chemical industry depends on the manufacturing costs, the product yield (amount of product produced), and the retail cost. These factors determine the profit a company would make.
Kc values are used to determine if the possible yield at a specific temperature is low or high.
NB: You must understand the meaning of Kc. The value of Kc is directly linked to the yield of the reaction.
3.5.3 Calculating the value of Kc for a closed system in dynamic equilibrium|
There are THREE types of calculations that you may be asked to perform. They will be covered in the following pages.
Type 1: The initial amount (mol) of each reactant is given
e.g. Worked example 2
Consider the following balanced reaction for the formation of sulphur trioxide, in the second step of the Contact Process:
2SO2(g) + O2(g) ⇋ 2SO3(g)
Exactly 8 mol SO2 and 10 mol O2 are sealed in a 500 cm3 reaction vessel which is initially empty. The compounds react to produce SO3. At equilibrium, the vessel contains 6 mol SO3.
1.1 Calculate the equilibrium concentrations of each of the compounds in the reaction vessel.
Solution
The Contact
Process is a three step process in the manufacturing of sulphuric acid.
Step 1:
Always start by drawing a table with a column for the descriptions and a column for each of the reactants and products. Fill in the formulas of the reactants and products and the number of moles of each in the balanced reaction equation in the top line. The table should always consist of 5 rows with names as indicated:
Step 2:
Now fill in all the values given in the question. If only reactants were added at the beginning of the reaction, the initial mol of the products is zero.
balanced equation | 2SO2(g) | O2(g) | 2SO3(g) |
initial mol | 8 | 10 | 0 |
mol used / produced | |||
final mol at equilibrium | 6 | ||
final [ ] at equilibrium | c = n = n | c = n = n | c = n = n |
Step 3:
Find the column with 2 values. Complete the mol used or produced line for this compound.
balanced equation | 2SO2(g) | O2(g) | 2SO3(g) |
initial mol | 8 | 10 | 0 |
mol used / produced | 6 | ||
final mol at equilibrium | 6 | ||
final [ ] at equilibrium | c = n = n | c = n = n | c = n = n |
Step 4:
Compare the ratio of moles according to which the reactants react and to which the products are produced, according to the balanced reaction equation: SO2 : O2 : SO3 = 2 : 1 : 2
Consider the actual mol used or produced for the compound which is known.
We have just calculated that 6 mol SO3 has been produced.
This means that the ratio of SO2 : O2 : SO3 = 2 : 1 : 2 (balanced equation) 6 :3:6
Fill in these values
Step 5:
For the reactants: initial mol – mol used = final mol at equilibrium.
For the products: initial mol + mol produced = final mol at equilibrium.
balanced equation | 2SO2(g) | O2(g) | 2SO3(g) |
initial mol | 8 | 10 | 0 |
mol used / produced | 6 | 3 | 6 |
final mol at equilibrium | 8 - 6 = 2 | 10 - 3 = 7 | 6 |
final [ ] at equilibrium | c = n = n | c = n = n | c = n = 6 = 12 |
Step 6:
Substitute the final mol values in the last line, and calculate the final concentrations.
These are the answers to Question 1.1 above.
balanced equation | 2SO2(g) | O2(g) | 2SO3(g) |
initial mol | 8 | 10 | 0 |
mol used / produced | 6 | 3 | 6 |
final mol at equilibrium | 2 | 7 | 6 |
final [ ] at equilibrium | 2 = 4 | 7 = 14 | 12 |
e.g. Worked example 3
1.2 Write an expression for the equilibrium constant for the forward reaction and calculate Kc for the forward reaction at this temperature.
Solution
1. Kc = [SO3]2 = 122 = 0,64
[SO2]2[O2] 42.14
Type 2: The initial amount (mol) of one reactant and the concentration of a reactant or product are given
e.g. Worked example 4
2 mol of NO2(g) and an unknown amount of N2O4(g) are sealed in a 2 dm3 container, that is fitted with a plunger, at a certain temperature. The following reaction takes place:
2NO2(g) ⇋ N2O4(g)
At equilibrium it is found that the NO2 concentration is 0,4 mol·dm–3. The equilibrium constant (Kc) at this temperature is 2.
1. Calculate the initial amount (in mol) of N2O4(g) that was sealed in the container.
Solution
Step 1:
Draw the table as before. Fill in the formulas of the reactants and products and the number of moles of each in the balanced reaction equation in the top line.
