A-level Chemistry/WJEC/Module 3/Equilibrium

Principles of Chemical Equilibrium

Pure dinitrogen tetroxide (N₂O₄) is a colourless gas that is widely used as a rocket fuel. Although N₂O₄ is colourless, when a container is filled with pure N₂O₄, the gas rapidly begins to turn a dark brown.

Nitrogen dioxide at −196 °C, 0 °C, 23 °C, 35 °C, and 50 °C. (NO₂) converts to the colourless dinitrogen tetroxide (N₂O₄) at low temperatures, and reverts to NO₂ at higher temperatures.

A chemical reaction is clearly occurring, and indeed, chemical analysis tells us that the gas in the container is no longer pure N₂O₄, but has become a mixture of dinitrogen tetroxide and nitrogen dioxide; N₂O₄ is undergoing a decomposition reaction to form NO₂. If the gaseous mixture is cooled, it again turns colourless and analysis tells us that it is again, almost pure N₂O₄; this means that the NO₂ in the mixture can also undergo a synthesis reaction to re-form N₂O₄. A more detailed analysis of the relative concentrations of N₂O₄ and NO₂ at various times gives us the data that are shown in the plot in Figure 10.2. Initially, only N₂O₄ is present. As the reaction proceeds, the concentration of N₂O₄ decreases and the concentration of NO₂ increases. However, if you examine the figure, after some time, the concentrations of N₂O₄ and NO₂ have stabilised and, as long as the temperature is not changed, the relative concentrations of the two gasses remain constant.

The reversible reaction of one mole of N₂O₄, forming two moles of NO₂, is a classic example of a chemical equilibrium.

When we write these chemical equations, we use a double arrow to signify that the reaction proceeded in both directions. Using this convention, the dissociation of dinitrogen tetroxide to form two molecules of nitrogen dioxide can be shown as:

N₂O₄ ⇌ 2 NO₂

If the temperature of our gas mixture is again held constant and the total pressure of the gas in the container is varied, analysis shows that the partial pressure of N₂O₄ varies as the square of the partial pressure of NO₂ (Figure 10.3; remember, the Ideal Gas Laws tell us that the partial pressure of a gas, p_gas, is directly proportional to the concentration of that gas in the container). Mathematically, the relationship between the partial pressures of the two gasses can be expressed by Equation 1.

${\frac {\left(p_{NO_{2}}\right)^{2}}{p_{N_{2}O_{4}}}}=K_{p}$
Equation 1.

The Equilibrium Constant

The constant K_p is called the equilibrium constant for the reaction. The equilibrium constant for this reaction tells us that, regardless of pressures (or concentrations), a mixture of the two gasses will undergo reaction such that the ratios of the partial pressures reach a constant value, given by the equilibrium constant, K_p. Once this constant ratio has been reached does this mean the reactions stop? Of course not; examine Figure 10.2. In the region of the plot where the concentrations of N₂O₄ and NO₂ are constant (the lines are level) N₂O₄ is still decomposing to form two molecules of NO₂ and two molecules of NO₂ are still reacting to synthesize a molecule of N₂O₄, but the lines are level because the rates of the two chemical reactions have become constant; N₂O₄ is decomposing at the same rate as two molecules of NO₂ are reacting to form N₂O₄. (Figure 10.4)

In theory, all chemical reactions are equilibria. In practice, however, most reactions are so slow in the reverse direction that they are considered "irreversible". When a reaction evolves a gas, forms a precipitate or proceeds with the generation of a large amount of heat or light (for example, combustion) the reaction is essentially irreversible. Many chemical reactions are, however, readily reversible and for these reactions the mathematical expression for the equilibrium constant can be written using a simple set of rules.

1. Partial pressures (or molar concentrations) of products are written in the numerator of the expression and the partial pressures (or concentrations) of the reactants are written in the denominator.
2. If there is more that one reactant or more that one product, the partial pressures (or concentrations) are multiplied together.
3. The partial pressure (or concentration) of each reactant or product is then raised to the power that numerically equals the stoichiometric coefficient appearing with that term in the balanced chemical equation.
4. Reactants or products that are present as solids or liquids do not appear in the equilibrium expression.

Thus, for the reaction of nitrogen with hydrogen gas to form ammonia:

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)

The expression for the equilibrium constant will have the partial pressure of ammonia in the numerator, and it will be squared, corresponding to the coefficient "2" in the balanced equation; $\left(p_{NH_{3}}\right)^{2}$. Because there are two reactants, the partial pressures for nitrogen and hydrogen will be multiplied in the denominator. The partial pressure of nitrogen will be raised to the "first power" (which is not shown) and the partial pressure of hydrogen will be cubed, corresponding to the coefficient "3"; $\left(p_{N_{2}}\right)\left(p_{H_{2}}\right)^{3}$. The final expression for the equilibrium constant is given in Equation 2.

