Any factor that can affect the rate of either the forward or reverse reaction relative to the other can potentially affect the equilibrium position. The following factors can change the chemical equilibrium position of a reaction:
It is important to understand what effect a change in one of these factors will have on a system that is in chemical equilibrium. However, performing an experiment every time to find out would waste a lot of time. Towards the end of the 1800s the French chemist Henry Louis Le Chatelier came up with principle to predict those effects. Le Chatelier's Principle helps to predict what effect a change in temperature, concentration or pressure will have on the position of the equilibrium in a chemical reaction. This is very important, particularly in industrial applications, where yields must be accurately predicted and maximised.
\(\color{darkgreen}{\Large\textbf{Le Chatelier's Principle}}\) When an external stress (change in pressure, temperature or concentration) is applied to a system in chemical equilibrium, the equilibrium will change in such a way as to reduce the effect of the stress. So if the concentration of one (or more) of the reactants or products is increased the equilibrium will shift to decrease the concentration. Or if the temperature is decreased the equilibrium will shift to increase the temperature by favouring the exothermic reaction. Le Chatelier's principle is that:
Each of these concepts is discussed in detail in the following pages. The following video gives an example of Le Chatelier's principle in action. If the concentration of a substance is changed, the equilibrium will shift to minimise the effect of that change.
For example, in the reaction between sulfur dioxide and oxygen to produce sulfur trioxide: \[\color{blue}{\text{2SO}_{2}\text{(g)}} + \color{blue}{\text{O}_{2}\text{(g)}} \rightleftharpoons \color{red}{\text{2SO}_{3}\text{(g)}}\]
Although not required by CAPS the common-ion effect is a useful concept for the students to know if there is time.
Common-ion effect In solutions the change in equilibrium position can come about due to the common-ion effect. The common-ion effect is where one substance releases ions (upon dissociating or dissolving) which are already present in the equilibrium reaction. If solid sodium chloride is added to an aqueous solution and dissolves, the following dissociation occurs: \[\text{NaCl}(\text{s}) \to \text{Na}^{+}(\text{aq}) + \text{Cl}^{-}(\text{aq})\] If that solution contains the following equilibrium: \[\text{HCl}(\text{ℓ}) + \text{H}_{2}\text{O}(\text{ℓ}) \rightleftharpoons \text{Cl}^{-}(\text{aq}) + \text{H}_{3}\text{O}^{+}(\text{aq})\] The added \(\text{Cl}^{-}\) ion (common-ion) interferes with the equilibrium by raising the concentration of the \(\text{Cl}^{-}\) ion. According to Le Chatelier's principle the reverse reaction speeds up as it tries to reduce the effect of the added \(\text{Cl}^{-}\). As a result the equilibrium position shifts to the left. If the temperature of a reaction mixture is changed, the equilibrium will shift to minimise that change.
For example the forward reaction shown below is exothermic (shown by the negative value for \(\Delta H\)). This means that the forward reaction, where nitrogen and hydrogen react to form ammonia, gives off heat, increasing the temperature (the forward reaction is exothermic). In the reverse reaction, where ammonia decomposes into hydrogen and nitrogen gas, heat is taken in by the reaction, cooling the vessel (the reverse reaction is endothermic). \(\color{blue}{\text{N}_{2}\text{(g)}} + \color{blue}{\text{3H}_{2}\text{(g)}} \rightleftharpoons {\color{red}{\text{2NH}_{3}\text{(g)}}} \qquad \Delta{H} = -92\) \(\text{kJ}\)
In the informal experiment on Le Chatelier's principle, the solution should be purple to start. To achieve this \(\text{CoCl}_{2}\) must be dissolved in ethanol and a few drops of water must be added. This solution is toxic, and all the usual laboratory precautions should be taken. The hot water will make the solution a deep blue, the cold water will make the solution a pink/red colour. If necessary the test tube can be gently shaken to ensure mixing. To determine the effect of a change in concentration and temperature on chemical equilibrium
The equation for the reaction that takes place is: \[\underset{\color{blue}{\text{blue}}}{\underbrace{{\color{blue}{{\text{CoCl}}_{4}^{2-}\text{(aq)}}}}} + 6{\text{H}}_{2}{\text{O(ℓ)}} \rightleftharpoons \underset{\color{red}{\text{pink}}}{\underbrace{{\color{red}{{\text{Co(H}}_{2}{\text{O)}}_{6}^{2+}\text{(aq)}}}}} + 4{\text{Cl}}^{-}\text{(aq)}\] Complete your observations in the table below, noting the colour changes that take place, and also indicating whether the concentration of each of the ions in solution increases or decreases.