Now fill in all the values given in the question. If only reactants were added at the beginning of the reaction, the initial mol of the products is zero. When the initial amount (mol) of a reactant or product is NOT zero and the value is not given (in this case the amount of N2 O4) is not given, let this amount equal x. Write the statement regarding ‘x’ down, “Let …. be x”.
Let the initial amount of N2O4 in the reaction vessel be x.
balanced equation | 2 NO2 | N2O4 |
initial mol | 2 | |
mol used / produced | ||
final mol at equilibrium | ||
final [ ] at equilibrium | 0,4 |
Step 2:
Use the concentration given and the formula for concentration (c = n/V ) to calculate the number of moles of NO2 in the equilibrium mixture. The volume of the container (2 dm3) was given. Show the calculation and substitute the calculated value in the table.
Step 3:
Find the column with Find the column with 2 values. Complete the mol used or produced line for this compound.
Step 4:
From the balanced equation ratio 2 : 1 determine the ratio of the moles used and produced. Therefore 2 : 1 becomes 1,2 (as calculated in Step 3): 0,6
Step 5:
Step 6:
Substitute the final mol values into the last line, and calculate the final concentration of the N2O4.
Step 7:
Write the expression for Kc and substitute all the known values – remember that the value of Kc was given. Solve for x.
The plunger is now pushed into the container and the volume of the container decreases.
2. How will this change influence the amount of nitrogen dioxide at equilibrium? Only write down increases, decreases or remains the same.
Solution
Decreases
3. Use Le Châtelier’s principle to explain your answer to Question 2.2.
Solution
When the volume of the container decreases, the pressure increases.
According to Le Châtelier’s Principle the reaction that decreases the pressure will be favoured ∴ the (forward) reaction that produces fewer gas moles, 2 mol (g) ⇌ 1 mol (g), is favoured.
Type 3: The initial mass of a reactant and the value of Kc are given and the mass of the reactant that remains unreacted, is asked
The thermal decomposition of calcium carbonate (CaCO 3) is an example of a heterogeneous equilibrium. The decomposition that takes place in a closed container can be represented by the following equation:
CaCO3(s) ⇌ CaO(s) + CO2(g)
Initially 5 g of CaCO3(s) is placed in a closed 500 cm3 container and then heated. Equilibrium is reached at 900°C.
Solutions
Step 1:
Convert the mass of the reactant to moles. ONLY moles may be used in the table.
Step 2:
Write the expression for Kc and substitute all the known values – remember that the value of Kc was given.
Step 3:
Use the calculated CO2(g) concentration and the formula for concentration (c = n/V ) to calculate the number of moles of CO2 in the equilibrium mixture. The volume of the container (500 cm3) was given. Convert the volume to dm3. Show all working.
Step 4:
Find the column with 2 values. Complete the mol used or produced line for this compound. (0 mol CO2 initially + produced mol CO2 = 0,0054 mol
∴ mol CO2 produced = 0,0054 mol)
Step 5:
From the balanced equation ratio 1 : 1 : 1 determine the ratio of the moles used and produced. Therefore 1 : 1 : 1 becomes 0,0054 (from Step 3): 0,0054 : 0,0054
Step 6
For the reactant CaCO3 ... initial mol – mol used = final mol at equilibrium
∴0,05 mol – 0,0054 = final mol at equilibrium = 0,0046 mol
∴0,0046 mol CaCO3 remains unreacted at equilibrium
M(CaCO3) = 40 + 12 + 3(16) = 100 g·mol–1
Hint : Mass was asked, so convert the mol calculated to mass n = m
M
3. Endothermic. ✓
4.1 Remains the same. Only temperature affects the value of Kc.
4.2 Increases. ✓✓
5. Remains the same. Adding a solid does not affect the equilibrium, it does not change the concentration of a reagent. ✓✓ (2) [37]
NB:
hint
The equation below represents an equilibrium reaction in a sealed 1dm3 container.
NO2(g) + NO(g) ⇋ N2O(g) + O2(g) ΔH > 0
Equilibrium was reached at a certain temperature and the value of Kc was 3,93. The concentration (in mol·dm–3) of each reactant and product in the container at equilibrium was:
One of the conditions affecting the equilibrium is changed and a new equilibrium is established. At the new equilibrium, the concentration of the NO2(g) is 0,12mol·dm–3.
1. Calculate the Kc value at the new equilibrium. (11)
2. Which condition, concentration or temperature, was changed? (1)
3. Give an explanation for the answer to Question 4.2. (1) [13]
Solutions
From the balanced reaction equation, 1 NO2 : 1 NO : 1 N2O : 1 O2
∴0,06 mol N2O + 0,06 mol O2 → 0,06 mol NO2 + 0,06 mol NO
So at equilibrium:
Temperature.
