Chemical Equilibria 1 PDF

MED-102 General Chemistry Chemical Equilibrium 1  LOBs covered Describe the concept of dynamic equilibrium Write equilibrium expressions Perform calculations of Kc from Kp and vice versa Predict in which direction a chemical system will shift to establish equi...

MED-102 General Chemistry Chemical Equilibrium 1  LOBs covered Describe the concept of dynamic equilibrium Write equilibrium expressions Perform calculations of Kc from Kp and vice versa Predict in which direction a chemical system will shift to establish equilibrium  What is chemical equilibrium? When we studied chemical kinetics, we answered the question “how fast?” Chemical equilibrium answers the question “how far?” – How far to completion does the reaction proceed? – How much product is formed? Chemical equilibrium – state where the concentrations of reactants and products do not appear to change The mixture of reactants and products at equilibrium is called the equilibrium mixture The concept of equilibrium is applicable only to reversible reactions. Many chemical reactions are reversible.  Dynamic equilibrium As a reaction proceeds forward to form products, the reverse reaction also begins forming reactants. At some point, the rates of the forward and the reverse reactions will become equal. At this point, the reaction appears to stop. WATCH: https://www.youtube.com/watch?v=wlD_ImYQAgQ  Dynamic Equilibrium In order for dynamic equilibrium to be established, the process must be reversible (have both forward and reverse reactions), and the system must be closed, i.e., no escape of matter is possible. At dynamic equilibrium, both the forward and the reverse reactions are taking place, but their rates are equal, so we see no change in the amount of reactants or products  Dynamic equilibrium When the top tap and the bottom tap are both open, then water is entering at the same rate as it is exiting. Under such conditions, the level of the water in the tank will not change at all.  The Equilibrium State Many reactions do not go to completion due to competing reverse reactions Let us look at an example: Equilibrium can be achieved from either end – We can start exclusively with either substance, N2O4, or NO2. The same equilibrium mixture will result at a given temperature. Forward and reverse reactions still go on, but their rates become equal, so we see no net change in the equilibrium mixture  The Equilibrium State – Revision Slide N2O4 is colorless while NO2 has a brown color. At 273 K, the equilibrium lies heavily to the left and the container looks colorless. At 373 K, the equilibrium lies heavily to the right, and the container looks dark brown. At temperatures between 273 K and 373 K, we get a different shade of brown color. If we set the temperature in the middle (323 K) and begin by placing only N2O4 in the container, after a while, the color will start turning brown, as NO2 is formed. When equilibrium is reached, the shade of brown will stop changing. If we set the temperature in the middle (323 K) and begin by placing NO2 in the container, after a while we will see the shade of brown becoming lighter as colorless N2O4 is formed, until equilibrium is reached, at which time the shade of brown will stop changing.  The Equilibrium Constant Kc Consider the general chemical reaction 𝑎A + 𝑏B 𝑐C + 𝑑D On the basis of experimental results, we can write the equilibrium equation [C]c [D]d Kc = [A]a [B]b where Kc is called the equilibrium constant The subscript ‘c’ refers to concentrations [ ] (molarities)  Equilibrium Constant Kc Looking at the equilibrium We find experimentally that [NO2 ]2 Kc = = 4.64 10−3 at 25 C [N2O4 ] This result is valid at the stated temperature This value means that the [N2O4] is much bigger than the [NO2]. We say in this case, that the equilibrium lies to the left.  The Equilibrium Constant Kp When gases are involved, it is easier to measure their partial pressures than their molarities ( PNO2 ) 2 Kp = ( PN 2O4 ) For an all-gaseous system, it is then preferable to define Kp, the equilibrium constant in terms of the partial pressures of the gases involved.  