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8.1 What is chemical equilibrium?
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8.3 Le Chatelier's principle
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Consider the reversible chemical reaction:
\[{\text{a}}\color{orange}{\text{A}} + {\text{b}}\color{orange}{\text{B}} \rightleftharpoons {\text{c}}\color{purple}{\text{C}} + {\text{d}}\color{purple}{\text{D}}\]
\(\color{orange}{\text{A}}\) and \(\color{orange}{\text{B}}\) are the reactants, \(\color{purple}{\text{C}}\) and \(\color{purple}{\text{D}}\) are the products and a, b, c, and d are the coefficients from the balanced reaction.
When the rate of the forward reaction equals the rate of the reverse reaction, the system is in chemical equilibrium. It is useful to know how much of each substance is in the container - in particular the amount of products compared to the amount of reactants. A simple ratio from the balanced chemical equation gives us a number called the equilibrium constant (\(\text{K}_{\text{c}}\)).
The subscript c refers to the concentration of all the substances at equilibrium.
\[\text{K}_{\text{c}} = \dfrac{{[\color{purple}{\text{C}}\text]}^{c}{[\color{purple}{\text{D}}\text]}^{d}}{{[\color{orange}{\text{A}}\text]}^{a}{[\color{orange}{\text{B}}\text]}^{b}}\]
The equation for the equilibrium constant uses the concentrations of the reactants and products. Square brackets mean the concentration of the compound written within the square brackets: [x] means the concentration of x.
\([\color{orange}{\text{A}}\text], [\color{orange}{\text{B}}\text], [\color{purple}{\text{C}}\text] \text{and} [\color{purple}{\text{D}}\text]\) are the molar concentrations of each substance present at equilibrium.
The equilibrium constant (\(\text{K}_{\text{c}}\)) is the ratio between the concentration of products and reactants in a chemical reaction.
When the concentration of the reactants is much larger than the concentration of the products \(\text{K}_{\text{c}}\) will be small (normally less than one). When the concentration of the reactants is much less than that of the products \(\text{K}_{\text{c}}\) will be large (normally greater than one).
For example:
\[2\color{orange}{\text{NO}{\text{(g)}}} + 2\color{orange}{\text{H}_{2}{\text{(g)}}} \rightleftharpoons \color{purple}{\text{N}_{2}{\text{(g)}}} + 2\color{purple}{\text{H}_{2}{\text{O(g)}}}\]
The \(\color{orange}{\textbf{reactants}}\) are \(\color{orange}{\text{NO}}\) and \(\color{orange}{\text{H}_{2}}\). The \(\color{purple}{\textbf{products}}\) are \(\color{purple}{\text{N}_{2}}\) and \(\color{purple}{\text{H}_{2}{\text{O}}}\).
It is important to look at the coefficients of the equation as well:
\[\color{orange}{\textbf{2}}\text{NO}(\text{g}) + \color{orange}{\textbf{2}}\text{H}_{2}(\text{g}) \rightleftharpoons \color{purple}{\textbf{1}}\text{N}_{2}(\text{g}) + \color{purple}{\textbf{2}}\text{H}_{2}\text{O}(\text{g})\]
\(\text{K}_{\text{c}}\) for this equation will therefore be written as follows:
\[\text{K}_{\text{c}}=\dfrac{\text{[N}_{2}{\text{]}}{\color{purple}{^{1}}}{\text{[H}}_{2}{\text{O]}}{\color{purple}{^{2}}}}{\text{[NO]}{\color{orange}{^{2}}}{\text{[H}}_{2}{\text{]}}{\color{orange}{^{2}}}}\]
In the expression for \(\text{K}_{\text{c}}\) the concentration of a product or reactant is taken to the power of its coefficient in the balanced reaction. So if the coefficient of a product \((\text{C})\) is 3 in the balanced equation, then the concentration of \(\text{C}\) \(([\text{C}])\) will be written \([\text{C}]^{3}\) in the expression for \(\text{K}_{\text{c}}\).
We leave out reactants or products that are either pure liquids or in the solid phase when calculating \(\text{K}_{\text{c}}\). For example:
\[\color{red}{\text{C(s)}} + \text{H}_{2}\text{O}(\text{g}) \rightleftharpoons \text{CO}(\text{g}) + \text{H}_{2}(\text{g})\]
\[\text{K}_{\text{c}}=\dfrac{{\text{[CO][H}}_{2}\text{]}}{{\text{[H}}_{2}\text{O]}}\]
When calculations involving the equilibrium constant are done, the following tips may help:
Always read the question carefully to be sure that you understand what you have been asked to calculate.
If the equilibrium constant is involved, make sure that the concentrations you use are the concentrations at equilibrium, and not the concentrations or quantities that are present at some other time in the reaction.
When you are doing more complicated calculations, it helps to draw up a RICE table (Table 8.1). This is gone into in more detail later in this section.