If the pressure of a gaseous reaction mixture is changed the equilibrium will shift to minimise that change.
When the volume of a system is decreased (and the temperature is constant), the pressure will increase. There are more collisions with the walls of the container. If there are fewer gas molecules there will be fewer collisions, and therefore lower pressure. The equilibrium will shift in a direction that reduces the number of gas molecules so that the pressure is also reduced. So, to predict in which direction the equilibrium will shift to change pressure you need to \(\color{darkgreen}{\text{look at the number of gas molecules in the balanced}}\) \(\color{darkgreen}{\text{reactions}}\).
Remember from Grade 11 that: \[\text{p} \propto \dfrac{\text{T}}{\text{V}}\] That is, if the temperature remains constant, and the volume is increased, the pressure will decrease. Figure 8.2 shows a decrease in pressure by an increase in the volume, and an increase in pressure by a decrease in the volume. For example, the equation for the reaction between nitrogen and hydrogen is shown below: \(\color{blue}{\text{N}_{2}\text{(g)}} + \color{blue}{\text{3H}_{2}\text{(g)}} \rightleftharpoons \color{red}{\text{2NH}_{3}\text{(g)}}\) The ratio in the balanced equation is \(1:3:2\). That is, for every \(\text{1}\) \(\text{molecule}\) of \(\color{blue}{\text{N}_{2}\text{ gas}}\) there are \(\text{3}\) \(\text{molecules}\) of \(\color{blue}{\text{H}_{2}\text{ gas}}\) and \(\text{2}\) \(\text{molecules}\) of \(\color{red}{\text{NH}_{3}\text{ gas}}\) (from the balanced equation). Therefore the ratio is \(\color{blue}{\textbf{4 molecules of reactant gas}}\) to \(\color{red}{\textbf{2 molecules of product gas}}\).
Consider illustration in Figure 8.2. Figure 8.2: (a) A decrease in the pressure of this reaction favours the reverse reaction (more gas molecules), the equilibrium shifts to the left. (b) An increase in the pressure of this reaction favours the forward reaction (fewer gas molecules), the equilibrium shifts to the right. Figure 8.2 shows how changing the pressure of a system results in a shift in the equilibrium to counter that change. In the original system there are 12 molecules in total: \({\color{skyblue}{6{\text {H}}_{2}}} + {\color{blue}{2{\text{N}}_{2}}} \rightleftharpoons {\color{red}{4{\text{NH}}_{3}}}\) If you decrease the pressure (shown by an increase in volume), the equilibrium will shift to increase the number of gas molecules. That shift is to the left and the number of \(\text{H}_{2}\) and \(\text{N}_{2}\) molecules will increase while the number of \(\text{NH}_{3}\) molecules will decrease: If you increase the pressure (shown by a decrease in volume), the equilibrium will shift to decrease the number of gas molecules. That shift is to the right and the number of \(\text{H}_{2}\) and \(\text{N}_{2}\) molecules will decrease while the number of \(\text{NH}_{3}\) molecules will increase: Note that the total number of nitrogen and hydrogen atoms remains the same in all three situations. Equations (a) and (b) are not balanced equations. Another example is the reaction between sulfur dioxide and oxygen: \(\color{blue}{\text{2SO}_{2}\text{(g)}} + \color{blue}{\text{O}_{2}\text{(g)}} \rightleftharpoons \color{red}{\text{2SO}_{3}\text{(g)}}\) In this reaction, \(\color{red}{\textbf{two molecules of product gas}}\) are formed for every \(\color{blue}{\textbf{three molecules of}}\) \(\color{blue}{\textbf{reactant gas}}\).
