Reaction direction
- irreversible reaction is a reaction, which goes all the way to completetion after all reactants have been consumed all products formed
- reversible reaction is a reaction, which goes to a point where products get made at the same rate they are being broken down back to reactants
- equilibrium is a state, in which new products are formed as well as broken down at the same time
- it is established when the rates of both reactants and products are the same
- reverse reaction is the reverse direction of a reaction
- forward reacion is the forward drection of a reaction
Reaction quotient $q$
- reaction quotient is a the ratio of the concentration of the products to the concentration of the reactants
- if the ratio is larger than 1, the reaction proceeds in the reverse direction
- if the ratio is smaller than 1, the reaction proceeds in the forward direction
- if the reaction is equal to 1, the reaction is in equilibrium
Calculation
$$aA \leftrightharpoons bB$$ $$r_1 = k_1[A]$$ $$r_2 = k_2[B]$$ $$\cfrac{r_2}{r_1} = \cfrac{k_2[B]}{k_1[A]}$$ $$q = \cfrac{[B]^b}{[A]^a}$$
Equilibrium constant $K_{eq}$
- equilibrium constant is a value of the reaction quotient with which the ratio $\cfrac{r_2}{r_1}$ is equal to 1
- if $Q > K_{eq}$, the reaction proceeds in the reverse direction
- if $Q < K_{eq}$, the reaction proceeds in the forward direction
- if $Q= K_{eq}$, the reaction is in equilibrium
- it can be calculated just like the reaction quotient but the concentration are specific for the reaction in equilibrium
- more generaly, equilibrium constant is defined as the ratio of activities of the reactants of the product
- if the equilibrium is larger than 1, the products are heavily favored in the equilibrium mixture
- irreversible reactions have the equilibrium constants extremely large
- if the equilibrium is smaller than 1, the reactants are heavily favored in the equilibrium mixture
- the inverse value of he equilibrium constant provides the same information about the equilibrium state, but is associated with a differently written reaction (if products and reactants were swapped)
Calculation
Method 1: Via concentrations ($K_c$)
$$aA + bB \leftrightharpoons cC + dD$$ $$K_{c} = \cfrac{a_{C,eq}^ca_{D,eq}^d}{a_{A,eq}^aa_{B,eq}^b}$$
- where:
- $a_{X,eq}$ is the activity of a substance in equilibrium
- $a_{X} = \cfrac{c_X}{c_{std}}$
- $c_{std}$ is the standard concentration (=1M)
- it makes the activity dimensionless
- $c_X = \cfrac{n_X}{V}$
- $c_{std}$ is the standard concentration (=1M)
- $a_{X} = \cfrac{c_X}{c_{std}}$
- $a_{X,eq}$ is the activity of a substance in equilibrium
Method 2: Via pressure ($K_p$)
$$aA + bB \leftrightharpoons cC + dD$$ $$K_{p} = \cfrac{a_{C,eq}^ca_{D,eq}^d}{a_{A,eq}^aa_{B,eq}^b}$$
- where:
- $a_{X,eq}$ is the activity of a substance in equilibrium
- $a_X = \cfrac{p_X}{p_{std}}$
- $p_{std}$ is the standard pressure (=100 000 Pa)
- it makes the activity dimensionless
- $p_X = \cfrac{n_XRT}{V}$
- $p_{std}$ is the standard pressure (=100 000 Pa)
- $a_X = \cfrac{p_X}{p_{std}}$
- $a_{X,eq}$ is the activity of a substance in equilibrium
Rescaling the stoichiometric coefficients
- if the stoichiometric coefficients were changed by a factor of $s$, the new equilibrium constant will be equal to $(K_{eq})^s$
Relationship between $K_c$ and $K_p$
$$K_p = K_c\left( RT\cfrac{c_{std}}{p_{std}} \right)^{\Delta{n}}$$
- where: $\Delta{n}$ is the change in stoichiometric coefficients
Equilibrium concentration
- equilibrium concentration is the conentration of a substance in the state of equilibrium
Calculation
- using an ICE table, the equilibrium concentrations can be calculated
- first, the initial concentrations are written down
- second, the changes in concentration are denoted using the stoichiometrich cofficients
- third, the formula for the equilibrium concentration is written down
- fourth, the equilibrium concentration formulas are substituted to the equilibrium constant formula
Example
$$aA \leftrightharpoons bB$$
- initial concentration:
- $A$: $c_0 = kM$
- $B$: $c_0 = 0M$
- change in concentration:
- $A$: $-ax$
- $B$: $bx$
- equilibrium concentration:
- $A$: $c_0 - ax$
- $B$: $c_0 + bx$
- substitution: $$K_c = \cfrac{(bx)^b}{(c_0-ax)^a}$$
Le Chatelier’s principle
When a system in equilibrium is disturbed by some change, it will respond so as to counteract the change.
Change in composition
If reactant is added to a mixture in equilibrium, the concentration of products increases and vice versa. If product is added to a mixture in equilibrium, the concentration of the reactant increases and vice versa.
Change in pressure
If fewer molecules are created by the forward reaction and the pressure is increased, the concentration of the product increases. If more molecules are created by the reverse reaction and the pressure is increased, the concentration of the reactant increases.
Change in temperature
If a reaction is exothermic and the temperature is increased, the concentration of the reactant increases and vice versa. If a reaction is endothermic and the temperature is increased, the concentration of the product increases.
Solubility equilibrium constant $K_{sp}$
- for a reaction $AB \leftrightharpoons A^+ + B^-$, the solubility equilibrium constant will be equal to $[A^+][B^-]$, because the concentration of $[AB]$ is by definition equal to 1
- solubility product is the concentration of a dissolved substance in a completely saturated solution
- it is calculated as $\sqrt{K_{sp}}$
- if represented in molar concentration, it can also be called molar solubility
Common-ion effect
If a compound containing an ion that already is present in a solution is added to a reaction in equilibrium, the concentration of reactants will increase.