Remember that the general form of the expression for \(\text{K}_{\text{c}}\) of a reaction at chemical equilibrium (a\(\color{orange}{\text{A}}\) + b\(\color{orange}{\text{B}}\) \(\rightleftharpoons\) c\(\color{purple}{\text{C}}\) + d\(\color{purple}{\text{D}}\)) is:
\[{K}_{c}=\dfrac{{[\color{purple}{\text{C}}\text]}^{c}{[\color{purple}{\text{D}}\text]}^{d}}{{[\color{orange}{\text{A}}\text]}^{a}{[\color{orange}{\text{B}}\text]}^{b}}\]
For the reaction:
\[9\text{X}(\text{g}) + \text{Y}_{3}(\text{g}) \rightleftharpoons 3\text{X}_{3}\text{Y}(\text{g})\]
Write an expression for the equilibrium constant \({\text{K}}_{\text{c}}\).
\[\text{K}_{\text{c}}=\dfrac{\text{[C]}^{c}{\text{[D]}}^{d}}{\text{[A]}^{a}{\text{[B]}}^{b}}\]
\(\text{X(g)}\) and \(\text{Y}_{3}(\text{g})\) are both reactants. They are gases and will be included in the expression.
\(\text{X}_{3}\text{Y}(\text{g})\) is a product. It is a gas and will be included in the expression.
\[\text{K}_{\text{c}}=\dfrac{\text{[X}_{3}{\text{Y]}}^{z}}{\text{[X]}^{x}{\text{[Y}}_{3}{\text{]}}^{y}}\]
The compound \(\text{X}\) has a coefficient of 9. The compound \(\text{Y}_{3}\) has a coefficient of 1. The compound \(\text{X}_{3}\text{Y}\) has a coefficient of 3.
\[\text{K}_{\text{c}}=\dfrac{\text{[X}_{3}{\text{Y]}}^{3}}{\text{[X]}^{9}{\text{[Y}}_{3}{\text{]}}^{1}}\]
For the reaction:
\[\text{S}(\text{s}) + \text{O}_{2}(\text{g}) \rightleftharpoons \text{SO}_{2}(\text{g})\]
Write an expression for the equilibrium constant.
Calculate the equilibrium concentration of \(\text{O}_{2}\) if:
\(\text{K}_{\text{c}}\) = 6 and \([\text{SO}_{2}]\) = \(\text{3}\) \(\text{mol·dm$^{-3}$}\) at equilibrium.
\[\color{red}{\text{S(s)}} + \text{O}_{2}(\text{g}) \rightleftharpoons \text{SO}_{2}(\text{g})\]
\(\text{O}_{2}\) and \(\text{SO}_{2}\) are gases and so will be part of the expression for \(\text{K}_{\text{c}}\). \(\text{S}\) is solid and so will not be part of the expression for \(\text{K}_{\text{c}}\).
\[\text{K}_{\text{c}}=\frac{{\text{[SO}}_{2}\text{]}}{{\text{[O}}_{2}\text{]}}\]
For the reaction:
\[\text{SO}_{2}(\text{g}) + \text{NO}_{2}(\text{g}) \rightleftharpoons \text{NO}(\text{g}) + \text{SO}_{3}(\text{g})\]
The concentrations of the compounds used are:
\([\text{SO}_{2}]\) = \(\text{0,2}\) \(\text{mol·dm$^{-3}$}\), \([\text{NO}_{2}]\) = \(\text{0,1}\) \(\text{mol·dm$^{-3}$}\), \([\text{NO}]\) = \(\text{0,4}\) \(\text{mol·dm$^{-3}$}\) and
\([\text{SO}_{3}]\) = \(\text{0,2}\) \(\text{mol·dm$^{-3}$}\)
Calculate the value of \(\text{K}_{\text{c}}\).
\[\text{K}_{\text{c}}=\dfrac{\text{[C]}^{c}{\text{[D]}}^{d}}{\text{[A]}^{a}{\text{[B]}}^{b}}\]
\(\text{SO}_{2}(\text{g})\) and \(\text{NO}_{2}(\text{g})\) are both reactants. They are gases and will be included in the expression.
\(\text{NO}(\text{g})\) and \(\text{SO}_{3}(\text{g})\) are both products. They are gases and will be included in the expression.
All four compounds have a coefficient of 1.
\[\text{K}_{\text{c}}=\frac{{\text{[NO]}}^{1}{\text{[SO}}_{3}\text{]}^{1}}{{\text{[SO}}_{2}{\text{]}}^{1}{\text{[NO}}_{2}\text{]}^{1}}\]
\[\text{K}_{\text{c}}=\dfrac{\text{(}\text{0,4}\text{)(}\text{0,2}\text{)}}{\text{(}\text{0,2}\text{)(}\text{0,1}\text{)}}= \text{4}\]
You do not need to fill in the coefficients in a \(\text{K}_{\text{c}}\) calculation when they are \(\text{1}\). They are shown here so that you do not forget where the coefficients are reflected in the equation.
