A little Quiz on Chemical Equilibrium

True or False

A certain amount of energy, called the activation energy, must be available if a reaction is to take place.
A reversible chemical reaction is one in which equilibrium is never established due to the constant decomposition of the products.
When the rate of the reverse reaction equals the rate of the forward reaction, equilibrium has been established.
Changes in temperature will change the value of an equilibrium constant.

Consider the chemical equilibrium of the following reaction:

2 SO_{2 (g)} + O_{2 (g)} <--> 2 SO_{3 (g)} + heat

A decrease in amount of O₂ will decrease the amount of SO₂ present.
A decrease in the volume will decrease the amount of SO₂ present.
A decrease in temperature will increase the amount of SO₂ present.
A decrease in the amount of SO₃ present will increase the amount of SO₂ present.

Consider the Equilibrium:

N_{2 (g)} + 3H_{2 (g)} <--> 2NH_{3 (g)}, K_p = 1 x 10^-4

The rate constant for the forward reaction is greater than that of the reverse reaction.
Since the reaction has a high activation energy, a catalyst is not needed.
The equilibrium constant is given by K_p = [N₂][H₂]³/[NH₃]²
Conducting the reaction under high pressures will increase the yield of ammonia.

According to Le Chatelier's Principle:

An increase in pressure always causes a change in the position of equilibrium for any reaction.
The greatest yield of ammonia in the exothermic reaction N₂ + 3 H₂ <--> 2 NH₃ is attained at high temperature.
The equilibrium constant is increased for the reaction A + B <--> C if the concentration of A is increased.
An increase of temperature causes a decrease in the value of the equilibrium constant for an exothermic reaction.

Consider the reaction:

4HCl(aq) + MnO₂ (s) <--> Cl₂ (g) + 2H₂O(l) + MnCl₂ (aq)

The equilibrium is displaced to the left if:

catalyst is added
pressure is lowered
temperature is lowered
H₂O(l) is added

Consider the reaction:

CH_{4 (g)} + 4Cl_{2 (g)} <--> CCl_{4 (liq)} + 4 HCl_(g) H⁰ = -398 kJ/mol

The equilibrium is displaced to the right if:

the temperature is raised
the pressure is lowered
some carbon tetrachloride is removed
some hydrogen chloride is added
none of these answers

The effect on equilibrium of increasing temperature of an exothermic reaction is to shift the equilibrium to the left.

Long Answer

What substances present in the chemical reaction would NOT be included in the equilibrium expression?

answer:

Pure phases like NaCl(s) in NaCl(s) <-> Na⁺_(aq) + Cl^-_(aq)
solvents

For

CO_{2 (g)} + H_{2 (g)} <--> CO_(g) + H₂O_(g)

If there are 1.43 moles each of CO and H₂O, 0.572 moles H₂ and 4.572 moles CO₂, in a 4.0 L container at equilibrium, what is K_C ?

Answer: 0.782

Consider the decomposition

2 SO₃ _(g) <-> 2 SO₂ _(g)+ O_2(g).

At equilibrium, there are 0.090 mol SO₂ , 0.110 mol O₂ , 0.100 mol SO₃ in a 25.0-L container. What is the value of K_C ?

Answer: 3.6 x 10^-3

For the reaction

3Fe_(s) + 4H₂O_(g) <--> Fe₃O_{4 (s)} + 4H_{2 (g)}
Write the expression for K_p

ANSWER: Kp = {P(H₂)}⁴ /{P(H₂O)}⁴

Write the expression for K_C in terms of K_p

ANSWER: K_C = K_p

What is the effect of adding Fe(s)?

ANSWER: no change

What is the effect of removing H₂ ?

ANSWER: shift to right

A mixture of 0.75 mol of N₂ and 1.20 mol of H₂ are placed in a 3.0 liter container. When the reaction

N_{2 (g)} + 3H_{2 (g)} <--> 2NH_{3 (g)}

reaches equilibrium, [H₂] = 0.100 M. What is the value of [N₂] and [NH₃] at equilibrium?

answer:

[NH₃] = 0.200 M
[N₂] = 0.150 M

A mixture of 2.5 moles H₂O and 100 g of C are placed in a 50-L container. At equilibrium for the reaction

C(s) + H₂O(g) <--> CO(g) + H₂(g),

[H₂] = 0.040 M. What is the value of [H₂O] at equilibrium?

Answer: [H₂O] = 0.010 M

The equilibrium constant, K_p , equals 1.78 at 250^oC for the decomposition reaction:

PCl₅ (g) <--> PCl₃ (g) + Cl₂ (g) Calculate the percentage of PCl₅ that dissociates if 0.05 mole of PCl₅ is placed in a closed vessel (constant volume) at 250^oC and at 2.00 atm pressure.

Solution:

V = (0.05 mol)(0.082 l atm/mol K)(523 K)/(2.00 atm) = 1.072 liters

K_c = 1.78/((0.082 l atm/mol K)(523K)) = 0.0414

K_c = (x/1.072)²/((0.05 - x)/1.072))

0.0445 = x²/(0.05 - x)

x = 0.03 (successive approx)

%dissociated = 0.03/0.05 x 100 = 60%

Consider the endothermic reaction:

N_{2 (g)} + O_{2 (g)} <--> 2NO(g), H⁰ = 192.5 kJ/mol

At 2000 K the equilibrium constant is 5.0 x 10^-4. At 2500 K the value of the equilibrium constant is:

solution:

ln(K₂/K₁) = -H/R (1/T₂ - 1/T₁)
K₂ = 5.1 x 10^-3

PJ Brucat // University of Florida

True or False

Write the expression for KC in terms of Kp

What is the effect of adding Fe(s)?

What is the effect of removing H2 ?

Write the expression for K_C in terms of K_p

What is the effect of removing H₂ ?