14.1-14.2: Chemical Equilibrium-The Equlibrium Constant

Chemical Equilibrium:

At dynamic equilibrium, Rate_f = Rate_r, the forward rate equals the reverse rate. Both sides are always reacting at a microscopic level at the same rate.

Chemical Equilibrium: the state reached by a reaction mixture when the rates of forward and reverse reactions have become equal.

In a Moles vs Time graph, notice the amounts of substances become constant at equilibrium:

In a Rate vs Time graph, notice that the forward rate is large at first but steadily decreases, while the reverse rate starts at zero and steadily increases, until both rates are equal at equilibrium:


Molar Composition:

Example:

For the reaction

CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)

When 1.000 mol CO and 3.000 mol H₂ are placed in a 10.00 L vessel at constant temperature and allowed to come to equilibrium, the mixture is found to contain 0.387 mol H₂O. How many moles of each substance are present in the equilibrium mixture?

Think about the starting amounts in moles, which are known. The change in the amount of moles as the reaction proceeds is an unknown, so we can label it "x". Each reactant will decrease by a factor of x (the factor is indicated by the coefficient of that particular reactant) and each product will increase by a factor of x (also found by the coefficient). The resulting amounts can be found by combining our known information into equations and solving for x.

It is easiest to do this by making a table. I, C, and E indicate Initial, Change, and Equilibrium.


If the initial amounts of products isn't stated, it is assumed to be 0. Also, if we know each product (with a coefficient of 1) increases by x, and we know the resulting amount of that product, then we know x. So here, x = 0.387 mol.

If the problem were to ask for concentrations, divide the results by the volume given, in this case 10.00L.

The Equilibrium Constant:

General Form of an Equation:
aA + bB ⇌ cC + dD

Where capital letters are reactants and products and lowercase letters are coefficients of the balanced chemical equation.

-The equilibrium constant can be obtained by multiplying the concentrations of products, dividing by the product of the concentrations of reactants, and raising each concentration term to a power equal to the coefficient in the chemical equation. This is much easier to see when it's written out:



-K_c can also be denoted as K_eq
-K_c has no units
-Concentrations must be entered in mols/Liter (M), otherwise convert to these units
-Must have an equilibrium to find K_c
-Ignore concentrations of pure liquids and solids, they are constant. Another way of writing concentrations is density (same dimensions of mass over volume) which will remain the same no matter how much of a solid or liquid you have.
-Just use the concentrations of gaseous and aqueous reactants and products.

Example:

CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)

K_c = [CH₄][H₂O] / [CO][H₂]³

Example:

N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

K_c = [NH₃]² / [NH₂][H₂]³

Example:

2A (g) + B(s) + 7C (aq) ⇌ 14D (l) + 3/2E (g)

K_c = [E]^3/2 / [A]²[C]⁷

-Manipulating the balanced chemical equation will change K_c:

-If you multiply the coefficients of a reaction through by a number, A, then K_c = K_c^A
-Suppose you multiply a reaction through by 2, then you must square K_c
-If you flip a reaction around so the reactants are now the products and the products are now the reactants, then you get the reciprocal, K_c = 1 / K_c

The Pressure Equilibrium Constant, K_p:

-Won't use this until later when we study Thermodynamics, but it is relevant now.
-K_p is another version of K_c
-The values plugged into K_p are partial pressures converted to atm instead of concentrations.
-Related to K_c via the ideal gas law (PV = nRT → n/V = P/RT)
-In other words, the molar concentration of a gas equals its partial pressure divided by RT, which is constant at a given temperature.


Example:

For the reaction
CH₄ (g) + 2O₂ (g) ⇌ CO₂ (g) + 2H₂O (g)

K_c = [CO₂][H₂O]² / [CH₄][O₂]²

K_p = P_CO2 · P_H2O² / P_CH4 · P_O2²

-Sometimes K_c = K_p, but not always, it depends on the coefficients of the chemical equation.

K_c = K_p when the sum of reactant coefficients equals the sum of the product coefficients.

Example:

For the same reaction
CH₄ (g) + 2O₂ (g) ⇌ CO₂ (g) + 2H₂O (g)

1 + 2 = 1 + 2

K_c = K_p

Equation to relate K_p to K_c:

K_p = K_c (RT)^Δn

R = the ideal gas constant (0.0821 atm.L/mol.K)
T = Temperature in degrees Kelvin
Δn = sum of the coefficients of the products – the sum of the coefficients of the reactants.

Note: when the sum of the reactant coefficients equals the sum of product coefficients, Δn will be zero. Anything raised to the zero power is 1, so RT^Δn will equal 1, and K_p will equal K_c times 1, or just K_c.