If a catalyst is added to a reaction, both the forward and reverse reaction rates will be increased. If both rates are increased then the concentrations of the reactants and products will remain the same. This means that a catalyst has no effect on the equilibrium position. However, a catalyst will affect how quickly equilibrium is reached. This is very important in industry where the longer a process takes, the more money it costs. So if a catalyst reduces the amount of time it takes to form specific products, it also reduces the cost of production. Concentration, pressure, and temperature all affect the equilibrium position of a reaction, and a catalyst affects reaction rates. However, only \(\color{red}{\textbf{temperature}}\) affects the value of \(\text{K}_{\text{c}}\).
So make sure that when comparing \(\text{K}_{\text{c}}\) values for different reactions, the different reactions took place at the same temperature. When a system is in chemical equilibrium, and there has been a change in conditions (e.g. concentration, pressure, temperature) the following steps are suggested:
Table salt is added to the (purple) solution in equilibrium: \(\underset{\color{blue}{\text{blue}}}{\underbrace{{\color{blue}{{\text{CoCl}}_{4}^{2-}}}}} + 6{\text{H}}_{2}{\text{O}} \rightleftharpoons \underset{\color{red}{\text{pink}}}{\underbrace{{\color{red}{{\text{Co(H}}_{2}{\text{O)}}_{6}^{2+}}}}} + 4{\text{Cl}}^{-}\)
Adding \(\text{NaCl}\) produces \(\text{Na}^{+}\) ions and \(\text{Cl}^{-}\) ions as the salt dissolves. Looking at the given equilibrium \(\text{Cl}^{-}\) is in the equation and the disturbance is the increase in concentration of the \(\text{Cl}^{-}\) ion.
By Le Chatelier's principle, the equilibrium position will shift to reduce the concentration of \(\text{Cl}^{-}\) ions.
The reverse reaction uses \(\text{Cl}^{-}\) ions and hence the rate of the reverse reaction will increase. The reverse reaction is favoured and the equilibrium will shift to the left.
The solution will appear more blue as more blue \(\text{CoCl}_{4}^{2-}\) ions are formed. Graphs can be used to represent data about equilibrium reactions. The following are some points to keep in mind when presented with a graph.
Rate-time graphs
For the reaction \(2\text{AB}(\text{g})\) \(\rightleftharpoons\) \(2\text{A}(\text{g}) + \text{B}_{2}(\text{g})\), \(\Delta\)H = \(\text{26}\) \(\text{kJ}\) the following graph can be plotted: What stress has occurred in this system? Label the graph with what is happening at each stage.
The axes are labelled rate and time. Therefore this is a rate-time graph.
Both rates are affected in the same way (increased) therefore the stress must be a catalyst or a change in temperature. (A change in pressure or concentration would favour one reaction direction only)
No, the forward rate is increased more than the reverse rate. Therefore the stress must be a change in temperature. (A catalyst would increase both rates equally)
The forward reaction is endothermic (\(\Delta\)H is positive). The forward reaction was favoured more than the reverse reaction. An increase in temperature will favour the reaction that cools the reaction vessel (the endothermic reaction). Therefore the stress must have been an increase in temperature.
What is responsible for the change at t = \(\text{10}\) \(\text{minutes}\) in the graph below?
The axes are labelled rate and time. Therefore this is a rate-time graph.
Both rates are affected in the same way (increased) therefore the stress must be a catalyst or a change in temperature. (A change in pressure or concentration would favour one reaction direction only)
Yes, both rates are increased by the same amount.