Kc was initially 3,93 (given) now it is 0,91. If the Kc value changed, the temperature must have changed – only a change in temperature can change the value of Kc. (1)
Consider the following reaction equation for the production of ammonia in the Haber process.
NB
N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H < 0
In each of the following cases:
1. The concentration of the nitrogen gas is increased. (1)
2. The pressure in the reaction system is decreased. (1)
3. FeO is added to the reaction system. (1)
4. The reaction system is heated. (5) [8]
Solutions
1. Remains constant because... ✓ (1)
2. ... Kc is only affected by a change in temperature. ✓ (1)
3. Kc Decreases ✓ (1)
4. Heating the reaction system favours the endothermic, reverse reaction ✓ which decreases the temperature ✓ of the system ∴ the concentration of the forward product, [NH3] ✓ decreases while the concentrations of the forward reactants, [N2] and [H2] increase ✓ ∴ Kc decreases. ✓ (5) [8]
The reaction represented by the equation below reaches equilibrium.
Co(H2O)62+(aq) + 4Cℓ–(aq) ⇋ CoCℓ42–(aq) + 6H2O(ℓ) ΔH>0
pink blue
Which ONE of the following changes to the reaction mixture will changes its colour from blue to pink?
Solution
1. D [2]
Dynamic equilibrium in a chemical system can be represented graphically by a graph of:
For each of these graphs:
e.g. Worked example 5
Equal amounts of SO2(g) and O2(g) react in a sealed container and reach equilibrium:
2SO2(g) + O2(g) ⇌ 2SO3(g) ∆H < 0
Pay special attention:
Steps to follow
Step 1: We need to interpret each part of the graph for each reactant and product, step by step as the reaction proceeds and
Step 2: identify what happens if there is a change in the rate/ concentration of each substance, which is visible by a change in the shape of the graph.
(i) When the reaction initially starts:
Initially
As the reaction takes place the
The forward reaction rate gradually decreases and the reverse reaction rate gradually increases.
(ii) The reaction reaches a state of dynamic chemical equilibrium
The forward and reverse reactions take place at the:
The concentrations of the reactants and of the products remain constant.
(iii) O2(g) is added to the system.
Adding O2 (g) to the system increases the concentration of the O2(g). O2(g) is a reactant in the forward reaction, therefore the rate of the forward reaction immediately shows a sharp increase.
According to Le Châtelier’s Principle:
The rate of the
(iv) SO3(g) is removed from the system.
Removing SO 3(g) to the system decreases the concentration of the SO3 (g). SO3(g) is a reactant in the reverse reaction, therefore the rate of the reverse reaction immediately shows a sharp decrease.
According to Le Châtelier’s Principle:
The rate of the
(v) The temperature of the equilibrium mixture is increased.
The heat of reaction is negative (∆H < 0) so the
Increasing the temperature of the system increases both the forward and the reverse reaction rates. The rate of the endothermic reaction always increases the most if the system’s temperature increases, so for this reaction, the reverse reaction rate increases the most.
According to Le Châtelier’s Principle:
The rate of the
(vi) The temperature of the equilibrium mixture is decreased.
The heat of reaction is negative (∆H < 0) so the
Decreasing the temperature of the system decreases both the forward and the reverse reaction rates. The rate of the endothermic reaction always decreases the most if the system’s temperature decreases, so for this reaction, the reverse reaction rate decreases the most.
According to Le Châtelier’s Principle:
The rate of the
(vii) The pressure of the equilibrium mixture is increased
From the balanced reaction equation, the number of gaseous moles is:
According to Le Châtelier’s Principle:
The rate of the
viii). The pressure of the equilibrium mixture is decreased (the volume is increased). The same reasoning is applied as in the previous example and the graphs are as follows:
(ix) The addition of a suitable catalyst
0,25 mol of A and 0,15 mol of B are introduced into a 1 dm3 vessel. A and B react and reach an equilibrium which can be represented by the following equation:
hint: Make use of the graph to read off the values
The graph shows the number of moles of A(g), B(g) and AB(g) vs time, under varying conditions.
Solutions
Nitrogen and oxygen gases react in a sealed container according to the following equation:
O2(g) + N2(g) ⇋ 2NO(g)
After the reaction reaches equilibrium, certain changes are made. The following graph of rate of reaction versus time represents the situation.