When you are doing more complicated calculations, it helps to draw up a RICE table (Table 8.1). A RICE table is an easy way of organising information in equilibrium calculations.
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R |
Reaction |
the balanced chemical equation |
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I |
Initial quantity |
the moles of reactants and products at the beginning of the reaction |
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C |
Change |
how much the moles of the reactants and products changed between the beginning of the reaction and equilibrium |
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E |
Equilibrium quantity |
the moles of the reactants and products at equilibrium |
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E |
Equilibrium concentration |
to calculate \(\text{K}_{\text{c}}\) you need the concentration of the reactants and products at equilibrium |
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Reaction |
\(\qquad\) | \(\qquad\) | \(\qquad\) | \(\qquad\) |
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Initial quantity (mol) |
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Change (mol) |
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Equilibrium quantity (mol) |
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Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
Table 8.1: A RICE table for mole values.
Here are some guidelines on how to use a RICE table:
Fill in the balanced chemical equation: \(\color{blue}{\text{aA}} + \color{blue}{\text{bB}} \rightleftharpoons \color{red}{\text{cC}} + \color{red}{\text{dD}}\)
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Reaction |
\(\color{blue}{\textbf{aA}}\) |
\(\color{blue}{\textbf{bB}}\) |
\(\color{red}{\textbf{cC}}\) |
\(\color{red}{\textbf{dD}}\) |
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Initial quantity (mol) |
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Change (mol) |
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Equilibrium quantity (mol) |
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Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
In the initial row fill in the number of moles of each substance present at the beginning of the reaction. For example, if there are \(\text{0,3}\) \(\text{moles}\) of A, \(\text{0,7}\) \(\text{moles}\) of B and no moles of product initially:
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Reaction |
\(\color{blue}{\textbf{aA}}\) |
\(\color{blue}{\textbf{bB}}\) |
\(\color{red}{\textbf{cC}}\) |
\(\color{red}{\textbf{dD}}\) |
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Initial quantity (mol) |
\(\text{0,3}\) |
\(\text{0,7}\) |
\(\text{0}\) |
\(\text{0}\) |
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Change (mol) |
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Equilibrium quantity (mol) |
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Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
In the change row fill in the number of moles of reactant used, or the number of moles of product formed, in terms of \(x\). Use the balanced equation for this. For example, if \(\text{d}x\) moles of \(\text{D}\) is produced, \(\text{b}x\) moles of B will be used:
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Reaction |
\(\color{blue}{\textbf{aA}}\) |
\(\color{blue}{\textbf{bB}}\) |
\(\color{red}{\textbf{cC}}\) |
\(\color{red}{\textbf{dD}}\) |
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Initial quantity (mol) |
\(\text{0,3}\) |
\(\text{0,7}\) |
\(\text{0}\) |
\(\text{0}\) |
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Change (mol) |
\(-\text{a}x\) |
\(-\text{b}x\) |
\(+\text{c}x\) |
\(+\text{d}x\) |
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Equilibrium quantity (mol) |
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Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
In the equilibrium row fill in the number of moles of each substance present at equilibrium in terms of \(x\). For example, there will be \(\text{0,3} - \text{a}x\) moles of \(\text{A}\) at equilibrium.
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Reaction |
\(\color{blue}{\textbf{aA}}\) |
\(\color{blue}{\textbf{bB}}\) |
\(\color{red}{\textbf{cC}}\) |
\(\color{red}{\textbf{dD}}\) |
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Initial quantity (mol) |
\(\text{0,3}\) |
\(\text{0,7}\) |
\(\text{0}\) |
\(\text{0}\) |
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Change (mol) |
\(-\text{a}x\) |
\(-\text{b}x\) |
\(+\text{c}x\) |
\(+\text{d}x\) |
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Equilibrium quantity (mol) |
\(\text{0,3} - \text{a}x\) |
\(\text{0,7} - \text{b}x\) |
\(+\text{c}x\) |
\(+\text{d}x\) |
|
Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
Use any extra information you have been given to calculate x, and then finish filling in the table.
For example, if the number of moles of \(\text{B}\) at equilibrium is \(\text{0,2}\):
\(\text{0,7} - \text{b}x = \text{0,2}\) and you can solve for \(x\).
In the last row fill in the concentration of each substance at equilibrium. You will have to calculate this value.
\(\text{1,4}\) \(\text{moles}\) of \(\text{NH}_{3}\)(g) is introduced into a sealed \(\text{2,0}\) \(\text{dm$^{3}$}\) reaction vessel. The ammonia decomposes when the temperature is increased to \(\text{600}\) \(\text{K}\) and reaches equilibrium as follows:
\[2\text{NH}_{3}(\text{g}) \rightleftharpoons \text{N}_{2}(\text{g}) + 3\text{H}_{2}(\text{g})\]
When the equilibrium mixture is analysed, the concentration of \(\text{NH}_{3}\)(g) is \(\text{0,3}\) \(\text{mol·dm$^{-3}$}\).