${\frac {\left(p_{NH_{3}}\right)^{2}}{p_{N_{2}}\left(p_{H_{2}}\right)^{3}}}=K_{p}$
Equation 2.

Example 10.1 Writing Equilibrium Expressions

For the chemical reactions shown below, write an expression for the equilibrium constant in terms of the partial pressures of the reactants and products.

PCl5(g) ⇌ PCl3(g) + Cl2(g)
2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)

Exercise 10.1 Writing Equilibrium Expressions

For the chemical reaction shown below, write an expression for the equilibrium constant in terms of the partial pressures of the reactants and products.

PCl3(g) + 3 NH3(g) ⇌ P(NH2)3(g) + 3 HCl(g)

Calculating Equilibrium Values

The numeric value of the equilibrium constant tells us something about the ratio of the reactants and products in the final equilibrium mixture. Likewise, the magnitude of the equilibrium constant tells us about the actual composition of that mixture.

In the three equilibrium systems shown in Figure 10.5, the first depicts a reaction in which the ratio of products to reactants is very small. Because the expression for the equilibrium constant is given by the pressure (or concentration) of products divided by the pressure (or concentration) of reactants, the equilibrium constant, K, for this system is also small. In the second example, the concentrations of reactants and products are shown to be equal, making the ratio (the equilibrium constant) equal to “1”. In the last example, the products are shown to dominate the equilibrium mixture, making the ratio ${}^{\left(p_{Products}\right)}\!\!\diagup \!\!{}_{\left(p_{Reactants}\right)}\;$very large. In these examples, the stoichiometric ratios of the reactants and products are one and there is only one reactant and only one product; if multiple reactants or products are involved, the relationship between their concentrations would be more complex, but that ratio is always given by the expression for K. This fact allows us to take data for an equilibrium reaction and, if K is known, calculate concentrations for reactants and products. Likewise, if all of the equilibrium concentrations are known, we can use these to calculate a value for the equilibrium constant.

In these types of problems, an ICE table is often useful. This table has entries for Initial concentrations (or pressures), Equilibrium concentrations and any Change between the initial and equilibrium states. For example, consider the reaction between carbon monoxide and chlorine to form phosgene, a deadly compound that was used as a gas warfare agent in World War I. (Figure 10.6)

CO(g) + Cl₂(g) ⇌ COCl₂(g)

A typical equilibrium problem might read as follows:

A mixture of CO and Cl₂ has initial partial pressures of 0.60 atm for CO and 1.10 atm for Cl₂. After the mixture reaches equilibrium, the partial pressure of COCl₂ is 0.10 atm. Determine the value of K.

The initial pressures for carbon monoxide and chlorine are placed in the first row and the equilibrium pressure for phosgene is placed in the last row. Initially, the pressure of phosgene was zero, so that goes in the first row; the change for phosgene is therefore "+0.10 atm".

	$P_{CO}$	$P_{Cl_{2}}$	$P_{COCl_{2}}$
Initial	0.60 atm	1.10 atm	0 atm
Change			+ 0.10 atm
Equilibrium			0.10 atm

Because one mole of CO is required to make one mole of COCl₂ the partial pressure of CO must have dropped by 0.10 atm (the Change) in order to make COCl₂ with a partial pressure of 0.10 atm, giving a final (Equilibrium) pressure of 0.50 atm for carbon monoxide. Likewise, one mole of chlorine is required to make one mole of COCl₂ making the Change for chlorine 0.10 atm and the Equilibrium partial pressure 1.00 atm. The completed table is shown below.

	$P_{CO}$	$P_{Cl_{2}}$	$P_{COCl_{2}}$
Initial	0.60 atm	1.10 atm	0 atm
Change	-0.10 atm	-0.10 atm	+ 0.10 atm
Equilibrium	0.50 atm	1.00 atm	0.10 atm

The equilibrium expression for the phosgene-forming reaction is given by Equation 3.

${\frac {p_{COCl_{2}}}{p_{CO}p_{Cl_{2}}}}=K_{p}$
Equation 3.

Substituting the values from the Table into this equation:

$K_{p}={\frac {p_{COCl_{2}}}{p_{CO}p_{Cl_{2}}}}={\frac {\left(0.10\right)}{\left(0.50\right)\left(1.00\right)}}=0.20atm-1$

Many textbooks differentiate between equilibrium constants calculated from partial pressures and molar concentrations by affixing subscripts; K_p and K_c.

Example 10.2 Determining Equilibrium Values

For the reaction shown below, all four gasses are introduced into a vessel, each with an initial partial pressure of 0.500 atm, and allowed to come to equilibrium; at equilibrium, the partial pressure of SO₃ is found to be 0.750 atm. Determine the value of K_p.