The addition of a catalyst (a change in temperature would affect both rates, but unequally). Concentration-time and mole-time graphs
Consider the following chemical equation and graph and answer the questions that follow. \[\text{CO}(\text{g}) + \text{Cl}_{2}(\text{g}) \rightleftharpoons \text{COCl}_{2}(\text{g})\]
The axes are labelled concentration and time. Therefore this is a concentration-time graph.
The concentration of all three compounds becomes constant at t = \(\text{15}\) \(\text{s}\). This means that the reaction has reached equilibrium.
If the concentrations of the reactants are steadily decreasing and the concentrations of the products are steadily increasing then the forward reaction is using reactants faster than the reverse reaction is using products. The rate of the forward reaction must be faster than the rate of the reverse reaction. If the concentrations of the reactants are steadily increasing and the concentrations of the products are steadily decreasing then the forward reaction must be using reactants slower than the reverse reaction is using products. The rate of the forward reaction must be slower than the rate of the reverse reaction. If the concentrations of the reactants and products are constant, the forward reaction must be using reactants at the same rate as the reverse reaction is using products. The rate of the forward reaction must be equal to the rate of the reverse reaction (the system is in equilibrium).
At \(t = \text{5} \text{ s}\) and at \(t = \text{23} \text{ s}\) the concentrations of the reactants are decreasing and the concentration of the product is increasing. Therefore the rate of the forward reaction is faster than the rate of the reverse reaction. At \(t = \text{17} \text{s}\) the concentrations of the reactants and products are constant (unchanging). Therefore the reaction is in equilibrium and the rate of the forward reaction equals the rate of the reverse reaction.
Reading off the graph you can see that:
\(\text{K}_{\text{c}} = \dfrac{\text{[COCl}_{2}{\text{]}}}{\text{[CO][Cl}_{2}{\text{]}}}\) At \(t = \text{17 s}\): \(\;\text{K}_{\text{c}} = \dfrac{\text{0,9}}{(\text{1,8})(\text{1,1})} =\) \(\text{0,45}\) At \(t = \text{27 s}\): \(\;\text{K}_{\text{c}} = \dfrac{\text{1,1}}{(\text{2,7})(\text{0,9})} =\) \(\text{0,45}\) At \(t = \text{42 s}\): \(\;\text{K}_{\text{c}} = \dfrac{\text{1,5}}{(\text{2,3})(\text{0,5})} =\) \(\text{1,30}\) Note that \(\text{K}_{\text{c}}\) is different at \(\text{42 s}\). This means that the stress at \(\text{30 s}\) must have been a change in temperature.
The concentration of \(\text{CO}\) increases sharply. The concentrations of the other compounds do not change dramatically. Therefore \(\text{CO}\) must have been added to the system. After this addition of \(\text{CO}\) there is a shift to reduce the amount of \(\text{CO}\), that is in the forward direction. Therefore the concentration of the product increases while the concentrations of the reactants decrease.
Only a change in temperature affects \(\text{K}_{\text{c}}\), therefore this will have no effect on \(\text{K}_{\text{c}}\).
Consider the following chemical equilibrium and graph and answer the questions that follow. \(\text{H}_{2}(\text{g}) + \text{I}_{2}(\text{g})\) \(\rightleftharpoons\) \(2\text{HI}(\text{g})\)
The axes are labelled concentration and time. Therefore this is a concentration-time graph.
The concentration of all three compounds becomes constant at t = \(\text{10}\) \(\text{s}\). This means that the system takes \(\text{10}\) \(\text{s}\) to reach equilibrium.
