Module 6.4: Chemical Equilibrium

6.4.1 Introduction

I Chemical equilibrium: a state at which the rate of forward and reverse reaction is equal

1. A state at which there is no tendency to form products or reactants
2. All equilibria (including chemical equilibria) are dynamic.

3. as reactions proceed, the reverse rate goes up; causing the reaction to “stop”.

II equilibrium constant (K): characteristic of the composition of a mixture at equilibrium

1. equilibrium constants are dependent on the temperature of the system.

2. the equilibrium constant can be calculated by $\textrm{K} = \frac{(\textrm{activities of products})^{\textrm{n}_{r}}}{(\textrm{activities of reactants})^{\textrm{n}_{r}}}$ (Eq. 106)

   1. if the substance is a gas, its activity is equal to its partial pressure.

   2. if the substance is pure solid or liquid, its activity equals to 1.

   3. if the substance is a solution, its activity is approximated by its concentration.

   4. all activities are unitless, and therefore the equilibrium constant is also unitless.

III reaction quotient (Q): characteristic of the composition of a mixture

1. the reaction quotient is also calculated the same way as K (Eq. 106)

2. unlike the equilibrium constant (K), there can be many combinations of Q

   1. it is representative of the composition of a certain mixture at a certain time.

   2. Q will approach the value of K, until the reaction is at equilibrium where Q = K.

   3. this is similar to the case of $\Delta \textrm{G}$ and $\Delta \textrm{G}^{\theta}$.

3. the reaction quotient can be calculated for any stage of the reaction with $\Delta \textrm{G}_{r}$ $\Delta \textrm{G}_{r} = \Delta \textrm{G}_{r}^{\theta} + \textrm{R} \textrm{T} \textrm{ln}(\textrm{Q})$ (Eq. 107)

   1. if $\Delta \textrm{G}_{r} = 0$, where the system is at equilibrium, this equation can be modified: $\Delta \textrm{G}_{r}^{\theta} = - \textrm{R} \textrm{T} \textrm{ln}(\textrm{K})$ (Eq. 108)

4. The composition of the system can be easily determined by looking at $\Delta \textrm{G}_{r}^{\theta}$.

   1. if $\Delta \textrm{G}_{r}^{\theta} \gt 0$, then $\textrm{K} \lt 1$ and reactants are favored at equilibrium.

   2. if $\Delta \textrm{G}_{r}^{\theta} \lt 0$, then $\textrm{K} \gt 1$ and products are favored at equilibrium.

5. the relationship of the equilibrium constant and $\Delta \textrm{H}$ & $\Delta \textrm{S}$ can also be made: $\textrm{ln}(\textrm{K}) = \frac{-\Delta \textrm{H}_{r}^{\theta}}{\textrm{R} \textrm{T}} + \frac{-\Delta \textrm{H}_{r}^{\theta}}{\textrm{R}}$ (Eq. 109)

   1. the relationship of K and temperature is dependent on the change of enthalpy, not the change of entropy.

6.4.2 Equilibrium Calculations

IV If a chemical equation is multiplied through by a factor N, then the equilibrium constant is $\textrm{K}_{n}$

V If a chemical equation is flipped, then the equilibrium constant is 1/K.

VI The equilibrium constant for an overall reaction is the product of the equilibrium constants for its component reactions.

VII To convert between equilibrium constants expressed in partial pressures and in concentrations:

$\textrm{K} = (\textrm{R} \textrm{T})^{\Delta \textrm{n}_{r}} \textrm{K}_{c}$ (Eq. 110)

6.4.3 Responses to changes

VIII Le Chatelier's principle: when a system in equilibrium is disturbed, the system will adjust to minimize the disturbance.

1. when the amount of a certain reactant or product is adjusted: the value of Q will change, and Q will approach the value of K by adjusting other reactants or products.

2. when the pressure of the system is adjusted: if the activities in the reaction quotient contain partial pressures, then see how the value of Q changes, and the value of Q will approach K by adjusting reactants or products.

3. when the temperature is changed: the van't Hoff equation can be used

   1. if there are gaseous reactants or products in the reaction, you must use K (in partial pressures) instead of $\textrm{K}_{c}$ (in concentration)

   2. as described by Eq. 109, the enthalpy of the reaction determines how temperature will affect equilibrium

      $\textrm{ln} \frac{\textrm{K}_{2}}{\textrm{K}_{1}} = \frac{\Delta \textrm{H}_{r}^{\theta}}{\textrm{R}} (\frac{1}{\textrm{T}_{1}} - \frac{1}{\textrm{T}_{2}})$ (Eq. 111)