SCH4U – Grade 12 Chemistry, University Preparation 7.3 The Equilibrium Constant Opposing Rates Law of Chemical Equilibrium Equilibrium Constant Equilibrium Constant & Temperature (pp.334-338) SCH4U – Grade 12 Chemistry, University Preparation Ms. Papaiconomou & Ms. Lorenowicz

We learned in previous lessons that… Reactions occur at a specific rate and can be influenced by different factors Reactions are reversible Some reactions are spontaneous/favourable This depends on enthalpy, temperature and entropy We can predict if a reaction is spontaneous using Gibb’s Free Energy (ΔG = ΔH - TΔS) ΔG > 0 reverse reaction is favourable ΔG < 0 forward reaction is favourable ΔG = 0 reaction is at equilibrium Equilibrium occurs when opposing changes occur simultaneously at the same rate

Lets evaluate the reversible reaction of N2O4(g) 2 NO2(g)

Lets evaluate the reversible reaction of N2O4(g) 2 NO2(g)

Equilibrium is reached from the forward and reverse direction N2O4(g) 2 NO2(g)

Law of Chemical Equilibrium At equilibrium, there is a constant ratio between the concentration of the products and reactants in any change.

The Equilibrium Constant Forward Reaction: N2O4(g) → 2 NO2(g) Forward Rate: kf[N2O4] Reverse Reaction 2 NO2(g) → N2O4(g) Reverse Rate: kr[NO2]2 At equilibrium: Forward Rate = Reverse Rate kf[N2O4] = kr[NO2]2

Keq = the equilibrium constant At equilibrium, there is a constant ratio between the concentration of the products and reactants in any change. You calculate the equilibrium constant by dividing the forward rate constant by the reverse rate constant kf = Keq kr So: kf = [NO2]2 = Keq kr [N2O4]

Equilibrium Constant for any equilibrium equation The equilibrium equation is usually expressed in terms of molar concentrations. So the equilibrium constant is both Keq and Kc For the chemical equation: aP + bQ cR+ dS so: Kc = [R]c[S]d [P]a[Q]b The equilibrium expression depends on the stoichiometry of the reaction and the concentration of the products & reactants. Usually, units are not included when using or calculating the value of Kc. Products(s) Reactant(s)

Sample Problem QUESTION: Write the equilibrium expression for the following reaction: N2(g) + O2(g) 2NO(g) ANSWER: Kc = [NO]2 [N2][O2]

Effect of temperature on the equilibrium constant The value of the equilibrium constant depends only on temperature. Changing the temperature changes the rates of the forward & reverse reactions differently because the forward & reverse reactions have different activations energies. The starting concentrations have no effect on the equilibrium constant.

Sample Problem QUESTION: A solution of phosphorus pentachloride gas was kept at 250 K in a 2.5L reaction flask. PCl5(g) PCl3(g) + Cl2(g) When the equilibrium mixture was analyzed, it was found to contain 0.025 mol of PCl5(g), 0.05 mol of PCl3(g) and 0.01 mol of Cl2(g). Calculate the equilibrium constant for this reaction.

Sample Problem - Analysis ANALYSIS OF QUESTION: You want to calculate the value of Kc  Write out the equilibrium expression You were given the volume of the container and the amount of the chemicals  Calculate concentration by taking the amounts (mol) and divide by the volume (L) to get the molar concentrations (mol/L)

Sample Problem - Solution PCl5(g) PCl3(g) + Cl2(g) [PCl5(g)] = 0.025 mol / 2.5 L = 0.01 mol/L [PCl3(g)] = 0.05 mol / 2.5 L = 0.02 mol/L [Cl2(g)] = 0.01 mol / 2.5 L = 0.004 mol/L Kc = [PCl3]1[Cl2]1 = [0.02][0.004] = 0.008 [PCl5]1 [0.01] So…The equilibrium constant for this reaction at 250 K is 0.008.

Homework Please re-read Section 7.3 (pp.334-338) and answer the following Questions: p.336 Q.1-5 p.338 Q.6-10 p.353 Q.1 Look ahead… Read Section 7.3 (pp.339-353) Check the website for homework/downloadable files/interesting youtube chemistry videos: http://papaiconomou.weebly.com