Reading off the graph you can see that at \(\text{10}\) \(\text{s}\): \([\text{H}_{2}]\) and \([\text{I}_{2}]\) = \(\text{0,75}\) \(\text{mol·dm$^{-3}$}\), \([\text{HI}]\) = \(\text{6,0}\) \(\text{mol·dm$^{-3}$}\) At \(\text{25}\) \(\text{s}\): \([\text{H}_{2}]\) and \([\text{I}_{2}]\) = \(\text{0,6}\) \(\text{mol·dm$^{-3}$}\), \([\text{HI}]\) = \(\text{4,8}\) \(\text{mol·dm$^{-3}$}\) At \(\text{45}\) \(\text{s}\): \([\text{H}_{2}]\) and \([\text{I}_{2}]\) = \(\text{1,5}\) \(\text{mol·dm$^{-3}$}\), \([\text{HI}]\) = \(\text{2,75}\) \(\text{mol·dm$^{-3}$}\)
\(\text{K}_{\text{c}} = \dfrac{\text{[HI]}^{2}}{\text{[H}_{2}{\text{][I}_{2}{\text{]}}}}\) At t = \(\text{10}\) \(\text{s}\): \(\text{K}_{\text{c}} = \dfrac{\text{6,0}^{2}}{(\text{0,75})(\text{0,75})} =\) \(\text{64,0}\) At t = \(\text{25}\) \(\text{s}\): \(\text{K}_{\text{c}} = \dfrac{\text{4,8}^{2}}{(\text{0,6})(\text{0,6})} =\) \(\text{64,0}\) At t = \(\text{45}\) \(\text{s}\): \(\text{K}_{\text{c}} = \dfrac{\text{2,75}^{2}}{(\text{1,5})(\text{1,5})} =\) \(\text{3,4}\) The different \(\text{K}_{\text{c}}\) at \(\text{45}\) \(\text{s}\) means that the event at t = \(\text{35}\) \(\text{s}\) must be a change in temperature.
The concentration of \(\text{HI}\) decreases sharply, as a result there is a slight decrease in the concentration of \(\text{H}_{2}\) and \(\text{I}_{2}\). Therefore \(\text{HI}\) must have been removed from the system. After this there is a shift to increase the amount of \(\text{HI}\), that is in the forward direction.
An increase in temperature caused the concentration of the product to decrease and the concentrations of the reactants to increase. This means that the reverse reaction has been favoured. An increase in temperature will favour the reaction that takes heat in and cools the reaction vessel (endothermic). Therefore the reverse reaction must be endothermic and the forward reaction must be exothermic. The reaction is exothermic. The following rules will help in predicting the changes that take place in equilibrium reactions:
The following simulation will help you to understand these concepts.
\(2\text{NO}(\text{g}) + \text{O}_{2}(\text{g})\) \(\rightleftharpoons\) \(2\text{NO}_{2}(\text{g})\) \(\quad \Delta{H} < 0\) How will:
Study the graph and answer the questions that follow:
The concentration of reactants decreases from the start of the reaction to equilibrium. Therefore A and B are reactants. The concentration of products increases from the start of the reaction to equilibrium. Therefore C is a product. \(\text{aA} + \text{bB}\) \(\rightleftharpoons\) \(\text{cC}\) We are told that all coefficients in the balanced equation equal 1. Therefore the general equation is: \(\text{A} + \text{B}\) \(\rightleftharpoons\) \(\text{C}\)
A and B (the reactants) have higher concentrations at equilibrium.
The reactants have higher concentrations than the products, therefore the equilibrium must favour the reactants.
[A] = \(\text{2,5}\) \(\text{mol·dm$^{-3}$}\) [B] = \(\text{2,0}\) \(\text{mol·dm$^{-3}$}\) [C] = \(\text{1,5}\) \(\text{mol·dm$^{-3}$}\)
\[\text{K}_{\text{c}}=\frac{\text{[C]}}{\text{[A][B]}}=\frac{\text{1,5}}{(\text{2,5})(\text{2,0})} = \text{0,3}\]
The forward reaction has \(\Delta{H} > 0\). This means that the forward reaction is endothermic. The reverse reaction must therefore be exothermic.
The endothermic reaction would be favoured by an increase in temperature (to lower the temperature). This is the forward reaction.
