Chemical Equilibrium

Chemical Equilibrium Keq

Tuesday, November 11, 2008 8:23 AM

=([C]c[D]d)/([A]a[B]b)

Equilibrium: the extent of a reaction - In stoichiometry we talk about theoretical yields, and the many reasons actual yields may be lower. - Another critical reason actual yields may be lower is the reversibility of chemical reactions: some reactions may produce only 70% of the product you may calculate they ought to produce. - Equilibrium is a state in which there are no observable changes as time goes by in terms of macroscopic observation. - Chemical equilibrium is reached when ○ The rates of the forward and reverse reactions are equal and ○ The concentration of the reactants and the products remain constant. ○ N2O4(g) ↔ 2NO2(g) - Physical equilibrium is reached when ○ H2O(l) ↔ H 2O(g) The Concept of Equilibrium - Consider the colorless frozen N2O4. At room temperature, it decomposes to brown NO2 - At some time, the color stops changing and we have a mixture of N2O4 and NO2. - Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse reactions. - At that point, the concentrations of all species are constant. (Reactions are still occurring but the rates of reactions are equal) - Using the collision model: ○ As the amount of NO2 builds up, there is a chance that two NO2 molecules will collide to form N2O4 ○ At the beginning of the reaction, there is no NO2, so the reverse reaction (2NO2(g) -> N2O4(g)) will not occur. - As the substance warms it begins to decompose: ○ N2O4(g) →2NO2(g) - When enough NO2 is formed, it can react to form N2O4. ○ 2NO2(g)↔N2O4(g) - At equilibrium, as much N2O4 reacts to form NO2 as NO2 reacts to re-form N2O4. - The double arrow implies the process is dynamic ○ N2O4(g)↔2NO2(g) - As the reaction progresses ○ [A] decreases to a constant ○ [B] increases from zero to a constant ○ When [A] and [B] are constant, equilibrium is achieved

Le Châtelier's Principle - If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

Changes in Concentration - N2(g) + 3H 2(g) ↔ 2NH3(g) … You can add NH3 to this reaction to add reverse pressure. - This will cause the equilibrium to shift stress to the left of the reaction. ○ aA + bB ↔ cC + dD Change

Shifts the Equilibrium

Increase the concentration of the products

Left

Decrease the concentration of the products

Right

Increase the concentration of reactants

Right

Decrease the concentration of the reactants

Left

Changes in Volume and Pressure - A(g) + B(g) ↔ C(g)

Homework Section 7.1 434-438 questions 1-5 pg 438 Qualitative Equilibrium Worksheet Read Sample Problem pg 433-436 Report Cards :'(

1a) The observable characteristics would be the concentrations of the reactants and the products initially and at equilibrium. 1b) The equilibrium is considered dynamic because eventually the reactants will be forming products at the same rate as products are forming reactants. 1c) The rates are equal in the system. 2a) Three systems that can be considered to be in 'equilibrium' - The amount of money that two companies are giving each other in terms of purchasing a product (monetary) and the amount of money that company will gain when the costs of selling and manufacturing that product are considered - The gravitational forces acting on an object that is held in the air for a short period of time: The forces holding it up are equal to the forces of gravity pushing it down. - The amount of oxygen entering someone's lungs in comparison to the amount of carbon dioxide leaving it. Also the amount of energy gained by both the trees and the person breathing. 3a) CH4(g) + Cl2(g) ↔ CH 3Cl (g) + HCl (g), because the limiting reagent is the 2.0mol of CH4(g) in this reaction only 2.0mol of CH 3Cl(g) can be produced. 3b) The percent reaction is (2.0mol-1.4mol)/2.0mol=0.3*100= 30%, meaning that only 70% of the expected reaction actually took place. Because less product than expected was produced, this means that overall the reaction shifted left, favoring the reactants. 4a) Change Type C2H4(g)

Br2(g)

Init. M

4.00mol/L

2.50mol/L

Equil M

4.00mol/L-x=2.5mol/L 2.50mol/L

C2H4Br2(g)

Qualitative Equilibrium Problems 1. In this problem, there may be a production of H2 and O2 being produced in the closed bottle, however since there is so much O2 in the bottle the rate at which the reactant of H2O is produced may be a much higher rate than the production of hydrogen and oxygen gas, thus never coming to equilibrium. 2. This can be considered a situation at equilibrium if the amount of students leaving the school was exactly equal to the amount of students entering the school in a new year or from new entries. If no one was leaving or entering, this would not be considered equilibrium. 3. 2SO2(g) + O2(g) ↔ 2SO3(g), this situation may be in equilibrium, if a catalyst was added to the reaction and the reaction shifted in terms of the amount of reactants produced and the products reduced, then it could be inducted that before adding the catalyst it was not in equilibrium. 4. This demonstrates that regardless of the way that a reaction begins, if given the same temperature and pressure and also a certain amount of time, equilibria that are the same can be attained. 5. This shows that the reactions is always occurring in that the iodine solid is constantly being reacted into solution just as often as the solution is being crystallized. Therefore, when the radioactive properties of the solution are found it can be assumed that the reaction was ongoing (re: dynamic). Le Châtelier's Principle Homework

Change

Shifts the Equilibrium

Increase Pressure

Side with fewest moles of gas

Decrease Pressure Side with greatest moles of gas Increase volume

Side with greatest moles of gas

Decrease volume

Side with fewest moles of gas

Changes in Temperature - (ONLY stress that changes the value of K which is the equilibrium constant) Change

-

Exothermic Reaction Endothermic Reaction

Increase temperature

K↓ (↑[react])

K↑ (↓[react])

Decrease temperature

K↑ (↓ [react])

K↓ (↑[react])

Chemistry Page 1

6. In this case, the amount of pressure in the syringe has been increased, and the concentration has increased (and the solution will darken accordingly. Because the N2O4 side has fewer moles, it is shifted leading to an increase in the amount of those in the syringe after the reaction stabilizes. Pressuring ice will not make it melt because the equilibrium has a side with neither more or less moles (H2O(l)↔H 2O(s)) and the reaction will not shift on either side more than the other. However, if there is equal volume and temperature on the ice as before, the increased pressure will cause it to melt. 7. In a. The increase in [H2] will cause the product side to increase. In b. The removal of N2(g) will cause the reaction to move to the right, therefore increasing the amount of products formed. In c. the decrease of pressure will cause the equilibrium constant to increase and decrease the amount of reactants. In d., there will be more products formed as they

Adding a Catalyst - Does not change K - Does not shift the position of an equilibrium system - System will reach equilibrium sooner. - Catalyst lowers Ea for both forward and reverse reactions Catalyst does not change equilibrium constant or shift equilibrium. Change

Shift Equilibrium Change Equilibrium Constant

Concentration

Yes

No

Pressure

Yes

No

Volume

Yes

No

Temperature

Yes

Yes

Catalyst

No

No

Chemistry Page 2

reactants. In d., there will be more products formed as they have less moles. In e., the catalyst will just increase the speed it takes to reach equilibrium. What is equal is the rate at which reactants are produced relative to products. 8. Because this reaction is endothermic, and increase in temperature will increase the amount of product. An increase in reactant concentration will increase product concentration, a decrease in product concentration will shift the reaction to the right, creating more yield.