Calculate the concentration of \(\text{N}_{2}\)(g) and \(\text{H}_{2}\)(g) in the equilibrium mixture.
Calculate the equilibrium constant for the reaction at \(\text{600}\) \(\text{K}\).
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Reaction |
|||
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Initial quantity (mol) |
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Change (mol) |
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Equilibrium quantity (mol) |
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Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
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Reaction |
\(\color{blue}{\textbf{2NH}_{3}}\) |
\(\color{red}{\textbf{1N}_{2}}\) |
\(\color{red}{\textbf{3H}_{2}}\) |
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Initial quantity (mol) |
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Change (mol) |
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Equilibrium quantity (mol) |
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|
Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
Make sure you know which compounds are the \(\color{blue}{\textbf{reactants}}\), and which are the \(\color{red}{\textbf{products}}\).
There are \(\text{1,4}\) \(\text{mol}\) of \(\text{NH}_{3}\), and no moles of \(\text{N}_{2}\) or \(\text{H}_{2}\) at the beginning of the reaction.
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Reaction |
\(\color{blue}{\textbf{2NH}_{3}}\) |
\(\color{red}{\textbf{1N}_{2}}\) |
\(\color{red}{\textbf{3H}_{2}}\) |
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Initial quantity (mol) |
\(\text{1,4}\) |
\(0\) |
\(0\) |
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Change (mol) |
|||
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Equilibrium quantity (mol) |
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Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
The mole ratio of \(\text{NH}_{3}(\text{g})\) : \(\text{N}_{2}\) : \(\text{H}_{2}\) is \(2:1:3\).
Therefore, for every \(\text{2}\) moles of \(\text{NH}_{3}\) used, \(\text{1}\) mole of \(\text{N}_{2}\) and \(\text{3}\) moles of \(\text{H}_{2}\) will be formed. If the number of moles of \(\text{NH}_{3}\) decreases by \(2x\), then the number of moles of \(\text{N}_{2}\) increases by \(1x\) and the number of moles of \(\text{H}_{3}\) increases by \(3x\).
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Reaction |
\(\color{blue}{\textbf{2NH}_{3}}\) |
\(\color{red}{\textbf{1N}_{2}}\) |
\(\color{red}{\textbf{3H}_{2}}\) |
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Initial quantity (mol) |
\(\text{1,4}\) |
\(0\) |
\(0\) |
|
Change (mol) |
\(-\text{2}x\) |
\(+x\) |
\(\text{+3}x\) |
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Equilibrium quantity (mol) |
|||
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Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
To get the number of moles of a substance at equilibrium you take the number of moles initially, and add the change in number of moles in terms of \(x\).
|
Reaction |
\(\color{blue}{\textbf{2NH}_{3}}\) |
\(\color{red}{\textbf{1N}_{2}}\) |
\(\color{red}{\textbf{3H}_{2}}\) |
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Initial quantity (mol) |
\(\text{1,4}\) |
\(0\) |
\(0\) |
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Change (mol) |
\(-\text{2}x\) |
\(+x\) |
\(\text{+3}x\) |
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Equilibrium quantity (mol) |
\(\text{1,4}\) \(-\text{2}x\) |
\(+x\) |
\(\text{+3}x\) |
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Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
You know that the concentration of \(\text{NH}_{3}\) is \(\text{0,3}\) \(\text{mol·dm$^{-3}$}\) at equilibrium and the reaction vessel has a volume of \(\text{2}\) \(\text{dm$^{3}$}\).
c = \(\dfrac{n}{V}\), therefore \(n = c \times V\)
\(n(\text{NH}_{3}\) at equilibrium) = \(\text{0,3}\) \(\text{mol·dm$^{-3}$}\) \(\times\) \(\text{2,0}\) \(\text{dm$^{3}$}\) = \(\text{0,6}\) \(\text{mol}\)
From the RICE table we can see that there are \(\text{1,4}\) \(-\text{2}x\) moles of \(\text{NH}_{3}\) at equilibrium. Therefore:
\(\text{0,6}\) \(\text{mol}\) = \(\text{1,4}\) \(\text{mol} - 2x\)
\(2x = \text{1,4}\) \(\text{mol}\) - \(\text{0,6}\) \(\text{mol}\) = \(\text{0,8}\) \(\text{mol}\)
Therefore, \(x = \text{0,4}\) \(\text{mol}\)
From the RICE table we can see that the number of moles of \(\text{N}_{2}\) at equilibrium \(= x\)
therefore \(n(\text{N}_{2}\)) = \(\text{0,4}\) \(\text{mol}\)
\(c(\text{N}_{2}\))\(= \dfrac{n}{V} = \dfrac{\text{0,4}\text{ mol}}{\text{2,0}\text{ dm$^{3}$}} =\) \(\text{0,2}\) \(\text{mol·dm$^{-3}$}\)
From the RICE table we can see that the number of moles of \(\text{H}_{2}\) at equilibrium \(= 3x\)
therefore \(n(\text{H}_{2}\)) = \(\text{3}\times \text{0,4}\) \(\text{mol}\) = \(\text{1,2}\) \(\text{mol}\)
\(c(\text{H}_{2}\)) \(= \dfrac{n}{V} = \dfrac{\text{1,2}\text{ mol}}{\text{2,0}\text{ dm$^{3}$}} =\) \(\text{0,6}\) \(\text{mol·dm$^{-3}$}\)
|
Reaction |
\(\color{blue}{\textbf{2NH}_{3}}\) |
\(\color{red}{\textbf{1N}_{2}}\) |
\(\color{red}{\textbf{3H}_{2}}\) |
|
Initial quantity (mol) |
\(\text{1,4}\) |
\(0\) |
\(0\) |
|
Change (mol) |
\(-\text{2}x\) |
\(+x\) |
\(\text{+3}x\) |
|
Equilibrium quantity (mol) |
\(\text{1,4}\) \(-\text{2}x\) |
\(+x\) |
\(\text{+3}x\) |
|
Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
\(\text{0,3}\) |
\(\text{0,2}\) |
\(\text{0,6}\) |
\(\text{NH}_{3}\) is the reactant, \(\text{N}_{2}\) and \(\text{H}_{2}\) are the products. They are all in the gas phase, and so are included in the expression for \(\text{K}_{\text{c}}\).
