The Concept of Dynamic Equilibrium – The Equilibrium Constant (K)

How does this chapter differentiate form the previous chapter of reaction rates? In the last chapter, we found that the speed of a chemical reaction was determined by kinetics. We also observed factors that affected reaction rates. In this chapter, we will determine how far a reaction will go, based on the equilibrium constant (K). – The equilibrium constant is experimentally determined. A large equilibrium constant = reaction nearly goes to completion (pretty much all of the reactants react to form products). Here the reaction lies far to the right at equilibrium (product concentrations are favored). A small equilibrium constant = reaction barely proceeds at all (Pretty much all reactant, remain unreacted). Here the reaction lies far to the left at equilibrium (reactant concentration is favored).

The Concept of Dynamic Equilibrium When a reaction is shown with a yield sign pointing in both directions, this means that the reaction is reversible. In the case above, as nitrogen gas begins to react with hydrogen gas to produce ammonia, the concentration of the reactants begins to decrease, and the rate of the forward reaction also slows down. At the same time the product concentration begins to increase, and the reverse reaction rate also increases. Eventually, the two rate will be equal. This is the point when a dynamic equilibrium has been reached

What is a dynamic equilibrium? A dynamic equilibrium for a chemical reaction is the condition in which the rate of the forward reaction equals the rate of the reverse reaction. The reason that rate is underlined above is to emphasize the point that at equilibrium it’s the concentrations of the reactants and products no longer change…but that does not mean that the concentrations of the reactants and products are the same!!! “Dynamic” means that the forward and reverse reactions are still occurring (the reaction doesn’t stop), but they are occurring at the same rate.

A Graph that Shows Equilibrium: Concentration vs. Time As you can see in the graph above, as the concentration of the reactant begin to decrease, the concentration of the products increase, until the rates at which the reactants and products are formed, are constant. This is the point where the lines are horizontal and a dynamic equilibrium has been established. The graphs vary depending on the reaction, but in the end, an equilibrium of reversible reactions will always be reached.

The Equilibrium Constant (K) The equilibrium constant (K) is a way to quantify the concentrations of the reactants and products at equilibrium. Let’s consider the reaction: aA + bB cC + dD Here A and B are reactants, C and D are products, and a, b, c, d are the stoichiometric coefficients. The equilibrium constant (K), at equilibrium, is defined as the ratio of the concentration of the products raised to their stochiometric coefficients, divided by the concentration of the reactants raised to their stochiometric coefficients.

Law of Mass Action Law of mass action is the relationship between the balanced chemical equation and the expression of the equilibrium constant: [C] c [D] d K = [A] a [B] b Let’s try a practice problem: Express the equilibrium constant for the following reaction: 2N 2 O 5 (g) 4NO 2 (g) + O 2 (g) [NO 2 ] 4 [O 2 ] K = [N ] 2

Let’s Try Another!!! Express the equilibrium constant for the combustion of propane as shown by the balanced equation: C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) [CO 2 ] 3 [H 2 0] 4 K = [C 3 H 8 ][O 2 ] 5

The Significance of the Equilibrium Constant A large K ( K > 1): The numerator is greater than the denominator. The forward reaction is favored (products are favored). A small K ( K < 1): The reverse reaction is favored (reactants are favored). If K = 1, neither direction is favored and the reaction proceeds about half-way.

Let’s Try a Practice Problem!!! The equilibrium constant for the reaction: A(g) B(g) is 10. A reaction mixture initially contains [A] = 1.1 M and [B] = 0.0 M. Which statement is true at equilibrium? (a) The reaction mixture will contain [A] = 1.0 M and [B] = 0.1 M (b) The reaction mixture will contain [A] = 0.1 M and [B] = 1.0 M (c) The reaction mixture will contain equal concentrations of A and B. (b) The reaction mixture will contain [A] = 0.1 M and [B] = 1.0 M, so that [B]/[A] = 10.

Relationships between the Equilibrium Constant and the Chemical Equation 1.) If you reverse the equation, invert the constant. K reverse = 1/ K forward = K’ 2.) If you multiply the coefficients in the equation by a factor, raise the equilibrium constant to that same factor. 3.) If you add two or more individual chemical equations to obtain an overall equation, multiply the corresponding equilibrium constants by each other to obtain an overall equilibrium constant.

Let’s Try a Practice Problem!!! The reaction A(g) 2B(g) has an equilibrium constant of K = What is the equilibrium constant for the reaction B(g) ½A(g)? (a)1 (b)10 (c)100 (d) (b) 10, here’s why. First, the equilibrium constant for the reverse reaction K’ = 1/K = 1/0.010 = 100. K’=100 for the reaction 2B(g) A(g). However the problem is asking us to multiply the coefficients by ½, so we must raise K’ to a factor of ½. 100 ½ = 10.

Let’s Try a Couple More Practice Problems!!! Consider the following chemical equation and the equilibrium constant at 25 o C: 2COF 2 (g) CO 2 (g) + CF 4 (g) K = 2.2x10 6 Calculate the equilibrium constant for the following reaction at 25 o C. 2CO 2 (g) + 2CF 4 (g) 4COF 2 (g) K’ = ? K’ = 1/K = (1/(2.2X10 6 )) 2 = 2.1X10 -13

Let’s Try Another!!! Predict the equilibrium constant for the first reaction shown here given the equilibrium constants for the second and third reactions: CO 2 (g) +3H 2 (g) CH 3 OH(g) + H 2 O(g) K 1 = ? CO(g) +H 2 O(g) CO 2 (g) + H 2 (g) K 2 = 1.0X10 5 CO(g) +2H 2 (g) CH 3 OH(g) K 3 = 1.4X10 7 Let’s rearrange the last two equations above, to predict the equilibrium constant of the first  next slide

CO 2 (g) +3H 2 (g) CH 3 OH(g) + H 2 O(g) K 1 = ? CO(g) +H 2 O(g) CO 2 (g) + H 2 (g) K 2 = 1.0X10 5 CO(g) +2H 2 (g) CH 3 OH(g) K 3 = 1.4X10 7 By reversing the second reaction and keeping the third reaction as it is, we can cancel substances that appear on both sides of the reaction, and we will end up with the first reaction: CO 2 (g) + H 2 (g) CO(g) +H 2 O(g) K 2 ’ = 1/(1.0X10 5 ) CO(g) +2H 2 (g) CH 3 OH(g) K 3 = 1.0X10 7 CO 2 (g) +3H 2 (g) CH 3 OH(g) + H 2 O(g) K 1 = K 2 ’ X K 3 = K 1 = 1.4X10 2

pgs #’s 21(a & b), 24, 28, & 30 Read pgs