The forward reaction is favoured, therefore the equilibrium would shift to the right. This means that the reactant concentrations ([A] and [B]) would decrease and the product concentration ([C]) would increase. Textbook Exercise 8.3
The following reaction has reached equilibrium in a closed container: \[\text{C}(\text{s}) + \text{H}_{2}\text{O}(\text{g}) \rightleftharpoons \text{CO}(\text{g}) + \text{H}_{2}(\text{g}) \quad \Delta{H} > 0\] The pressure of the system is then decreased. How will the concentration of the \(\text{H}_{2}\)(g) and the value of \(\text{K}_{\text{c}}\) be affected when the new equilibrium is established? Assume that the temperature of the system remains unchanged.
(IEB Paper 2, 2004)
According to Le Chatelier's principle, the reaction that increases the number of gas molecules (and therefore the pressure) will be favoured. That is the forward reaction, and so the concentration of hydrogen will increase. Only temperature affects the value of \(\text{K}_{\text{c}}\), therefore \(\text{K}_{\text{c}}\) remains the same. b) Hydrogen concentration increases and equilibrium constant stays the same
A learner, noticing that the colour of the gas mixture in the syringe is no longer changing, comments that all chemical reactions in the syringe must have stopped. Is this assumption correct? Explain.
No. The learner is not correct. A dynamic chemical equilibrium has been reached and the products are changing into reactants at the same rate as reactants are changing into products. This results in no colour change being observed in the syringe.
The gas in the syringe is cooled. The volume of the gas is kept constant during the cooling process. Will the gas be lighter or darker at the lower temperature? Explain your answer.
The forward reaction is exothermic (\(\Delta H < 0\)). A decrease in temperature will favour the forward reaction. \(\text{N}_{2}\text{O}_{4}(\text{g})\) is colourless, therefore the gas will be lighter at the lower temperature.
The volume of the syringe is now reduced (at constant temperature) to \(\text{75}\) \(\text{cm$^{3}$}\) by pushing the plunger in and holding it in the new position. There are \(\text{0,032}\) \(\text{moles}\) of \(\text{N}_{2}\text{O}_{4}\) gas present once the equilibrium has been re-established at the reduced volume (\(\text{75}\) \(\text{cm$^{3}$}\)). Calculate the value of the equilibrium constant for this equilibrium. (IEB Paper 2, 2004)
The volume is reduced, so the pressure is increased. This means that the foward reaction will be favoured to reduce the number of gas molecules. There are \(\text{0,01}\) moles of \(\text{NO}_{2}\) and \(\text{0,03}\) moles of \(\text{N}_{2}\text{O}_{4}\) initially. For every 2x moles of \(\text{NO}_{2}\) used, 1x moles of \(\text{N}_{2}\text{O}_{4}\) are produced.
There are \(\text{0,032}\) \(\text{mol}\) \(\text{N}_{2}\text{O}_{4}\) at the new equilibrium. Therefore \(\text{0,03}\) \(\text{mol}\) + x = \(\text{0,032}\) \(\text{mol}\) x = \(\text{0,032}\) - \(\text{0,03}\) \(\text{mol}\) = \(\text{0,002}\) \(\text{mol}\) n(\(\text{NO}_{2}\)) at equilibrium = \(\text{0,01}\) \(\text{mol}\) - \(\text{2}\) \(\times\) \(\text{0,002}\) \(\text{mol}\) = \(\text{0,006}\) \(\text{mol}\) C(\(\text{NO}_{2}\)) at equilibrium = \(\dfrac{\text{0,006}\text{ mol}}{\text{0,075}\text{ dm$^{3}$}}\) = \(\text{0,08}\) \(\text{mol·dm$^{-3}$}\) C(\(\text{N}_{2}\text{O}_{4}\)) at equilibrium = \(\dfrac{\text{0,032}\text{ mol}}{\text{0,075}\text{ dm$^{3}$}}\) = \(\text{0,43}\) \(\text{mol·dm$^{-3}$}\)
\(\text{K}_{\text{c}}=\dfrac{\text{[N}_{2}\text{O}_{4}\text{]}}{\text{[NO}_{2}\text{]}^{2}}=\dfrac{\text{0,43}}{(\text{0,08})^{2}}=\text{67,19}\)
How many moles of gas \(\text{X}_{2}\text{Y}_{3}\) are formed by the time the reaction reaches equilibrium at \(\text{30}\) \(\text{seconds}\)?