\(\text{K}_{\text{c}}\) \(= \dfrac{\text{[N}_{2}\text{]}\text{[H}_{2}\text{]}^{3}}{\text{[NH}_{3}\text{]}^{2}}\)
\(\text{K}_{\text{c}}\) \(= \dfrac{\text{(}\text{0,2}\text{)}\text{(}\text{0,6}\text{)}^{3}}{\text{(}\text{0,3}\text{)}^{2}} =\) \(\text{0,48}\)
Hydrogen and iodine gas react according to the following expression:
\(\text{H}_{2}(\text{g}) + \text{I}_{2}(\text{g})\) \(\rightleftharpoons\) \(2\text{HI}(\text{g})\)
When \(\text{0,496}\) \(\text{mol}\) \(\text{H}_{2}\) and \(\text{0,181}\) \(\text{mol}\) \(\text{I}_{2}\) are heated at \(\text{450}\)\(\text{°C}\) in a \(\text{1}\) \(\text{dm$^{3}$}\) container, the equilibrium mixture is found to contain \(\text{0,00749}\) \(\text{mol}\) \(\text{I}_{2}\). Calculate the equilibrium constant for the reaction at \(\text{450}\)\(\text{°C}\).
|
Reaction |
|||
|
Initial quantity (mol) |
|||
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Change (mol) |
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Equilibrium quantity (mol) |
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|
Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
|
Reaction |
\(\color{blue}{\textbf{1H}_{2}}\) |
\(\color{blue}{\textbf{1I}_{2}}\) |
\(\color{red}{\textbf{2HI}}\) |
|
Initial quantity (mol) |
|||
|
Change (mol) |
|||
|
Equilibrium quantity (mol) |
|||
|
Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
There are \(\text{0,496}\) \(\text{moes}\) of \(\text{H}_{2}\), \(\text{0,181}\) \(\text{moles}\) of \(\text{I}_{2}\) and no moles \(\text{HI}\) at the beginning of the reaction.
|
Reaction |
\(\color{blue}{\textbf{1H}_{2}}\) |
\(\color{blue}{\textbf{1I}_{2}}\) |
\(\color{red}{\textbf{2HI}}\) |
|
Initial quantity (mol) |
\(\text{0,496}\) |
\(\text{0,181}\) |
0 |
|
Change (mol) |
|||
|
Equilibrium quantity (mol) |
|||
|
Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
The mole ratio of \(\text{H}_{2}\) : \(\text{I}_{2}\) : \(\text{HI}\) is \(1:1:2\).
Therefore, for every \(\text{1}\) mole of \(\text{H}_{2}\) used, \(\text{1}\) mole of \(\text{I}_{2}\) is used and \(\text{2}\) moles of \(\text{HI}\) will be formed. If the number of moles of \(\text{H}_{2}\) decreases by x (\(-\text{1}\)x), then the number of moles of \(\text{I}_{2}\) also decreases by x (\(-\text{1}\)x) and the number of moles of \(\text{HI}\) increases by 2x (\(\text{+2}\)x).
|
Reaction |
\(\color{blue}{\textbf{1H}_{2}}\) |
\(\color{blue}{\textbf{1I}_{2}}\) |
\(\color{red}{\textbf{2HI}}\) |
|
Initial quantity (mol) |
\(\text{0,496}\) |
\(\text{0,181}\) |
0 |
|
Change (mol) |
-x |
-x |
+2x |
|
Equilibrium quantity (mol) |
|||
|
Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
To get the number of moles of a substance at equilibrium you take the number of moles initially, and add the change in number of moles in terms of x.
