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Chapter 14 Worksheet 1 (ws14.1)
Chemical Equilibrium, the Equilibrium Constant (Keq), and the Reaction Quotient (Q)
As a reaction proceeds in a closed system, the concentrations of reactants and products change until, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. At that point there is no net change in the concentrations of reactants or products. For example, for the simple reaction: A = B:
Regardless of the initial concentrations, the concentrations of reactants and products present at equilbrium are related through the equilibrium constant (Keq) as follows:
For the reaction: aA + bB ⇌ cC + dD
If all reactants and products are in solution then:
Keq = Kc = [pic]
(The subscript “eq” indicates “equilibrium” concentration. The subscript “c” stands for “concentration”. The exponents are the coefficients from the balanced equation.)
If all reactants and products are ideal gases then either partial pressures or concentrations can be used in the equilibrium constant expression. In terms of partial pressures:
Keq = Kp = [pic] (The subscript “p” stands for “pressure”.)
The numerical value of the two equilibrium constants above will be different. Remember the ideal gas law shows that the partial pressure of a gas is directly proportional to its molarity:
P = [pic] = MRT (M is molarity and R = 0.0821 L-atm/mol-K).
Using this relationship, it is easy to show that (see pages 605-606 in your textbook) :
Kp = Kc(RT)Δn where Δn = moles of gaseous products – moles of gaseous reactants
For reactions involving both gases and aqueous solutions, molar concentrations and partial pressures can appear in the equilibrium constant expression.
Important points about equilibrium constants:
1. Every reaction has a characteristic equilibrium constant that depends only on temperature. The value of the equilibrium constant is independent of the initial concentrations of reactants and products.
2. Equilibrium constants have no units. This is because each concentration or pressure is actually a ratio of the concentration or pressure to their standard values. Standard concentration for a substance in solution is 1 M and standard pressure for a gas is 1 atm.
3. The concentrations of pure solids or pure liquids do not appear explicitly in the equilibrium constant expression. The concentrations of pure solids and liquids are constant (density divided by molar mass). These constant concentrations are simply incorporated into the equilibrium constant.
Alternative explanation: The standard concentration for any pure substance is simply that of the pure substance itself (which doesn’t change). Thus, the ratio that would appear in the equilibrium constant expression is always one (see number 2 above).
4. All reactants and products must be present at equilibrium. Even though pure solids and liquids do not appear in an equilibrium constant expression, if they participate in a reaction they must be present to establish equilibrium.
5. The numerical value of the equilibrium constant expresses the tendency for reactants to be converted to products. If Keq >> 1 we say that “products are favored”. If Keq 1000) means that the reaction will nearly go to completion. A very small equilibrium constant ( rater, reaction will proceed from left to right ( → ; more products)
If Q > Keq, then ratef < rater, reaction will proceed from right to left ( ← ; more reactants)
Note: ratef is the rate of the forward reaction and rater is the rate of the reverse reaction.
1. At 1000 K, the value of Kp for the reaction 2SO3(g) = 2SO2(g) + O2(g) is 0.338. Calculate the value for Q, and predict the direction in which the reaction will proceed toward equilibrium if the initial pressures of reactants are PSO3 = 2 x 10-3 atm; PSO2 = 5 x 10-3 atm; PO2 = 3 x 10-2 atm.
2. Chemists always want to find ways to increase the yield of a reaction. It would appear that equilibrium is the chemist’s worst enemy! (At least when a reaction has a small equilibrium constant. Read important point 5.) Two common ways for a chemist to increase the yield of a reaction are described below. Explain why they increase the yield.
a. Method 1: Add more of one of the reactants.
b. Method 2: Remove one of the products as it is being made.
3. At 573 K, the equilibrium constant for: N2(g) + 3 H2(g) ⇌ 2 NH3(g) is 4.34 x 10-3. At 700 K,
Kc = 1.04 x 10-4.
a. Is this an endothermic or exothermic reaction (in the forward direction)? (See important points 1 and 8.) Explain your answer.
b. The formation of ammonia from nitrogen gas and hydrogen gas is an important industrial process (called the “Haber process”). As you can see from the information above, the equilibrium constant for this reaction decreases as the temperature increases. Why, then, is the reaction performed at very high temperature? How do you think the yield of ammonia is maximized?
What is SN1 Prime reaction mechanism in organic chemistry? SN1′ (Substitution Nucleophilic Unimolecular Prime) Mechanism In SN 1 ′ reactions, the word "SN" stands for "nucleophilic substitution", "1" means that the rate- determining step is unimolecular in nature, and prime indicates that there is a double bond in the vicinity of