Quantitative Changes in Equilibrium Systems Chapter 7

Testing to see if an equilibrium has been established. The reaction quotient (Q), uses the equilibrium expression with the available concentrations to determine if the reaction has attained equilibrium or not. For the general chemical reaction: aA + bB  cC + dD

Predicting the direction of an equilibrium Reaction Quotient Description of equilibrium Q=K eq The system is at equilibrium Q>K eq The system is “product rich”. It must “shift” to the left to achieve equilibrium. Q<K eq The system is “reactant rich”. It must “shift” to the right to achieve equilibrium.

Quantitative analysis of equilibria Using Le Châtelier’s Principle & the Equilibrium Law one can quantitatively assess;  The equilibrium constant (K eq )  The position or progress of the equilibrium system with the reaction quotient (Q)  The equilibrium concentrations

Determining the position or status of an equilibrium system. Calculate the reaction quotient using the equilibrium concentrations. Compare the reaction quotient value to the equilibrium constant value. Determine whether the reaction is reactant or product rich and adjust the equilibrium to establish the equilibrium.

At 500 o C, the value of K eq for reaction between nitrogen and hydrogen to produce ammonia is The equilibrium concentrations for the chemicals are; [N 2 ] = 0.10 mol/L, [H 2 ] = 0.30 mol/L & [NH 3 ]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium?

Calculating equilibrium concentrations To determine the equilibrium concentrations, one must use:  A balanced equation for the equilibrium reaction  An equilibrium expression and constant, K eq  An Initial concentration, Change in concentration, and Equilibrium concentration (ICE) table.

Calculating equilibrium concentrations In cases where the K eq value is very small the addition and subtraction of the change value (x) may be insignificant and thereby omitted.  Use the ratio of the smallest initial concentration and the K eq to assess the affect of the change.

Calculating equilibrium concentrations  If clue > 500, the addition or subtraction of “x” is insignificant and may be ignored.  If 100<clue<500, the addition or subtraction of “x” is probably insignificant, but should be checked.  If clue<100, the addition or subtraction of “x” is significant and must be included in the calculations.

Calculating equilibrium concentrations In cases where the K eq expression results in the resolution of a quadratic equation, the use of the quadratic formula my prove helpful.  When the quadratic formula produces two answers select the one that is viable. Remembering that you can not have a negative concentration!

Example – At high temperatures, as with lightning, nitrogen and oxygen will react to produce nitrogen monoxide. A chemist puts moles of N 2(g) and mol of O 2(g) in a 1.0 L flask at high temperature, where the K eq = 4.2 x What is the concentration of the NO (g) in the mixture at equilibrium? Strategy 1. Divide the smallest initial concentration by the K eq to determine if the change in concentration can be ignored. 2. Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression. 3. Set up an ICE table letting “x” represent the change in concentrations. 4. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”. 5. Calculate the required value(s).

1.Divide the smallest initial concentration by the K eq to determine if the change in concentration can be ignored.

2.Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression. N 2(g) + O 2(g)  NO (g) N 2(g) + O 2(g)  2 NO (g)

3.Set up an ICE table letting “x” represent the change in concentrations. Concentration (mol/L) N 2(g) + O 2(g)  2 NO (g) Initial concentration Change in concentration -x 2x Equilibrium concentration – x ≈ – x ≈ x

4.Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”. The negative value is impossible as one can not have a negative concentration.

5.Calculate the required value(s). What is the NO (g) concentration at equilibrium? The NO (g) concentration at equilibrium is 1.2 x mol/L.

Example – In a 1.00 L flask, 2.00 mol of H 2(g) is combined with 3.00 mol of I 2(g) which produces HI (g). The K eq for the reaction is 25 at 1100 K. What is the What is the concentration of each gas in the mixture at equilibrium? Strategy 1. Divide the smallest initial concentration by the K eq to determine if the change in concentration can be ignored. 2. Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression. 3. Set up an ICE table letting “x” represent the change in concentrations. 4. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”. 5. Calculate the required value(s).

1.Divide the smallest initial concentration by the K eq to determine if the change in concentration can be ignored. The clue is much less than 500 so we can not ignore the changes in H 2(g), I 2(g) and HI (g).

2.Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression. H 2(g) + I 2(g)  HI (g) H 2(g) + I 2(g)  2 HI (g)

3.Set up an ICE table letting “x” represent the change in concentrations. Concentration (mol/L) H 2(g) + I 2(g)  2 HI (g) Initial concentration Change in concentration -x 2x Equilibrium concentration 2.00 – x3.00 – x2x

4.Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.

4.3 and 1.7 are determined through the quadratic formula calculation, but only one is correct! 3.00 – 4.3 = negative value so 1.7 mol/L is the correct value. (A negative concentration is impossible!) Therefore the final equilibrium concentrations are: [H 2 ] eq = 2.00 – 1.7 = 0.3 mol/L [I 2 ] eq = 3.00 – 1.7 = 1.3 mol/L [HI] eq = 2( 1.7) = 3.4 mol/L

SUMMARY OF LESSONS 1,2 AND 3 EQUILIRIUM EXPRESSION Solids and liquids are not included in equilibrium expression ONLY gases and aqueous. units: use M for K c and or atm for K p SIZE OF K K >1 → products favored in equilibrium system K < 1 → reactants favored in equilibrium system ENDOTHERMIC K increases with increase temperature. K decreases with decrease temperature EXOTHERMIC K decreases with increase temperature. K increases with decrease temperature

SUMMARY OF LESSONS 1,2 AND 3 HOW “K” CHANGES AS THE REACTION CHANGES 1.Reverse rxn: K →1/K 2.Multiply rxn by n: K → K n 3.Add rxs: K3 = K1 x K2 REACTION QUOTIENT (Q): a snapshot of the reaction; shows which way reaction proceeds (L, R, no shift) Q > K → left Q < K → right Q = K → at equilibrium (no shift) CALCULATIONS 1.given [ ] eq → find K 2.given [ ]o + other information → find [ ] eq → K 3.given K → [ ] eq using one of three methods a)perfect squares b)Approximation: 100 rule c)Quadratic equation

SUMMARY OF LESSON 1,2 AND 3 Le CHATELIER’S PRINCIPLE: reaction shifts to restore equilibrium (opposes stress on the system) FACTORS AFFECTING EQUILIBRIUM Concentration; Temperature; Pressure Concentration: add/remove reactants/products part of K expression; pure liquids or solids usually do not affect equilibrium increase concentration of A, move away from A ( A  ) Decrease concentration of A, move towards A (  A) Temperature: endothermic: heat written as a reactant; exothermic: heat written as a product Increase temp always favours the endothermic reaction Pressure: Must be gaseous and must have unequal number of moles. P ↑ move in a direction with less gas mol if possible; convert V changes to P FACTORS NOT AFFECTING EQUILIBRIUM Catalyst, Pure liquid or solid (water sometimes can shift rxn). Inert gases does not affect equilbrium