Relating Kc to Kp Using the ideal gas law we define the partial pressures of all gases involved: n K p = K c ( RT ) Δn is the change in the number of moles of gas ‘products – reactants’ For reasons beyond the scope of this course, equilibrium constants do not have units.  Examples Kp = Kc ( RT ) H2 (𝑔) + I2 (𝑔) 2 HI(𝑔) K p = Kc K p = K c ( RT ) 5-Minute Break  Heterogeneous Equilibria In homogeneous equilibria, reactants and products are all in the same phase (e.g. all gases or all (aq) solutions) In heterogeneous equilibria, reactants and products are in different phases: CaCO3(s) CaO(s) + CO2(g) For the system above, we have two solids and a gas. It is therefore a heterogeneous equilibrium.  Heterogeneous Equilibria CaCO3(s) CaO(s) + CO2(g) [CaO][CO2 ] Kc = [CaCO3 ] Pure liquids and pure solids should not be included in the equilibrium expression – their concentration is shown below to be constant throughout a chemical reaction n  m 1 m 1 1 Concentration = Molarity = =   = = = constant V  MM  V V MM MM K c = [CO 2 ]  Using the Equilibrium Expression 2 H2(g) + O2(g) 2 H2O(g) [H2O]2 Kc = 2 = 2.4  10 47 at 500 C [H 2 ] [O2 ] This value indicates that the numerator (top) is much bigger than the denominator (bottom), which means that the equilibrium lies heavily to the right side 2 H2O(g) 2 H2(g) + O2(g) [H 2 ]2 [O2 ] −48 Kc = 2 = 4.2  10 at 500 C [H 2O] This value indicates that the fraction Kc is very small. Therefore, the bottom of the fraction is much bigger than the top. The equilibrium lies heavily to the left side  Using the Equilibrium Expression Predicting the direction of reaction H2(g) + I2(g) 2 HI(g) Kc = 57.0 at 700 K Let us now assume that we have a mixture of H2(g), I2(g), and HI(g) at 700 K with the following initial concentrations: [H 2 ]0 = 0.10 M, [I 2 ]0 = 0.20 M, [HI]0 = 0.40 M Using these concentrations, we can write the reaction quotient, Qc: [HI]2 (0.40)2 Qc = = = 8.0 [H 2 ][I2 ] (0.10)(0.20)  Using the Equilibrium Expression Predicting the direction of reaction H2(g) + I2(g) 2 HI(g) Kc = 57.0 at 700 K The reaction quotient, Qc is simply Kc with a set of initial concentrations [HI]2 (0.40)2 Qc = = = 8.0 [H 2 ][I2 ] (0.10)(0.20) If Qc < Kc the reaction will move to the right to establish equilibrium If Qc = Kc the reaction is at equilibrium If Qc > Kc the reaction will move to the left to establish equilibrium WATCH: https://www.youtube.com/watch?v=2PM1yc_z4Bk  Summary for Revision Chemical equilibrium is the branch of chemistry dealing with reversible reactions, where both the forward and reverse reactions take place. Dynamic equilibrium is achieved when the rate of the forward reaction is equal to the rate of the reverse reaction. At dynamic equilibrium, there is no change in the amounts of reactants or products. Dynamic equilibrium can be reached from either end. One can start with only reactants, or only products, but the same equilibrium mixture will be produced. The equilibrium mixture composition has a dependence on the temperature of the reaction. We can define the equilibrium state by using equilibrium constants Kc or Kp. Kc uses molarities whereas Kp uses partial pressures of gases. Kc and Kp are related mathematically. In writing Kc expressions, we include only concentrations of gases or aqueous solutions. In writing Kp expressions, we include only partial pressures of gases. We do not include pure solids or pure liquids in Kc or Kp expressions. Equilibrium constants do not have units. Temperature is the only variable that affects their values. Kc and Kp are defined with equilibrium concentrations and partial pressures, respectively. The reaction quotient Q (Qc or Qp), is defined with instantaneous concentrations or partial pressures. Its value must be compared with the value of Kc or Kp, and this will guide us in deciding in which direction the reaction must move to establish equilibrium. The value of Kc or Kp tells us whether the equilibrium mixture contains mostly reactant or mostly product.