Most chemical reactions never go to completion where all the reactants are converted into products. Instead these reactions reach a chemical equilibrium.
Concept #1: Understanding Chemical Equilibrium.
A reaction reaches equilibrium once the rate of the forward reaction equals the rate of the reverse reaction. Once at equilibrium there is no net change in the concentration of reactants or products.
Whereas the rate constant, k, is affected by factors dealing with the rate of reaction, the equilibrium constant, K, is affected by only temperature.
Practice: Which one of the following statements does not describe the equilibrium state?
a. While at equilibrium, a dynamic process is still occurring.
b. The concentration of the reactants is equal to the concentration of the products.
c. The concentration of the reactants and products reach a constant level.
d. At equilibrium, the net concentration of all species is not changing.
e. All are true.
Associated with any reaction at equilibrium is the equilibrium constant K. Its numerical value determines if reactants or products are more greatly favored within a reaction.
Concept #2: Understanding the Equilibrium Constant.
The equilibrium constant K is a ratio of products to reactants. It only deals with gaseous or aqueous compounds.
Example #1: Write the equilibrium expression for the following reaction:
Practice: State which is greater in amount: reactants or products, based on the given equilibrium constant, K.
Practice: The decomposition of nitrogen monoxide can be achieved under high temperatures to create the products of nitrogen and oxygen gas.
6 NO(aq) ⇌ 3 N2(aq) + 3 O2(aq)
a) What is the equilibrium equation for the reaction above?
b) What is the equilibrium expression for the reverse reaction.
Practice: The equilibrium constant, K, for 2 NO (g) + O2 (g) ⇌ 2 NO2 (g) is 6.9 x 102.
What is the [NO] in an equilibrium mixture of gaseous NO, O2, and NO2 at 500 K that contains 1.5 x 10 –2 M O2 and 4.3 x 10 –3 M NO2?
When dealing with equilibrium units in terms of pressure the equilibrium constant of Kp must be used, but if the equilibrium units are in molarity then we must use the equilibrium constant of Kc.
Concept #3: The Different Types of Equilibrium Constants.
Kp and Kc are related to one another by the following equation below:
Example #2: For the reaction above, the Kc = 4.9 x 10-9 at 25oC. Which of the following statements is true?
Practice: Methane (CH4) reacts with hydrogen sulfide to yield hydrogen gas and carbon disulfide, a solvent used in the manufacturing rayon and cellophane. What is the value of Kc at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.20 atm methane, 0.15 atm hydrogen sulfide, 0.30 atm carbon disulfide and 0.10 atm hydrogen gas?
CH4 (g) + 2 H2S (g) ⇌ 4 H2 (g) + CS2 (g)
Determination of the value for Δn can help us determine if Kp is greater than, less than or equal to Kc.
Practice: In which of the given reactants is Kp greater than, less than and equal to Kc?
a) SO3 (g) + NO (g) ⇌ SO2 (g) + NO2 (g)
b) P4 (s) + 5 O4 (g) ⇌ P4O10 (s)
c) 4 NH3 (g) + 3 O2 (g) ⇌ 2 N2 (g) + 6 H2O (g)
Practice: Given the hypothetical reaction 2 A (s) + ? B (g) ⇌ 3 C (g), Kp = 0.0105 and Kc = 0.45 at 250 degrees Celsius. What is the value of the coefficient of B?
The old concept of Hess’s Law can be applied to our new concept of the Equilibrium Constant K.
Example #3: The equilibrium constant K for the first reaction is 6.83 x 10-12 at 1000 K. Calculate K for the second reaction.
Reversing a reaction gives us the reciprocal of K, while multiplying a reaction by a value makes it the power for K.
Example #4: Calculate the rate constant, Kc, for the reaction below:
H (g) + Br (g) ----> HBr (g)
Use the following information to calculate Kc.
H2 (g) ---> 2 H (g) Kc = 11.8
Br2 (g) ---> 2 Br (g) Kc = 1.15 x 10-5
H2 (g) + Br2 (g) ---> 2 HBr (g) Kc = 2.78 x 103
Example #5: Hess's Law & Equilibrium Constant K
Example #6: Hess's Law & Equilibrium Constant K