|
Reaction |
\(\color{blue}{\textbf{1H}_{2}}\) |
\(\color{blue}{\textbf{1I}_{2}}\) |
\(\color{red}{\textbf{2HI}}\) |
|
Initial quantity (mol) |
\(\text{0,496}\) |
\(\text{0,181}\) |
0 |
|
Change (mol) |
-x |
-x |
+2x |
|
Equilibrium quantity (mol) |
\(\text{0,496}\) - x |
\(\text{0,181}\) - x |
+2x |
|
Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
You know that number of moles of \(\text{I}_{2}\) at equilibrium is \(\text{0,00749}\) \(\text{mol}\). From the RICE table we can see that there are \(\text{0,181}\) - x moles of \(\text{I}_{2}\) at equilibrium. Therefore:
\(\text{0,00749}\) \(\text{mol}\) = \(\text{0,181}\) \(\text{mol}\) - x
x = \(\text{0,181}\) \(\text{mol}\) - \(\text{0,00749}\) \(\text{mol}\) = \(\text{0,1735}\) \(\text{mol}\)
The volume of the container is \(\text{1}\) \(\text{dm$^{3}$}\).
From the RICE table we can see that the number of moles of \(\text{H}_{2}\) at equilibrium = \(\text{0,496}\) \(\text{mol}\) - x. Therefore:
n(\(\text{H}_{2}\)) = \(\text{0,496}\) \(\text{mol}\) - \(\text{0,1735}\) \(\text{mol}\) = \(\text{0,3225}\) \(\text{mol}\)
C(\(\text{H}_{2}\)) \(= \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0,3225}\text{ mol}}{\text{1}\text{ dm$^{3}$}} =\) \(\text{0,3225}\) \(\text{mol·dm$^{-3}$}\)
C(\(\text{I}_{2}\)) \(= \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0,00749}\text{ mol}}{\text{1}\text{ dm$^{3}$}} =\) \(\text{0,00749}\) \(\text{mol·dm$^{-3}$}\)
From the RICE table we can see that the number of moles of \(\text{HI}\) at equilibrium = 2x. Therefore:
n(\(\text{HI}\)) = \(\text{2}\) x \(\text{0,1735}\) \(\text{mol}\) = \(\text{0,347}\) \(\text{mol}\)
|
Reaction |
\(\color{blue}{\textbf{1H}_{2}}\) |
\(\color{blue}{\textbf{1I}_{2}}\) |
\(\color{red}{\textbf{2HI}}\) |
|
Initial quantity (mol) |
\(\text{0,496}\) |
\(\text{0,181}\) |
\(\text{0}\) |
|
Change (mol) |
-x |
-x |
+2x |
|
Equilibrium quantity (mol) |
\(\text{0,496}\) - x |
\(\text{0,181}\) - x |
+2x |
|
Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
\(\text{0,3225}\) |
\(\text{0,00749}\) |
\(\text{0,347}\) |
\(\text{K}_{\text{c}}\) \(= \dfrac{\text{[HI]}^{2}}{\text{[H}_{2}{\text{][I}}_{2}{\text{]}}}\)
\(\text{K}_{\text{c}}\) \(= \dfrac{(\text{0,347})^{2}}{(\text{0,3225})(\text{0,00749})}\) = \(\text{49,85}\)
The formula for \(\text{K}_{\text{c}}\) has:
the concentration of the products in the numerator and the concentration of reactants in the denominator
a \(\text{K}_{\text{c}}\) value \(\color{red}{\textbf{greater than 1}}\) means that the equilibrium lies to the right
\(\color{red}{\textbf{the equilibrium favours the products}}\)
a \(\text{K}_{\text{c}}\) value \(\color{blue}{\textbf{between 0 and 1}}\) means that the equilibrium lies to the left
\(\color{blue}{\textbf{the equilibrium favours the reactants}}\)
This has been summarised in Table 8.2. Note that \(\text{K}_{\text{c}}\) cannot be 0 or less.
Yield describes the quantity of product in the container relative to the maximum product possible.
|
\(\text{K}_{\text{c}}\) value |
Equilibrium |
Yield |
|
\(\color{red}{\text{K}_{\text{c}} > 1}\) |
\(\color{red}{\text{to the right}}\) |
equilibrium favours \(\color{red}{\textbf{the products}}\) |
|
\(\color{blue}{0 <\text{K}_{\text{c}}< 1}\) |
\(\color{blue}{\text{to the left}}\) |
equilibrium favours \(\color{blue}{\textbf{the reactants}}\) |
Table 8.2: The meaning of \(\text{K}_{\text{c}}\) values.
A high \(\text{K}_{\text{c}}\) value means that:
the concentration of products is high
the equilibrium lies far to the right
the reaction has a high yield
A low \(\text{K}_{\text{c}}\) value (close to 0) means that:
the concentration of reactants is high
the equilibrium lies far to the left
the reaction has a low yield
Write the equilibrium constant expression, \(\text{K}_{\text{c}}\), or balanced chemical equation for the following reactions:
\(\text{NO}\) and \(\text{Cl}_{2}\) are both reactants. \(\text{NOCl}\) is the product. They are all in the gas state and will be included in the equilibrium constant expression.