From the graph: \(\text{0,5}\) \(\text{mol}\)
Calculate the value of the equilibrium constant at t = \(\text{50}\) \(\text{s}\).
Draw up a RICE table.
There are \(\text{0,5}\) \(\text{mol}\) of \(\text{X}\) present at equilibrium. Therefore x = \(\text{0,5}\) \(\text{mol}\) From the graph n(\(\text{X}\)) = \(\text{3}\) \(\text{mol}\) n(\(\text{Y}\)) = \(\text{4}\) - \(\text{3}\) x \(\text{0,5}\) \(\text{mol}\) = \(\text{2,5}\) \(\text{mol}\)
C(\(\text{X}\)) = \(\dfrac{\text{3}\text{ mol}}{\text{2}\text{ dm$^{3}$}} =\)\(\text{1,5}\) \(\text{mol·dm$^{-3}$}\) C(\(\text{Y}\)) = \(\dfrac{\text{2,5}\text{ mol}}{\text{2}\text{ dm$^{3}$}} =\)\(\text{1,25}\) \(\text{mol·dm$^{-3}$}\) C(\(\text{X}_{2}\text{Y}_{3}\)) = \(\dfrac{\text{0,5}\text{ mol}}{\text{2}\text{ dm$^{3}$}} =\)\(\text{0,25}\) \(\text{mol·dm$^{-3}$}\)
\({K}_{c}=\dfrac{\text{[X}_{2}\text{Y}_{3}\text{]}}{\text{[X]}^{2}\text{[Y]}^{3}}=\dfrac{\text{0,25}}{(\text{1,5})^{2}(\text{1,25})^{3}}=\text{0,057}\)
At \(\text{70}\) \(\text{s}\) the temperature is increased. Is the forward reaction endothermic or exothermic? Explain in terms of Le Chatelier's Principle.
Exothermic. The amount of product decreases (and amount of reactants increases) when the temperature is increased indicating that the reverse reaction is favoured. Le Chatelier's principle states that an increase in temperature will favour the endothermic reaction (cooling the reaction vessel). Therefore the reverse reaction must be endothermic and the forward reaction is exothermic.
How will this increase in temperature affect the value of the equilibrium constant?
A change in temperature is the only thing that affects the equilibrium constant. The amount of product decreased and the amount of reactants increased, therefore the equilibrium constant will decrease.
State how long (in minutes) it took for the reaction to reach equilibrium for the first time.
The number of moles of each reactant and the product is constant (have reached equilibrium) after t = \(\text{5}\) \(\text{minutes}\)
Write down an expression for the equilibrium constant, \(\text{K}_{\text{c}}\), for this particular reaction.
\(\text{K}_{\text{c}} = \dfrac{\text{[AB}_{2}{\text{]}}^{2}}{\text{[A}_{2}{\text{][B}}_{2}{\text{]}}^{2}}\)
Calculate the concentration of each of the reactants and the product using figures from the graph between \(\text{5}\) \(\text{minutes}\) and \(\text{10}\) \(\text{minutes}\) and hence calculate the equilibrium constant \(\text{K}_{\text{c}}\) for this reaction at \(\text{298}\) \(\text{K}\)
The concentration values are read off the graph. C(\(\text{A}_{2}\)) \(= \dfrac{\text{2,0}\text{ mol}}{\text{2}\text{ dm$^{3}$}} =\) \(\text{1,0}\) \(\text{mol·dm$^{-3}$}\) C(\(\text{B}_{2}\)) \(= \dfrac{\text{1,2}\text{ mol}}{\text{2}\text{ dm$^{3}$}} =\) \(\text{0,6}\) \(\text{mol·dm$^{-3}$}\) C(\(\text{AB}_{2}\)) \(= \dfrac{\text{0,8}\text{ mol}}{\text{2}\text{ dm$^{3}$}} =\) \(\text{0,4}\) \(\text{mol·dm$^{-3}$}\) \(\text{K}_{\text{c}} = \dfrac{\text{[AB}_{2}{\text{]}}^{2}}{\text{[A}_{2}{\text{][B}_{2}{\text{]}}^{2}}} = \dfrac{\text{0,4}^{2}}{(\text{1,0})(\text{0,6})^{2}} =\) \(\text{0,44}\)
State what a low value of \(\text{K}_{\text{c}}\) indicates about the yield of product for a reaction.
