Question

explain how Q and K are related to the diraction in which a reaction proceed?

How is the value of the activation energy related to the rate of a reaction? In what two ways. Can the activation energy be lowered?

Answer 1

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Where Keq is constant at a given temperature, i.e. it is dependant on Temperature only

If Keq > 1, this favours products, since this relates to a higher amount of C + D

If Keq < 1, this favours reactants, since this relates to a higher amount of A + B

If Keq = 1, this is in equilibrium, therefore, none is favoured, both are in similar ratios

Note that the concentrations MUST be in equilibrium. If these are not in equilibrium, then the reaction will take place until there is equilibrium achieved.

For this, we use Q, the reaction quotient of products/reactants, it allows us to understand the ratio distribution and the direction/shit of equilibrium

Q is defined as:

Q = [C]^c * [D]^d / ([A]^a * [B]^b)

In this Case, the concentrations are NOT in equilibrium

Therefore:

If Q < Keq; this has much more reactants than products, therefore expect reactants to form more product in order to achieve equilibrium

If Q > Keq; this has much more products than reactants, therefore expect products to form more reactants in order to achieve equilibrium

If Q = Keq; this has the same ratio in equilibrium for reactants and products. Expect no reaction. It is safe to assume this is already in equilibrium.

Q2.

activation energy is the energy required to "start" a reaction which forms a product

the collisions must have enough energy ( acitvation energy ) in order to proceed forwards

recall Ahrrenius equation:

According to Arrhenius, we can relate the rate constants as following:

K = A*exp(-Ea/(RT))

Where:

K = rate constant at Temperature “T”

A = Frequency Factor

E = Activation Energy in J/mol

R = ideal gas constant, 8.314 J/mol-K

T = absolute temperature

k is related directly to Ea

as Ea increases, K decreases

Q3.

we can decrease Ea via:

- addition of a catalyst

- increase on Temperature; typically will favour collision energy / kinetic energy of molecules rate goes faster