Chapter 18 – Rates of Reactions and Equilibrium Every biological and non-biological chemical reaction in nature eventually reaches a state called equilibrium.

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Presentation transcript:

Chapter 18 – Rates of Reactions and Equilibrium Every biological and non-biological chemical reaction in nature eventually reaches a state called equilibrium. Equilibrium occurs when the rate of the forward reaction, equals the rate of the reverse reaction. The time it takes for a reaction to reach equilibrium varies depending on some factors.

Rates of Reactions and Equilibrium Rates of Chemical Reactions The rate, or speed, of a chemical reaction is measured in units of a mass / time. Reaction rate = mass / time = g/min or g/sec

Rates of Reactions and Equilibrium Rates of Chemical Reactions The rate of a chemical reaction is affected by temperature, surface area of the reactants, and the presence of a catalyst. A catalyst is any substance that speeds up a chemical reaction without being used up.

Rates of Reactions and Equilibrium Energy Diagrams

Rates of Reactions and Equilibrium Chemical Equilibrium Occurs when the rate of the forward reaction (  ) equals the rate the reverse reaction (  ). Reactants Products

Rates of Reactions and Equilibrium Chemical Equilibrium Even though a chemical reaction is at equilibrium, the concentration of the products and the reactants do not have to be the same.

Rates of Reactions and Equilibrium Chemical Equilibrium Equilibrium Expression (K eq ) aA + bB  cC + dD K eq = [products] x / [reactants] y K eq = [C] c [D] d / [A] a [B] b Solids and liquids are not included in the K eq expression.

Rates of Reactions and Equilibrium Chemical Equilibrium Equilibrium Expression (K eq ) If Keq >1, then the products are favored. If Keq = 1, then neither products nor reactants are favored If Keq <1, then the reactants are favored. K eq = [products] x / [reactants] y

Rates of Reactions and Equilibrium Chemical Equilibrium Write the K eq for the for the reaction between sodium metal and water. 2Na (s) + 2H 2 O (l)  2NaOH (aq) + H 2(g) K eq =

Rates of Reactions and Equilibrium Le Chatelier’s Principle Once a chemical reaction has reached equilibrium, it will adjust to remain at equilibrium if it is stressed.

Rates of Reactions and Equilibrium Le Chatelier’s Principle Chemical equilibrium is like a balanced seesaw, if it is thrown off balanced by something, it will readjust to the balanced state.

Rates of Reactions and Equilibrium Le Chatelier’s Principle Stress Factors Change in concentration of a reactant or product. Change in the temperature of the system. Change in pressure on the system.

Rates of Reactions and Equilibrium Le Chatelier’s Principle 2Na (s) + 2H 2 O (l)  2NaOH (aq) + H 2(g) + heat If more water is added to this chemical reaction, will the equilibrium shift to the products side (right) or the reactant side (left)?

Rates of Reactions and Equilibrium Le Chatelier’s Principle 2Na (s) + 2H 2 O (l)  2NaOH (aq) + H 2(g) + heat If hydrogen gas is removed from this chemical reaction, will the equilibrium shift to the products side (right) or the reactant side (left)?

Rates of Reactions and Equilibrium Le Chatelier’s Principle 2Na (s) + 2H 2 O (l)  2NaOH (aq) + H 2(g) + heat If the temperature is increased on this chemical reaction, will the equilibrium shift to the products side (right) or the reactant side (left)?

Rates of Reactions and Equilibrium Le Chatelier’s Principle 2Na (s) + 2H 2 O (l)  2NaOH (aq) + H 2(g) + heat If pressure is increased on this chemical reaction, will the equilibrium shift to the products side (right) or the reactant side (left)?

Rates of Reactions and Equilibrium Le Chatelier’s Principle 2Na (s) + 2H 2 O (l)  2NaOH (aq) + H 2(g) + heat If this chemical reaction is cooled, will the equilibrium shift to the products side (right) or the reactant side (left)?

Rates of Reactions and Equilibrium Solubility Product Constant (K sp ) of salts. o We can write an equilibrium expression for the solvation of a salt. Na 2 S (s)  2Na +1 (aq) + S -2 (aq) K sp = [Na +1 ] 2 [S -2 ]

Rates of Reactions and Equilibrium Solubility Product Constant (K sp ) of salts. o Write the Solubility Product Constant expression for the following; AgCl (s)  Ag +1 (aq) + Cl -1 (aq)

Rates of Reactions and Equilibrium Solubility Product Constant (K sp ) of salts. o Write the Solubility Product Constant expression for the following; PbCl 2(s)  Pb +2 (aq) + 2Cl -1 (aq)

Rates of Reactions and Equilibrium Solubility Product Constant (K sp ) of salts. o Write the Solubility Product Constant expression for the following; Al 2 S 3(s) 

Rates of Reactions and Equilibrium Solubility Product Constant (K sp ) of salts. o What does the magnitude of K sp tell us about the solubility of a salt? K sp of PbCl 2(s) is 1.7 x x = [Pb +2 ] [Cl -1 ] 2 The Product of the concentration of the [Pb +2 ] and [Cl -1 ] ions are very small. Therefore, PbCl 2 must not dissociate very much in water.

Rates of Reactions and Equilibrium Solubility Product Constant (K sp ) of salts. o Write the dissociation reaction of Cobalt (II) sulfide in water. o Write the K sp expression for cobalt(II)sulfide. o The K sp of cobalt (II) sulfide is 5.0 x Comment on the relative solubility of this salt.

