Chemical Equilibria

Question 1

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) + heat
How can the k^c of this reaction be decreased?

Answer 1

Increasing the temperature
* A change in temperature is the only factor that can change the equilibrium constant k_c
* The reaction is exothermic because heat is on the right side of the equation
* Since heat is a product, increasing the temperature will push the equilibrium towards the reactants which in turn will increase the concentration of the reactants
* Since reactants are in the denominator of the k^c formula, increasing the temperature will decrease the equilibrium constant k^c of this reaction

Question 2

How can an exothermic reaction at equilibirum decrease its equilibirium constant k_c?

Answer 2

An exothermic reaction can decrease the k_c only by increasing temperature

Question 3

How can an endothermic reaction at equilibirum increase its equilibirium constant k^c?

Answer 3

An endothermic reaction can increase the k^c only by increasing temperature

Question 4

In which direction does a reaction shift if Q^c > k^c?

Answer 4

The reaction will shift to the left
too much product is present (numerator is too big) so the rxn must shift to the left (increasing the denominator)

Question 5

Given the following exothermic equilibrium reaction:
P_4(s) + 10 Cl_2(g) ⇌ 4 PCl_5(g) + heat
What would occur if:
The volume of the container were increased?
The temperature of the reaction were decreased?
More P_4(s) were added?

Answer 5

If volume of the container were increased: the reaction will shift the left towards the reactants because increasing the volume will decrease the pressure, causing the reaction to shift to whichever side gives more gas molecules

If temperature of the reaction were decreased: this will cause the reaction to shift to the right towards the products in order to establish equilibrium, which will increase the amount of gaseous PCl₅

If more P_4(s) were added: P_4(s) is a pure solid and does not appear in the equilibrium expression, so adding more P_4(s) will not cause a shift in the reaction or disturb the equilibrium

Question 6

Molar solubility

Answer 6

Measures the concentration of a solid that dissolves in solution
The larger the K_sp, the more the products are favored in the equilibrium expression, increasing the dissolves ions in water which correlates to a higher molar solubility

Question 7

According to Le Chatelier’s principle, decreasing volume of a reaction container at equilibrium shifts the equilibrium to:

Answer 7

The side with FEWER gas molecules

Question 8

Increasing the pressure for a reaction at equilibrium shifts the reaction to the side with:

Answer 8

Fewer molecules/moles

Question 9

the _________ the value of K_eq, the higher the concentration of products at equilibrium

Question 10

if K_eq < 1:

Answer 10

• there is a greater concentration of reactants than products at equilibrium
• positive ΔG° value
• reactants are lower in energy and more stable than products (aka more reactants)

Question 11

K_eq > 1

Answer 11

• there is a greater concentration of products than reactants at equilibrium
• negative ΔG° value
• products are lower in energy and more stable than reactants (aka more products)

Question 12

if Q < K_eq:

Answer 12

there is a higher concentration of reactants than there is at equilibrium and the reaction proceeds in the forward direction

Question 13

if Q > K_eq:

Answer 13

there is a higher concentration of products than there is at equilibrium and the reaction proceeds in the reverse direction

Question 14

when is the solubility constant K_sp used?

Answer 14

to figure out the direction in which the reaction will proceed, the saturation of the solution and whether or not precipitation will occur

Question 15

if Q > K_sp:

Answer 15

the reaction mixture is supersaturated and will therefore proceed in the reverse reaction - precipitation of a solid occurs

Question 16

if Q < K_sp:

Answer 16

the reaction mixture is unsaturated and will therefore proceed in the forward direction - no precipitation is formed