\(\text{NO}\) has a coefficient of 2, \(\text{Cl}_{2}\) has a coefficient of 1 and \(\text{NOCl}\) has a coefficient of 2.
\(2\text{NO}(\text{g}) + \text{Cl}_{2}(\text{g})\) \(\rightleftharpoons\) \(2\text{NOCl}(\text{g})\)
\(\text{K}_{\text{c}}\) \(= \dfrac{\text{[NOCl]}^{2}}{\text{[NO]}^{2}\text{[Cl}_{2}\text{]}}\)
\({K}_{c}=\dfrac{\text{[H}_{2}\text{O(g)]}^{2}}{\text{[H}_{2}\text{(g)]}^{2}\text{[O}_{2}\text{(g)]}}\)
The \(\text{H}_{2}\text{O}(\text{g})\) must be the product and have a coefficient of \(\text{2}\).
The \(\text{H}_{2}(\text{g})\) must be a reactant and have a coefficient of \(\text{2}\).
The \(\text{O}_{2}(\text{g})\) must be a reactant and have a coefficient of \(\text{1}\).
\(2\text{H}_{2}(\text{g}) + \text{O}_{2}(\text{g})\) \(\to\) \(2\text{H}_{2}\text{O}(\text{g})\)
The following reaction takes place:
\(\text{Fe}^{3+}(\text{aq}) + 4\text{Cl}^{-}(\text{aq})\) \(\rightleftharpoons\) \(\text{FeCl}_{4}^{-}(\text{aq})\)
\(\text{K}_{\text{c}}\) for the reaction is \(\text{7,5} \times \text{10}^{-\text{2}}\). At equilibrium, the concentration of \(\text{FeCl}^{-}_{4}(\text{aq})\) is \(\text{0,95} \times \text{10}^{-\text{4}}\) \(\text{mol·dm$^{-3}$}\) and the concentration of \(\text{Fe}^{3+}(\text{aq})\) is \(\text{0,2}\) \(\text{mol·dm$^{-3}$}\). Calculate the concentration of chloride ions at equilibrium.
We know the value of \(\text{K}_{\text{c}}\) and the concentration of two of the three substances at equilibrium. So we use the expression for \(\text{K}_{\text{c}}\) to find the concentration of chloride ions.
\(\text{FeCl}_{4}^{-}(\text{aq})\) is the product and has a coefficient of \(\text{1}\).
\(\text{Fe}^{3+}(\text{aq})\) is a reactant and has a coefficient of \(\text{1}\).
\(\text{Cl}^{-}(\text{aq})\) is a reactant and has a coefficient of \(\text{4}\).
\(\text{K}_{\text{c}}\) \(= \dfrac{\text{[FeCl}_{4}^{-}{\text{]}}}{\text{[Fe}^{3+}{\text{][Cl}^{-}\text{]}^{4}}}\)
\([\text{Cl}^{-}]^{4}\) \(= \dfrac{\text{[FeCl}_{4}^{-}{\text{]}}}{\text{[Fe}^{3+}{\text{][K}_{\text{c}}\text{]}}}\)
\([\text{Cl}^{-}]^{4}\) \(= \dfrac{\text{0,95} \times \text{10}^{-\text{4}}}{(\text{0,2})(\text{7,5} \times \text{10}^{-\text{2}})}\) = \(\text{0,0063}\)
\([\text{Cl}^{-}] =\)\(\text{0,28}\) \(\text{mol·dm$^{-3}$}\)
Ethanoic acid \((\text{CH}_{3}\text{COOH})\) reacts with ethanol \((\text{CH}_{3}\text{CH}_{2}\text{OH})\) to produce ethyl ethanoate and water. The reaction is:
\(\text{CH}_{3}\text{COOH}(\text{aq}) + \text{CH}_{3}\text{CH}_{2}\text{OH}(\text{aq})\) \(\rightleftharpoons\) \(\text{CH}_{3}\text{COOCH}_{2}\text{CH}_{3}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})\)
At the beginning of the reaction, there is \(\text{0,5}\) \(\text{mol}\) of ethanoic acid and \(\text{0,5}\) \(\text{mol}\) of ethanol. At equilibrium, \(\text{0,3}\) \(\text{mol}\) of ethanoic acid was left unreacted. The volume of the reaction container is \(\text{2}\) \(\text{dm$^{3}$}\). Calculate the value of \(\text{K}_{\text{c}}\).