Chemical Thermodynamics The concept of chemical thermodynamics deals with how the enthalpy change and entropy change of a chemical reaction are related. Some reactions occur automatically at a certain temperature and pressure. We call these reaction spontaneous. Spontaneous reactions are not spontaneous in reverse.

Chemical Thermodynamics Spontaneous Reactions – Proceed on their own without any assistance. o Ice melting (when T > 0°C). o The rusting of iron. o Which one of these is spontaneous? Water gets warmer if a hot object is placed into it. The decomposition of water into H 2 and O 2 at room temperature. The sublimation of dry ice at -100°C. (The freezing point of dry ice is -78°C.

Chemical Thermodynamics Entropy – A measure of the randomness of a system. o ΔS = S final - S initial (The entropy of a system depends only on the initial and final states of the system.) o For an isothermal process: ΔS = q T

Chemical Thermodynamics What is the entropy change of 50.0 g of liquid mercury freezing? The freezing point of Hg = °C and ΔH fusion = 2.29 kJ/mol.

Chemical Thermodynamics Second Law of Thermodynamics – Spontaneous reactions, ones that are irreversible, always lead to an increase in the entropy of the system (ΔS = +). In other words, processes that occur naturally on their own always lead to more ‘disorder’.

Chemical Thermodynamics Entropy and Phase Changes – Predict whether each of the following phase changes produced an increase or decrease in the entropy of the system; o CO 2(s)  CO 2 (g) o CaO (s) + CO 2(g)  CaCO 3(s) o HCl (g) + NH 3(g)  NH 4 Cl (s) o 2SO 2(g) + O 2(g)  2SO 3(g)

Chemical Thermodynamics Third Law of Thermodynamics – The entropy of a pure crystalline solid at absolute zero is zero. At absolute zero, all molecular motion, rotational and translational, would cease.

Chemical Thermodynamics Entropy Changes in Chemical Reactions Standard molar entropies (ΔS°) – The value of the molar entropy of a substance at 298 K and 1 atm. They are experimentally determined. The ΔS° for an element in their free states is not zero. The ΔS° for gases are higher than the ΔS° of liquids or solids of the same element. ΔS° usually increases with increasing molar mass. ΔS° usually increases with increasing number of atoms in a a compound.

Chemical Thermodynamics Entropy Changes in Chemical Reactions ΔS° = Σ nΔS° (products) - Σ mΔS° (reactants) The standard entropy change of a chemical reaction is equal to the difference of the the standard entropy change of the products minus the standard entropy change of the reactants.

Chemical Thermodynamics Using the standard entropies in appendix C of your textbook, calculate the standard enthalpy change, (ΔS°), for the following reaction at 298K; Al 2 O 3(s) + 3H 2(g)  2Al (s) + 3H 2 O (g)

Chemical Thermodynamics Gibbs Free Energy The problem - Some endothermic reactions are spontaneous while others are not. Some exothermic reactions are spontaneous reactions while others are not. The entropy and enthalpy of a chemical Reaction determine its spontaneity. So how can we tell if a reaction will be spontaneous or not?

Chemical Thermodynamics Gibbs Free Energy (ΔG) – The amount of useable energy either released or absorbed in a chemical reaction. ΔG = ΔH – TΔS If ΔG > 0, reaction is not spontaneous If ΔG = 0, reaction is at equilibrium If ΔG < 0, reaction is spontaneous

Chemical Thermodynamics Calculate the standard free energy change for the formation of NO (g) from N 2(g) and O 2(g) at 298K. 2NO (g)  N 2(g) + O 2(g) Given that ΔH° = kJ and ΔS° = 24.7 J/K. Is the reaction spontaneous under these conditions?

Chemical Thermodynamics A particular reaction has ΔH° = 24.6 kJ and ΔS° = 132 J/K at 298 K. Calculate ΔG° to determine the spontaneity of the chemical reaction.

Chemical Thermodynamics A particular reaction has ΔH° = 24.6 kJ and ΔS° = 132 J/K at 298 K. Calculate ΔG° to determine the spontaneity of the chemical reaction.

Chemical Kinetics Rates of chemical reactions o The speed at which a chemical reaction occurs is called its ‘reaction rate’. o The rate of a chemical reaction is measured as the change in concentration of reactant or products per unit of time. Rate = Δ Concentration / Time = Molarity / Time = M / s ** In chemistry [ ] means the molar concentration **

Chemical Kinetics Rates of chemical reactions o The decomposition of N 2 O 5 proceeds according to the following reaction: 2N 2 O 5(g)  4NO 2(g) + O 2(g) If the rate of decomposition of N 2 O 5 at a particular instant in a reaction vessel is 4.2 x M/s, what is the rate of appearance of (a) NO 2 and (b) O 2 ?

Chemical Kinetics Rates Laws of Chemical Reactions aA + bB  cC + dD Rate = k [A] m [B] n k = rate constant [A] = initial molar concentration of A [B] = initial molar concentration of B m = reaction order exponent n = reaction order exponent

Chemical Kinetics Rates Laws of Chemical Reactions NH 4 + (aq) + NO 2 - (aq)  N 2(g) + 2H 2 O (l) Rate = k [NH 4 + ] [NO 2 - ] Since NH 4 + and NO 2 - affect the rate proportionally, we do not need to raise to concentration of the ions to any power. The order of both reactants is 1. TrialInitial [NH 4 + ]Initial [NO 2 - ]Rate (M/s) x x x x x x 10 -7