A value of \(\text{K}_{\text{c}}\) between \(\text{0}\) and \(\text{1}\) indicates that the equilibrium lies to the left. Therefore the yield of product is low (more reactants than products).
Why is it not possible to calculate \(\text{K}_{\text{c}}\) using figures from the graph during the first \(\text{5}\) \(\text{minutes}\)
\(\text{K}_{\text{c}}\) is the equilibrium constant, therefore it cannot be calculated except when the reaction is in equilibrium. During the first \(\text{5}\) \(\text{minutes}\) the reaction is not in equilibrium.
State Le Chatelier's principle.
When an external stress (change in pressure, temperature or concentration) is applied to a system in chemical equilibrium, the equilibrium will change in such a way as to reduce the effect of the stress.
At \(\text{10}\) \(\text{minutes}\) the temperature of the flask was increased. Using Le Chatelier's principle, determine if the production of \(\text{AB}_{2}\) is exothermic or endothermic?
\(\text{AB}_{2}\) is produced through the forward reaction. Le Chatelier's principle states that if the temperature is increased the equilibrium will change to decrease the temperature of the vessel. That is the reaction that takes in heat (endothermic) will be favoured. The number of moles of the reactants decreases after the temperature increase, while the number of moles of the product increases. Therefore the temperature increase favours the forward reaction. This means that the forward reaction is endothermic.
Increasing the pressure of the flask with no change to temperature.
Adding a catalyst to the flask.
Increasing the temperature of the flask. In industrial processes, it is important to get the product as quickly and as efficiently as possible. The less expensive the process the better. The Haber process is a good example of an industrial process which uses the equilibrium principles that have been discussed. The equation for the process is as follows: \[\text{N}_{2}(\text{g}) + 3\text{H}_{2}(\text{g}) \rightleftharpoons 2\text{NH}_{3}(\text{g}) + \text{energy}\] Since the forward reaction is exothermic, to produce a lot of product and favour the forward reaction the system needs to be colder. However, cooling a system slows down all chemical reactions and so the system can't be too cold. This process is carried out at a much higher temperature to ensure the speed of production. Because high temperature favours the reverse reaction, the \(\text{NH}_{3}\) product is actually removed as it is made (product concentration decreased) to prevent ammonia being used in the reverse reaction. The decrease of product concentration favours the forward reaction. High pressure is also used to ensure faster reaction time and to favour the production of \(\text{NH}_{3}\). The forward reaction is favoured by higher pressures because there are \(\text{2}\) gas molecules of product for every \(\text{4}\) gas molecules of reactant. Refer to Chapter 14 for more information on the Haber process and other industrial applications. Textbook Exercise 8.4
What happens to the value of \(\text{K}_{\text{c}}\) as the temperature increases?
The value of \(\text{K}_{\text{c}}\) decreases
Which reaction is being favoured when the temperature is \(\text{300}\) \(\text{℃}\)?
A value of \(\text{K}_{\text{c}}\) between 0 and 1 means that there are more reactants than products. Therefore the reverse reaction is favoured.
According to this table, which temperature would be best if you wanted to produce as much ammonia as possible? Explain.
\(\text{25}\) \(\text{℃}\). At this temperature the forward reaction is favoured and so the the maximum yield of ammonia is achieved. |