Solutions
A fertiliser company produces ammonia on a large scale at a temperature of 450°C. The balanced equation below represents the reaction that takes place in a sealed container.
N2(g) + 3H2(g) ⇌ 2NH3(g) ∆H < 0
To meet an increased demand for fertiliser, the management of the company instructs their engineer to make the necessary adjustments to increase the yield of ammonia. In a trial run on a small scale in the laboratory, the engineer makes adjustments to the TEMPERATURE, PRESSURE and CONCENTRATION of the equilibrium mixture. The graphs below represent the results obtained.
Solutions
[17]
2 mol of NO2(g) and an unknown amount of N2O4(g) are sealed in a 2 dm3 container, that is fitted with a plunger, at a certain temperature. The following reaction takes place:
2NO2(g) ⇌ N2O4(g)
At equilibrium it is found that the NO2 concentration is 0,4 mol·dm–3. The equilibrium constant at this temperature is 2.
Solutions
1.
2NO2 | N2O4 | |
Initial number of mole (mol) | 2 | x |
Number of moles used/formed (mol) | -1,2 | +0,6 |
Number of moles at equilibrium(mol) | 0,8 | x + 0,6 |
Equilibrium concentration (mol·dm -3) | 0,4 | (x + 0,6) 2 |
2. Decreases (1)
3. When the pressure is increased the system will try to decrease the pressure. The forward reaction (2 mol to 1 mol) is favoured. (2) [11]
Many industrial processes involve reversible chemical reactions that reach equilibrium.
Two factors that are important from an economic point of view are the yield (amount of products produced relative to the amount of reactants) and the reaction rate (rate at which the products are formed).
hint: The success of chemical processes in both the laboratory and the industry depends on: the amount of product that is produced and how cost effective this production is.
NB: Remember Standard atmospheric pressure (1 atm) is the atmospheric pressure at sea level.
Haber Process | Step 2 of the Contact Process | |
Nitrogen, N2(g) Hydrogen, H2(g) | Reactants | Sulphur dioxide, SO2(g) Oxygen, O2(g) |
| Source of reactants |
|
Ammonia, NH3(g) | Products | Sulphur trioxide, SO3(g) |
N2(g) + 3H2(g) ⇋ 2NH3(g) | Reaction equation | 2SO2(g) + O2(g) ⇋ 2SO3(g) |
4 mol (g) : 2 mol (g) | Gas mol ratio | 3 mol (g) : 2 mol (g) |
∆H<0 ∴ forward reaction is exothermic | Heat of reaction | ∆H<0 |
| Increase the reaction rate by |
|
| Increase the yield by |
|
Temperature: approx. 450 °C | Optimal Reaction conditions | Temperature: approx. 450 °C |
Fe (iron) OR FeO (iron(II)oxide) | Catalyst | V2O5 (vanadium pentoxide) |
NB
Name the industrial process for the production of ammonia (1) [1]
Solution
Sulphuric acid is an important substance used in the manufacture of fertilisers. The equation below represents one of the steps in the industrial preparation of sulphuric acid.
2SO2(g) + O2(g) ⇋ 2SO3(g) ∆H < 0
Solutions
SO2 | O2 | SO3 | |
Molar ratio | 2 | 1 | 2 |
Initial quantity (mol) | 0,6 | 0,5 | 0,4 |
Change (mol) | 0,2 | 0,1 | 0,2 |
Quantity at equilibrium | 0,8 | 0,6 | 0,2 |
Concentration (mol·dm–3) | 0,4 | 0,3 | 0,1 |
NB
This how marks are allocated to the answer of this Kc calculation:
The reaction below represents the catalysed step in the contact process:
2SO2(g) + O2(g) ⇋ 2SO3(g) ∆H <0
The reaction takes place in a closed container and reaches equilibrium at 427°C. How will a HIGHER temperature affect each of the following? Write down only INCREASES, DECREASES or REMAINS THE SAME.
Solutions
1. Increases ✓✓ (2)
2. Decreases ✓✓ (2)
3.
SO2 | O2 | SO3 | |
Molar ratio | 2 | 1 | 2 |
Initial quantity (mol) | 4 | x 32 | 0 |
Change (mol) | 3 | 1,5 | 3 ratio |
Quantity at equilibrium | 1 | x - 1,5 32 | 3 |
Concentration (mol ·dm –3 | c = n = 1 = 0,5 V 2 | (x - 48) 64 | 3/2 = 1,5 (divide by 2) |
(8) [12]