0.20 Reaction equilibrium in the gas phase  (Page 6/8)

Dynamic equilibrium and reaction rates

Since we have studied rate laws for simple reactions, we should be able to use this information to understand the competition between the forward and reverse reactions, assuming both are elementary processes. Remember than an elementary process is a reaction which occurs in a single step. In the case of an elementary process, the rate law can be predicted from the coefficients in the reaction equation.

Let's apply this reasoning.to the reaction in Equation (4), N ₂ O ₄ (g) → 2 NO ₂ (g), where we determined the equilibrium constant in Equation (9). If the forward and reverse reactions are both elementary processes, then the rate laws for the forward and reverse reactions will be:

Rate _forward =k _forward [N ₂ O ₄ ]

Rate _reverse =k _reverse [NO ₂ ] ²

Assuming that the reaction comes to dynamic equilibrium, these two rates must be equal to each other. So let's set the right sides of the two equations equal to each other (and we'll let the subscript f stand for forward and the subscript r stand for reverse .

k _f [N ₂ O ₄ ]=k _r [NO ₂ ] ²

This is easily rearranged into a somewhat familiar form:

$\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}=\frac{{\left[\text{N}{\text{O}}_2\right]}^2}{\left[{\text{N}}_2{\text{O}}_2\right]}$

This looks very much like the equilibrium constant in Equation (9), except that this equation shows gas concentrations instead of gas pressures. We can use the Ideal Gas Law to make the connection, since for example [NO ₂ ] = n _NO2 /V = P _NO2 /RT . If we substitute this into Equation (11), we get

$\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}=\frac{{\left[\text{N}{\text{O}}_2\right]}^2}{\left[{\text{N}}_2{\text{O}}_2\right]}=\frac{{\text{P}}_{\text{N}{\text{O}}_2}^2}{{\text{P}}_{{\text{N}}_2{\text{O}}_4}}=\frac{\text{1}}{\text{R}\text{T}}$

Just multiplying by RT , we can now see that

${\text{K}}_{\text{p}}=\text{R}\text{T}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}=\frac{{\text{P}}_{\text{N}{\text{O}}_2}^2}{{\text{P}}_{{\text{N}}_2{\text{O}}_4}}$

The equilibrium constant K _p is the ratio of the forward and reverse rate constants, times the factor RT . Equation (12) comes from two assumptions: that the reaction is in dynamic equilibrium, so that we can set the forward and reverse reaction rates equal to each other, and that the forward and reverse reactions are elementary processes, so that we can write the rate laws easily. Since the result agrees with the experimental rate constant in Equation (9), we can conclude that these assumptions are valid. Therefore, the reaction equilibrium is a dynamic equilibrium, with equal forward and reverse reaction rates.

Although we demonstrated this outcome for a reaction in which the forward and reverse reactions are both elementary processes, it turns out that the result is general. This is trickier to show, but we can conclude that reactions in equilibrium are in dynamic equilibrium.

Observation 3: temperature dependence of the reaction equilibrium

We have previously observed that phase equilibrium, and in particular vapor pressure, depends on the temperature, but we have not yet studied the variation of reaction equilibrium with temperature. We focus our initial study on the reaction

H ₂ (g) + I ₂ (g) → 2 HI (g) (3)

and we measure the equilibrium partial pressures at a variety of temperatures. From these measurements, we can compile the data showing the temperature dependence of the equilibrium constant K _p for this reaction in Table 4.

Note that the equilibrium constant increases dramatically with temperature, more than tripling over this temperature range. As a result, at equilibrium, the pressure of HI must also increase dramatically as the temperature is increased.