Fill in a RICE table with the initial moles, the change in moles in terms of \(x\) and the number of moles at equilibrium in terms of \(x\):
|
Reaction |
\(\color{blue}{\textbf{1CH}_{3}\textbf{COOH}}\) |
\(\color{blue}{\textbf{1CH}_{3}\textbf{CH}_{2}\textbf{OH}}\) |
\(\color{red}{\textbf{1CH}_{3}\textbf{COOCH}_{2}\textbf{CH}_{3}}\) |
\(\color{red}{\textbf{1H}_{2}\textbf{O}}\) |
|
Initial quantity (mol) |
\(\text{0,5}\) |
\(\text{0,5}\) |
\(\text{0}\) |
\(\text{0}\) |
|
Change (mol) |
\(-x\) |
\(-x\) |
\(+x\) |
\(+x\) |
|
Equilibrium quantity (mol) |
\(\text{0,5} - x\) |
\(\text{0,5} - x\) |
\(+x\) |
\(+x\) |
|
Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
Calculate \(x\):
n(\(\text{CH}_{3}\text{COOH}\)) = \(\text{0,3}\) \(\text{mol}\) at equilibrium. Therefore:
\begin{align*} \text{0,5} \text{ mol} - x &= \text{0,3} \text{ mol} \\ x &= \text{0,5} - \text{0,3}\text{ mol} \\ &= \text{0,2} \text{ mol} \end{align*}Substitute \(x\) into the RICE table:
|
Reaction |
\(\color{blue}{\textbf{1CH}_{3}\textbf{COOH}}\) |
\(\color{blue}{\textbf{1CH}_{3}\textbf{CH}_{2}\textbf{OH}}\) |
\(\color{red}{\textbf{1CH}_{3}\textbf{COOCH}_{2}\textbf{CH}_{3}}\) |
\(\color{red}{\textbf{1H}_{2}\textbf{O}}\) |
|
Initial quantity (mol) |
\(\text{0,5}\) |
\(\text{0,5}\) |
\(\text{0}\) |
\(\text{0}\) |
|
Change (mol) |
\(-\text{0,2}\) |
\(-\text{0,2}\) |
\(\text{0,2}\) |
\(\text{0,2}\) |
|
Equilibrium quantity (mol) |
\(\text{0,3}\) |
\(\text{0,3}\) |
\(\text{0,2}\) |
\(\text{0,2}\) |
|
Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
Calculate the concentration of each of the reactants and products at equilibrium using \(c=\dfrac{n}{V}\) where \(V = \text{2} \text{ dm$^{3}$}\).
\begin{align*} c(\text{CH}_{3}\text{COOH}) &= c(\text{CH}_{3}\text{CH}_{2}\text{OH}) \\ &= \dfrac {\text{0,3} {\text{ mol}}}{\text{2} {\text{ dm}}^{3}} \\ &= \text{0,15} \text{ mol·dm$^{-3}$} \end{align*} \begin{align*} c(\text{CH}_{3}\text{COOCH}_{2}\text{CH}_{3}) &= c(\text{H}_{2}\text{O}) \\ &= \dfrac {\text{0,2} {\text{ mol}}}{\text{2} {\text{ dm}}^{3}} \\ & = \text{0,1} \text{ mol·dm$^{-3}$} \end{align*}|
Reaction |
\(\color{blue}{\textbf{1CH}_{3}\textbf{COOH}}\) |
\(\color{blue}{\textbf{1CH}_{3}\textbf{CH}_{2}\textbf{OH}}\) |
\(\color{red}{\textbf{1CH}_{3}\textbf{COOCH}_{2}\textbf{CH}_{3}}\) |
\(\color{red}{\textbf{1H}_{2}\textbf{O}}\) |
|
Initial quantity (mol) |
\(\text{0,5}\) |
\(\text{0,5}\) |
\(\text{0}\) |
\(\text{0}\) |
|
Change (mol) |
\(-\text{0,2}\) |
\(-\text{0,2}\) |
\(\text{0,2}\) |
\(\text{0,2}\) |
|
Equilibrium quantity (mol) |
\(\text{0,3}\) |
\(\text{0,3}\) |
\(\text{0,2}\) |
\(\text{0,2}\) |
|
Equilibrium concentration (\(\text{mol·dm$^{-3}$}\)) |
\(\text{0,15}\) |
\(\text{0,15}\) |
\(\text{0,1}\) |
\(\text{0,1}\) |
\(\text{CH}_{3}\text{COOH}\) and \(\text{CH}_{3}\text{CH}_{2}\text{OH}\) are reactants, and \(\text{CH}_{3}\text{COOCH}_{2}\text{CH}_{3}\) and \(\text{H}_{2}\text{O}\) are products.
However, \(\text{H}_{2}\text{O}\) is a pure liquid and is not included in the \(\text{K}_{\text{c}}\) expression. So the equilibrium constant is:
\({K}_{c}=\dfrac{\text{[CH}_{3}{\text{COOCH}}_{2}{\text{CH}}_{3}{\text{]}}}{\text{[CH}_{3}{\text{COOH][CH}}_{3}{\text{CH}}_{2}{\text{OH]}}} = \dfrac{(\text{0,1})}{(\text{0,15})(\text{0,15})} = \text{4,44}\)
|
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8.1 What is chemical equilibrium?
|
Table of Contents |
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8.3 Le Chatelier's principle
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