Chapter 17 Equilibrium
collision model (17.1) Chapter 17 Review p.541 Key Terms activation energy (Ea) (17.) catalyst (17.) enzyme (17.) equilibrium (17.3) Chemical equilibrium (17.3) law of chemical equilibrium (17.5) equilibrium expression (17.5) Equilibrium constant (17.5) equilibrium position (17.5) homogeneous equilibria (17.6) heterogeneous equilibria (17.6) Le Chatelier's principle (17.7) solubility product (Ksp) (17-9)
Summary 1. Chemical reactions can be described by the collision model, which assumes that molecules must collide to react. In terms of this model, a certain threshold energy, called the activation energy (Ea), must be overcome for a collision to form products.. A catalyst is a substance that speeds up a reaction without being consumed. A catalyst operates by providing a lower-energy pathway for the reaction in question. Enzymes are biological catalysts. 3. When a chemical reaction is carried out in a closed vessel, the system achieves chemical equilibrium, the state where the concentrations of both reactants and products remain constant over time. Equilibrium is a highly dynamic state; reactants are converted continually into products, and vice versa, as molecules collide with each other. At equilibrium, the rates of the forward and reverse reactions are equal.
Summary (continued) 4. The law of chemical equilibrium is a general description of the equilibrium condition. It states that for a reaction of the type aa + be cc + dd the equilibrium expression is given by c d where K is the equilibrium constant. K = [ C [ D [ A a [ B b 5. For each reaction system at a given temperature, there is only one value for the equilibrium constant, but there are an infinite number of possible equilibrium positions. An equilibrium position is defined as a particular set of equilibrium concentrations that satisfy the equilibrium expression. A specific equilibrium position depends on the initial concentrations. The amount of a pure liquid or a pure solid is never included in the equilibrium expression.
Summary (cont d) 6. Le Chatelier's principle allows us to predict the effects of changes in concentration, volume, and temperature on a system at equilibrium. This principle states that when a change is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to compensate for the imposed change. 7. The principle of equilibrium can also be applied when an excess of a solid is added to water to form a saturated solution. The solubility product (Ksp) is an equilibrium constant defined by the law of chemical equilibrium. Solubility is an equilibrium position, and the Ksp value of a solid can be determined by measuring its solubility. Conversely, the solubility of a solid can be determined if its Ksp value is known.
Active Learning Questions Consider an equilibrium mixture of four chemicals (A, B, C, and D, all gases) reacting in a closed flask according to the following equation: A + B C+ D a. You add more A to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original set-up at equilibrium, and add more D to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.
The collision model successfully explains why reactions go faster when the concentrations of reactants and temperatures. 1. decrease; decrease. increase; increase 3. increase; decrease 4. decrease; increase
Catalysts speed up the rates of chemical reactions by lowering the. 1. temperature. activation energy 3. collisional frequency 4. kinetic energy
Chlorofluorocarbons (CFCs) are notorious as major contributors to ozone layer damage. They break down in the presence of UV radiation and release atomic Cl into the stratosphere. Atomic Cl acts as a in the breakdown of ozone, and as such is not consumed in the reaction. 1. reactant. modifier 3. catalyst 4. enzyme
Chemical equilibrium is a dynamic process in which the of the forward reaction equals that of the reverse reaction. 1. rate. reactivity 3. heat 4. enthalpy
In the previous chapter we learned that the partial dissociation of weak acids such as acetic acid could be written thus: HC H 3 O (aq) = H + (aq) + C H 3 O - (aq) Since we now know that the double arrow indicates that this reaction is in equilibrium, write the equilibrium constant expression for the dissociation of acetic acid. K = [H + [C 3 1.. [HC H H O 3 O K [CH3O = [H + + [HC H O 3 K = 3. 4. [H [HC + [C H 3 H O 3 O [HCH3O = + [H K
The For the reaction below, calculate the value of the equilibrium constant, given the equilibrium concentrations. N O 4 (g) = NO (g) [N O 4 = 0. 055 M [NO = 0.060 M 1. K = 15. K = 1.5 3. K = 1.1 4. K = 0.065
Write out the equilibrium expression for the reaction: CaF (s) = Ca + (aq) + F - (aq) + - [Ca [F 1. K =. [CaF K = + [Ca [F [CaF - - [F 3. K = [Ca + 4. [CaF + + - K = [Ca [F
In the equilibrium reaction: N (g) + 3H (g) = NH 3 (g) If the system is at equilibrium, and the total pressure is increased by reducing the volume, the position of equilibrium. 1. stays constant. shifts to the right 3. shifts to the left 4. cannot be predicted
When solid sodium hydroxide dissolves in water, the process is exothermic: NaOH(s) = Na + (aq) + OH - (aq) + heat If you add NaOH to water and it doesn t all dissolve, would you heat it or cool it to get more to dissolve?
NO is formed at high temperatures from the combustion of atmospheric N in the internal combustion engine: N (g) + O (g) + heat = NO(g) In order to cut down on the production of NO, a greenhouse gas and a precursor of NO, which is an ingredient of smog, would you increase or decrease the combustion temperature of the engine?
At a given temperature, K = 50 for the reaction: H (g) + I (g) = HI(g) Calculate the equilibrium concentration of H given: [I = 1.5 x 10 - M and [HI = 5.0 x 10-1 M 1. 1.5 x 10 - M. 3.0 x 10 - M 3. 5.0 x 10-1 M 4. 3.3 x 10-1 M
Write out the K sp expression for the reaction: Ag SO 4 (s) = Ag + (aq) + SO 4 - (aq) 1. [Ag + Ksp = [SO4. K sp = [Ag + [SO4-3. K sp + [Ag [SO 4 = [Ag SO 4 4. K sp = [Ag +
If a saturated solution of PbCl is prepared by dissolving some of the salt in distilled water and the concentration of Pb + is determined to be 1.6 x 10 - M, what is the value of K sp? 1..6 x 10-4..0 x 10-4 3. 3. x 10-4. 1.6 x 10-5