Chapter 15 and Chapter 17Chemical Equilibrium and Ksp• Chapter 14 focused on the rate of chemical reactions. The assumption in the calculations was that only the forward reaction occurs.• However, many chemical processes are reversible so backwards reactions also occur.• Chemical equilibrium occurs when the rate of the forward reaction is the same as the reverse reaction, thus there is no net change in the chemical concentrations.Simple Equilibrium• Let’s assume that the there is a reaction between two gases so that “A” converts to “B” and “B” can convert back to “A”, so we have:A  BB  A• The forward reaction rate is: kf[A]• The reverse reaction rate is: kr[B]• Equilibrium conditions are denoted by two arrows and not just a double-headed arrow.A BSimple Equilibrium• Equilibrium occur when the forward and reverse reactions occur at the same rate, so:kf[A] = kr[B]kf/kr= [B]/[A] = a constant• Therefore, the ratios of the rate constants for the forward and reverse reaction gives the ratios of the chemicals at equilibrium.Additional Points on Equilibrium• An equilibrium can be reached from starting on either side of the reaction. (e.g. all chemical A or all chemical B or a mixture). The end equilibrium will still be the same.• The point at which equilibrium is reached will vary with the reaction conditions.– In particular, temperature can dramatically affect the equilibrium.• At equilibrium, the chemicals continue to react and transform into each other, but there is no NET change in the amounts of each chemical present over time.Demonstration #1NO2and N2O4equilibrium• A sealed glass contain has both NO2and N2O4in it. These chemicals can convert between each other.– NO2is a brown colored gas.– N2O4is a clear to gray gas (or a liquid below 21°C).• At higher temperatures, the decomposition of N2O4is relatively fast, so the equilibrium between the two chemicals is formed with more NO2present than N2O4.• At cool temperatures, the breakdown of N2O4is slower, so an equilibrium is established where N2O4is the dominant form.Demonstration #1NO2and N2O4EquilibriumNO2and N2O4will interconvert with both temperature and pressure. Below we see how you can favor one gas over the other by dipping a sealed tube of gas into ice water and then warm water. https://www.youtube.com/watch?v=zVZXq64HSV4Demonstration #2Cobalt Complexes• This is another demonstration of a shifting equilibrium between two chemical forms at different temperatures.Co(H2O)6+ 4ClCoCl42−+ 6H2Opink color blue color• At room temperature, the solution is a purple color since both chemical forms are present.• Warming this solution will shift the equilibrium towards the blue color.• Cooling the solution will shift the equilibrium to the pink form.Demonstration #2Cobalt ComplexesYou can prepare a solution of Co(H2O)6+ 4Cl-by adding CoCl2to water. Below an aqueous solution of CoCl2is prepared and poured into a U-shaped tube. Heat is applied to one side so you can see the formation of CoCl42-.https://www.youtube.com/watch?v=BGfYf8OQzukDemonstration #2Cobalt Complexes• It is important to note in both these demonstrations that no chemicals were added. The shift in the equilibrium was simply caused by affecting the reaction rates by changing the temperature.• Also, both of the reactions were completely reversible. We could restore the initial conditions exactly by reversing the process.– Remember that truly reversible processes are rare. Chemical equilibria in reactions or phase changes are some of the really reversible processes.• Another important point is that the equilibrium constant (explained in a second) varies with temperature.The Equilibrium ConstantLet’s take the generic reaction of:aA + bB cC + dDThe forward reaction rate is: kf[A]a[B]bThe reverse reaction rate is: kr[C]c[D]dAt equilibrium, the forward reaction rate must equal the reverse reaction rate. Therefore we can set the two equations equal to each other.kf[A]a[B]b= kr[C]c[D]dkf/kr= [C]c[D]d/[A]a[B]b= Keqwhere Keqis the equilibrium constant.• This is called the equilibrium-constant expression.• The equilibrium expression is the concentration of the products divided by the concentration of the reactants.– The equilibrium expression depends on the stoichiometry of the reaction equation.• It can also be formulated as the ratio of the forward and reverse reaction rate constants.– This formulation is rarer since it is harder to determine two reaction rate constants compared to chemical concentrations.• This is also sometimes called the “law of mass action”.• Formally, the equilibrium constant has no units.[C]c[D]d[A]a[B]bKeq==productsreactantskfkr=• There are only two valid units for the concentration measures:1) For aqueous solutions, the only valid unit is molar (M)2) For gases, concentration can be expressed in pressure (atmospheres) or molar (M, which is still mol/L)The value of K will be different depending if the gas concentration is expressed as molar or pressure value. You will see a Kpto denote a pressure-based K value. Unless otherwise specified, Keqare molar-based values since it can accommodate both aqueous and gaseous system.[C]c[D]d[A]a[B]bKeq==productsreactantskfkr=• The equilibrium constant is not affected by the initial amount of the chemicals present. • The equilibrium constants are affected by temperature (like in the demonstrations).– Temperature is about the only thing that can change a value of Keq. – Temperature changes the reaction rate constants by the Arrhenius equation, which then change the Keqvalue.[C]c[D]d[A]a[B]bKeq==productsreactantskfkr=• Looking at the numerical value of K indicates whether the reactants or products will be more abundant– When Keqis greater than 1, then the equilibrium favors the products.• In other words, when “products” > “reactants”, then K > 1– When Keqis less than 1, then the equilibrium favors the reactants.• Or, when “reactants” > “products” then K < 1[C]c[D]d[A]a[B]bKeq==productsreactantskfkr=Example #1Hydrofluoric acid is a weak acid that only partly ionizes in water. What is the equilibrium constant for the ionization reaction:HF(aq) H+(aq) + F−(aq)if the concentration of HF is 0.013 M and the concentrations of H+and F−are both 0.00286 M each at equilibrium?Example #1 solutionA balanced chemical equation is always a good starting point for chemistry