Questions Related to Concepts of Chemical Equilibrium

Questions Related to Concepts of Chemical Equilibrium General Statements:

A certain amount of energy, called the activation energy, must be available if a reaction is to take place. It is this activated nature of Chemistry that makes Vapor pressure, reaction rate constants and Equilibrium constants all have exponential {exp(-1/T)} temperature dependences
A reversible chemical reaction can reach equilibrium by balancing the forward and reverse reaction rates. It is this balance that determines the equilibrium concentrations of the reactants and products in the Reaction.
The equilibrium constant for a reaction is the ratio of the forward to reverse rate constants, and is a function of temperature, only.

According to Le Chatelier's Principle:

A change in temperature will change the value of the equilibrium constant differently depending on the exothermicity of the reaction. Exothermic reactions will have an increase in their equilibrium constant when the temperature is lowered.

An increase in pressure may causes a change in the equilibrium advancement of a reaction, but only if there is a net change in the number of moles of gas in going from reactant to product.

Consider the chemical equilibrium of the following reaction:

2 SO_{2 (g)} + O_{2 (g)} <--> 2 SO_3
(g) + heat

An increase in amount of O₂ will decrease the equilibrium amount of SO₂ present.
A decrease in the volume will decrease the equilibrium amount of SO₂ present.
A increase in temperature will increase the equilibrium amount of SO₂ present.
Removal of SO₃ from the reactor will decrease the equilibrium 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 less than that of the reverse reaction.
The value of the activation barrier cannot be determined from the data given.
The equilibrium constant is given by K_C = [NH₃]² / [N₂][H₂]³
Conducting the reaction under high pressures will increase the yield of ammonia.
The synthesis of ammonia is an exothermic reaction, so the rate of product formation may increase with temperature, but the equilibrium constant, and thus the yield of the the reaction, decreases with increasing temperature.

Consider the reaction:

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

The equilibrium is unaffected if a catalyst is added
The equilibrium is shifted to the right if the pressure is lowered
The equilibrium is shifted to the right if temperature is lowered
The equilibrium is shifted to the left if H₂O(l) is added

Consider the reaction:

CH_{4 (g)} + 4Cl_{2 (g)} <--> CCl_4
(liq) + 4 HCl_(g) DH⁰ = -398 kJ/mol

The equilibrium is shifted to the left if the temperature is raised

The equilibrium is shifted to the left if the pressure is lowered

The equilibrium is unaffected if some carbon tetrachloride is removed

The equilibrium is shifted to the left if some hydrogen chloride is added

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 and allowed to come to equilibrium subject to the following reaction: C(s) + H₂O(g) <--> CO(g) + H₂(g), The equilibrium concentration of Hydrogen is found to be [H₂] = 0.040 M. What is the equilibrium concentration of water, [H₂O] ?

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 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) DH⁰ = 192.5 kJ/mol At 2000 K the equilibrium constant is 5.0 x 10^-4. What is the value of the equilibrium constant at 2500 K ?

solution:

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

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