0.20 Reaction equilibrium in the gas phase  (Page 8/8)

To understand this effect, we rewrite the equilibrium constant in Equation (10) to explicitly show the volume of the container. This is done by applying Dalton’s Law of partial pressures, so that each partial pressure is given by the Ideal Gas Law:

${\text{K}}_{\text{p}}=\frac{\text{n}{\left(\text{N}{\text{H}}_3\right)}^2{\left(\frac{\text{RT}}{\text{V}}\right)}^2}{\text{n}\left({\text{N}}_2\right)\left(\frac{\text{RT}}{\text{V}}\right)·\text{n}{\left({\text{H}}_2\right)}^{\text{3}}{\left(\frac{\text{RT}}{\text{V}}\right)}^3}=\frac{\text{n}{\left(\text{N}{\text{H}}_3\right)}^2}{\text{n}\left({\text{N}}_2\right)·\text{n}{\left({\text{H}}_2\right)}^{\text{3}}{\left(\frac{\text{RT}}{\text{V}}\right)}^2}$

Therefore,

${\text{K}}_{\text{p}}{\left(\frac{\text{RT}}{\text{V}}\right)}^2=\frac{\text{n}{\left(\text{N}{\text{H}}_3\right)}^2}{\text{n}\left({\text{N}}_2\right)·\text{n}{\left({\text{H}}_2\right)}^{\text{3}}}$

This form of the equation makes it clear that, when the volume increases, the left side of the equation decreases. This means that the right side of the equation must decrease also, and in turn, n(NH ₃ ) must decrease while n(N ₂ ) and n(H ₂ ) must increase. The equilibrium is thus shifted from products to reactants when the volume increases for Reaction (2).

The effect of changing the volume must be considered for each specific reaction, because the effect depends on the stoichiometry of the reaction. One way to determine the consequence of a change in volume is to rewrite the equilibrium constant as we have done in Equation (15).

Finally, we consider changes in temperature. We note that K _p increases with T for endothermic reactions and decreases with T for exothermic reactions. As such, the products are increasingly favored with increasing temperature when the reaction is endothermic, and the reactants are increasingly favored with increasing temperature when the reaction is exothermic. On reflection, we note that when the reaction is exothermic, the reverse reaction is endothermic. Putting these statements together, we can say that the reaction equilibrium always shifts in the direction of the endothermic reaction when the temperature is increased.

All of these observations can be collected into a single unifying concept known as Le Châtelier’s principle. This principle can be stated as follows: when a reaction at equilibrium is stressed by a change in conditions, the equilibrium will be reestablished in such a way as to counter the stress. This statement is best understood by reflection on the types of “stresses” we have considered in this section. When a reactant is added to a system at equilibrium, the reaction responds by consuming some of that added reactant as it establishes a new equilibrium. This offsets some of the stress of the increase in reactant. When the temperature is raised for a reaction at equilibrium, this adds thermal energy. The system shifts the equilibrium in the endothermic direction, thus absorbing some of the added thermal energy, countering the stress.

The most challenging of the three types of stress considered in this section is the change in volume. By increasing the volume containing a gas phase reaction at equilibrium, we reduce the partial pressures of all gases present and thus reduce the total pressure. Recall that the response of Reaction (2) to the volume increase was to create more of the reactants at the expense of the products. One consequence of this shift is that more gas molecules are created, and this increases the total pressure in the reaction flask. Thus, the reaction responds to the stress of the volume increase by partially offsetting the pressure decrease with an increase in the number of moles of gas at equilibrium.

Le Châtelier’s principle is a useful mnemonic for predicting how we might increase or decrease the amount of product at equilibrium by changing the conditions of the reaction. From this principle, we can predict whether the reaction should occur at high temperature or low temperature, and whether it should occur at high pressure or low pressure.

Review and discussion questions

1. In the data given for equilibrium of Reaction (3) on Page 3, there is no volume given. Show that changing the volume for Reaction (3) does not change the number of moles of reactants and products present at equilibrium, i.e. changing the volume does not shift the equilibrium.

   H ₂ (g) + I ₂ (g) ↑ 2 HI (g) (3)

2. For Reaction (4)

   N ₂ O ₄ (g) ↑ 2 NO ₂ (g)(4)

   the number of moles of NO ₂ at equilibrium increases if we increase the volume in which the reaction is contained. Explain why this must be true in terms of dynamic equilibrium, give a reason why the rates of the forward and reverse reactions might be affected differently by changes in the volume.
3. We could balance Reaction (2) by writing

   2 N ₂ (g) + 6 H ₂ (g) ↑ 4 NH ₃ (g)(2’)

   Write the form of the equilibrium constant for the reaction balanced as in Reaction (2’). What is the value of the equilibrium constant? (Refer to Table 3.) Of course, the pressures at equilibrium do not depend on whether the reaction is balanced as in Reaction (2) or as in Reaction (2’). Explain why this is true, even though the equilibrium constant can be written differently and have a different value.
4. Show that the equilibrium constant K _p in Equation (10) for Reaction (2) can be written in terms of the concentrations or particle densities, e.g. [N ₂ ]= n _N2 /V, instead of the partial pressures. In this form, we call the equilibrium constant K _c . Find the relationship between K _p and K _c , and calculate the value of K _c .
5. For each of these reactions, predict whether increases in temperature will shift the reaction equilibrium more towards products or more towards reactants.
   1. 2 CO (g) + O2 (g) ↑ 2 CO2 (g)
   2. O ₃ (g) + NO(g) ↑ NO2 (g) + O2 (g)
   3. 2O ₃ (g) ↑ 3O2 (g)
6. Plot the data in Table 4 on a graph showing K _p on the y-axis and T on the x-axis. The shape of this graph is reminiscent of the graph of another physical property as a function of increasing temperature. Identify that property, and suggest a reason why the shapes of the graphs might be similar.
7. Using Le Châtelier’s principle, predict whether the specified “stress” will produce an increase or a decrease in the amount of product observed at equilibrium for the reaction:

   2 H ₂ (g) + CO(g) ↑ CH3OH(g) ΔH˚ = -91 kJ/mol

   1. Volume of container is increased.
   2. Helium is added to container.
   3. Temperature of container is raised.
   4. Hydrogen is added to container.
   5. CH3OH is extracted from container as it is formed.