SO₂(g) + NO₂(g) ⇌ SO₃(g) + NO(g)

Exercise 10.2 Determining Equilibrium Values

For the reaction shown above, the initial partial pressures of SO₃ and NO are 0.500 atm under conditions where the equilibrium constant is, K_p = 9.00. The equilibrium partial pressure for SO₂ is found to be 0.125 atm. Calculate the equilibrium partial pressure for SO₃.

Le Chatelier's Principle: Stress and Equilibria

Consider a simple chemical system that is at equilibrium, such as dinitrogen tetroxide: nitrogen dioxide. The Law of Mass Action states that when this system reaches equilibrium, the ratio of the products and reactants (at a given temperature) will be defined by the equilibrium constant, K_p. Now imagine that, after equilibrium has been reached, more dinitrogen tetroxide is introduced into the container. In order for the ratio to remain constant (as defined by K_p) some of the N₂O₄ that you added must be converted to NO₂. The addition of reactants or products to a system at equilibrium is referred to as a "stress". The response of the system to this stress is dictated by Le Chatelier's Principle, which states that, if a "stress" is applied to a chemical reaction at equilibrium, the system will readjust in the direction that best reduces the stress imposed on the system. Again, stress refers to a change in concentration, a change in pressure or a change in temperature, depending on the system being examined. If temperature is changed, the numeric value K_p will change; if only pressure or concentration changes are involved, K_p does not change. We will consider temperature and pressure effects in General Chemistry, but for now, remember; in a reaction at equilibrium, the introduction of more products will shift the mass balance towards more reactants, but the ratio of Products/Reactants (as defined by the equilibrium expression) does not change, hence, K_p is unchanged.

Study Points

• If two opposing chemical reactions proceed simultaneously at the same rate, the processes are said to be in equilibrium. The two opposing reactions are shown linked with a double arrow (⇌). An example of opposing chemical reactions and their equilibrium expressions are:

N₂(g) + 3 H₂(g) → 2 NH₃(g) and 2 NH₃(g) → N₂(g) + 3 H₂(g):

The equilibrium would be written as: N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)

• An equilibrium constant (K) is a numerical value that relates the concentrations of the products and reactants for a chemical reaction that is at equilibrium. The numeric value of an equilibrium constant is independent of the initial concentrations of reactants, but is dependent on the temperature.
• Because equilibrium constants are written as the concentrations (or partial pressures) of products divided by the concentrations (or partial pressures) of reactants, a large value of K means that there are more products in the equilibrium mixture than there are reactants. Likewise, a small value of K means that, at equilibrium, there are more reactants than products.
• Equilibrium constants that are based on partial pressures are often written as K_p, while equilibrium constants based on molar concentrations are written as K_c.
• An expression for an equilibrium constant can be written from a balanced chemical equation for the reaction. The Law of Mass Action states the following regarding equilibrium expressions:
  1. Partial pressures (or molar concentrations) of products are written in the numerator of the expression and the partial pressures (or concentrations) of the reactants are written in the denominator.
  2. If there is more that one reactant or more that one product, the partial pressures (or concentrations) are multiplied together.
  3. The partial pressure (or concentration) of each reactant or product is then raised to the power that numerically equals the stoichiometric coefficient appearing with that term in the balanced chemical equation.
  4. Reactants or products that are present as solids or liquids do not appear in the equilibrium expression.

• As an example of an equilibrium expression, consider the reaction of nitrogen and hydrogen to form ammonia. The partial pressure of ammonia will be in the numerator, and it will be squared. Because there are two reactants, the partial pressures for nitrogen and hydrogen will be multiplied in the denominator. The partial pressure of nitrogen will be raised to the “first power” (which is not shown) and the partial pressure of hydrogen will be cubed.

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g):
${\frac {\left(p_{NH_{3}}\right)^{2}}{p_{N_{2}}\left(p_{H_{2}}\right)^{3}}}=K_{p}$

• If equilibrium values for a given reaction are known, the equilibrium constant can be calculated simply by substituting those values in the equilibrium expression. Quite often, however, initial and equilibrium values are only given for selected reactants and products. In these cases, initial and equilibrium values are arranged in an ICE Table, and the changes between initial and equilibrium states are calculated based on reaction stoichiometry.
• Le Chatelier's Principle states that, if a "stress" is applied to a chemical reaction at equilibrium, the system will readjust in the direction that best reduces the stress imposed on the system. In this context, stress refers to a change in concentration, a change in pressure or a change in temperature, although only concentration is considered here. If temperature is changed, the numeric value K will change; if pressure or concentration changes are involved, K does not change. In a reaction at equilibrium, the introduction of more products will shift the mass balance towards more reactants, and the introduction of more reactants will lead to the formation of more products, but the ratio of Products/Reactants (as defined by the equilibrium expression) does not change, hence, K is unchanged.