Equilibrium Constant for the Sum of Reactions:

-If a given chemical equation can be obtained by taking the sum of other equations, the equilibrium constant for the given equation equals the product of the equilibrium constants of the other equations.
-Note: use Hess's Law

Example:

Equation 1:
CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)
Equation 2:
CH₄ (g) + 2H₂S (g) ⇌ CS₂ (g) + 4H₂ (g)
Hess's Law:
CO + 3H₂ + CH₄ + 2H₂S ⇌ CH₄ + H₂O + CS₂ + 4H₂
Equation 3:
CO (g) + 2H₂S (g) ⇌ H₂O (g) + CS₂ (g) + H₂ (g)

K_c for Equation 1 = K₁
K₁ = [CH₄][H₂O] / [CO][H₂]³

K_c for Equation 2 = K₂
K₂ = [CS₂][H₂]⁴ / [CH₄][H₂S]²

K_c for Equation 3 = K₁ · K₂
K₃ = [CS₂][H₂O][H₂] / [CO][H₂S]²
(K₃ can be verified by looking at Equation 3 and writing K_c directly from it)

13.7-13.9: Mechanisms-Catalysis

Mechanisms:

Reaction Mechanism: The set of elementary reactions whose overall effect is given by the net chemical equation. An elementary reaction is a single molecular event, such as a collision of molecules, resulting in a reaction.

Example:

Consider the reaction of nitrogen dioxide with carbon monoxide:

NO₂ (g) + CO (g) → NO (g) + CO₂ (g) (net chemical equation)
NO₂ + NO₂ → NO₃ + NO (elementary reaction)
NO₃ + CO → NO₂ + CO₂ (elementary reaction)

Note: Elementary reactions don't have phase labels. Also, NO₃ is a molecule, not the nitrate ion, and is highly unstable. It is both created and destroyed by the reaction and is called a reaction intermediate. In the last step, the CO molecule needs something desperately energetic to smack into in order to react, and the NO₃ provides that.

Following Hess's Law, obtain the overall equation by adding the elementary reactions and canceling species that occur on both sides:

NO₂ + NO₂ → NO₃ + NO (elementary reaction)
NO₃ + CO → NO₂ + CO₂ (elementary reaction)
-------------------------------------
NO₂ +NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂ (overall)

Molecularity:

Molecularity: the number of molecules on the reactant side of an elementary reaction. Products are not considered in molecularity, only reactants of the elementary reaction.

Example:

Unimolecular: an elementary reaction that involves one reactant molecule.
O₂ → 2O

Bimolecular: an elementary reaction that involves two reactant molecules.
NO₂ + CO → NO + CO₂

Termolecular: an elementary reaction that involves three reactant molecules.
Br₂ +Cl₂ + C → CBr₂Cl₂ (generally exceptionally slow because highly unlikely statistically, remember Civil War colliding bullets example).

Determining the Rate Law from an Elementary Reaction:

Example:

What is the rate law for an overall reaction? In order to do this, slow step and fast step must be labeled. Need to know the slow step because it moderates the reaction; the reaction can't proceed faster than the slowest step. So the rate law for the slow step is the rate law for the overall reaction.
Note: These are both elementary reactions, the (slow step) and (fast step) imply that. Also, k is unique for each step. Once the slow step is identified, use the general form to assume the rate equation for this step:

Rate = k₁[NO₂][F₂]

Note: The components in the rate law must be part of the overall reaction. Intermediates cannot be part of the rate law.

This is pretty straightforward for reactions with an initial slow step. It's more complicated for reactions with an initial fast, equilibrium step.

Example:

Mechanism:


When we build the rate law from our slow step, we have an intermediate which needs to be eliminated:
Rate = k₂[NO₂][NO₃]

Use the fast step to find something plausible to replace [NO₃] with:

Rate_f = k₁[N₂O₅]
Rate_r = k_-1[NO₂][NO₃]

Where Rate_f is the forward rate and Rate_r is reverse rate (k₁ and k_-1 also mean forward and reverse rate). Also, at equilibrium, by definition, the forward and reverse rates are equal to each other. They are at dynamic equilibrium, where the forward reaction and the reverse reaction are happening at the same time, all the time.

So:
k₁[N₂O₅] = k_-1[NO₂][NO₃]
Solve for the component you want to eliminate, the intermediate [NO₃]:
[NO₃] = k₁[N₂O₅] / k_-1[NO₂]
Substitute this into the rate law from the slow step for [NO₃]:
Rate = k₂[NO₂] x (k₁[N₂O₅] / k_-1[NO₂])
([NO₂] will cancel)
Rate = (k₂k₁ / k_-1) [N₂O₅]
Where all k represent a constant and can be combined into a single k to get:
Rate = k[N₂O₅]

Catalysis:

Catalysts: cause a reaction to proceed faster without being consumed by the reaction. Catalysts do this by lowering the activation energy, E_a.



The solid line represents the reaction without a catalyst. It requires more energy to reach the activated state and thus has a higher E_a. The dotted line represents the reaction with the addition of a catalyst. It requires much less energy and thus has a much lower E_a.

Homogeneous Catalysis: the use of a catalyst in the same phase as the reacting species. An example would be a gas phase catalyst and gas phase reacting species; they are all in the same phase (gas).

Heterogeneous Catalysis: the use of a catalyst that exists in a different phase from the reacting species, usually a solid catalyst in contact with a gaseous or liquid solution of reactants.