Chemical Equilibrium Lecture Notes PDF

Chemical Equilibrium Prof. Dr. Yasser Mostafa Abdallah professor of Physical Chemistry Fall 2024  Equilibrium Is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: the rates of the forward...

Chemical Equilibrium Prof. Dr. Yasser Mostafa Abdallah professor of Physical Chemistry Fall 2024  Equilibrium Is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal. the concentrations of the reactants and products remain constant Physical equilibrium: H2O (l) H2O (g) Chemical equilibrium: N2O4 (g) 2NO2 (g) colorless dark brown  Chemical Equilibrium The reactions considered until now have had reactants react completely to form products. These reactions “went” only in one direction. Some reactions can react in either direction. They are “reversible”. When this occurs some amount of reactant(s) will always remain in the final reaction mixture.  Reversible reactions When the products are formed, they react to produce the starting reactants. e.g. Liquid + heat → vapor Vapor + cooling → liquid 2 NO2 (g) Cooling N2O4 (g) N2O4 (g) heating 2 NO2 (g) ➣ A reversible chemical reaction is one in which the products formed, react to produce the original reactants. Both the forward and reverse reactions occur simultaneously. The forward reaction is called the reaction to the right, and the reverse reaction is called the reaction to the left. A double arrow is used in the equation to indicate that the reaction is reversible.  Consider the hypothetical reaction: A + B → C + D (forward reaction) C + D → A + B (reverse reaction) At equilibrium state: Rate of the forward reaction = Rate of the reverse reaction As the reaction proceeds, the number of A and B molecules available for reaction decreases and the rate of reaction slows down (forward reaction). If the reaction is reversible, the speed of the reverse reaction is zero at first (Backward reaction) and gradually increases as the concentrations of C and D increase. As the number of A and B molecules decreases the forward rate slows down because A and B cannot find one another.  Consider A B If the reaction begins with just A, as the reaction progresses: [A] decreases to a constant concentration, [B] increases from zero to a constant concentration. When [A] and [B] are constant, equilibrium is reached. Rate of loss of A = kf [A]; decreases to a constant, Rate of formation of B = kr [B]; increases from zero to a constant. When k [A] = k [B] equilibrium is reached.  Law of Mass Action (Equilibrium Expression) For a reaction: jA + kB  lC + mD The law of mass action is represented by the Equilibrium Expression: where K is the Equilibrium Constant. l C D m K = j A B k Writing Equilibrium Constant Expressions The concentrations of the reactants species in the condensed phase are expressed in M. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.  Chemical Kinetics and Chemical Equilibrium kf A + 2B ⇋ AB2 ratef = kf [A][B]2 kr rater = kr [AB2] At equilibrium The rate of forward reaction = rate of reversed reaction ratef = rater kf [A][B]2 = kr [AB2] kf [AB2] = Kc = kr [A][B]2 Where Kc is the equilibrium constant Kf , Kr are the velocity constants of the forward and reversed reaction. N2O4 (g) 2NO2 (g) [NO2]2 K= = 4.63 x 10-3 [N2O4] aA+bB cC+dD [C]c[D]d K= [A]a[B]b  CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq) [CH3COO-][H3O+] Kc = [H2O] = constant [CH3COOH][H2O] [CH3COO-][H3O+] Kc = [CH3COOH] Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N2O4 (g) 2NO2 (g) 2 PNO [NO2]2 2 Kc = Kp = [N2O4] PN O 2 4  a A (g) + b B (g) c C (g) + d D (g) Kp = Kc(RT)n n = moles of gaseous products – moles of gaseous reactants Δn = (c + d) – (a + b)  CaCO3 (s) CaO (s) + CO2 (g) Kc = [CO2] Kp = PCO 2 NOTE  Examples One of the environmentally important reactions involved in acid rain production has the following equilibrium expression. From the expression, what would be the balanced chemical reaction? Note: all components are in the gas phase. K = [SO3]/([SO2][O2]1/2) A. SO3(g)  SO2(g) + 2O2(g) B. SO3(g)  SO2(g) + 1/2O2(g) C. SO2(g) + 2O2(g)  SO3(g) D. SO2(g) + 1/2O2(g)  SO3(g)  The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. CO (g) + Cl2 (g) COCl2 (g) [COCl2] 0.14 Kc = = = 216 [CO][Cl2] 0.012 x 0.054 Kp = Kc(RT)n n = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K Kp = 216 x (0.0821 x 347)-1 = 7.7  The equilibrium constant Kp for the reaction 2NO2 (g) 2NO (g) + O2 (g) is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO2 = 0.270 atm? 2 PNO PO2 Kp = 2 PNO2 2 PNO PO = Kp 2 2 2 PNO PO2 = 158 x (0.270)2/(0.400)2 = 72 atm  Consider the following equilibrium at 295 K: NH4HS (s) NH3 (g) + H2S (g) The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction? Kp = PNH PH2S = 0.265 x 0.265 = 0.0702 3 Kp = Kc(RT)n 0.0702 = Kc (0.082 * 295)2 Kc= 0.0702/ 585.16 = 1.19 x10-4 A mixture of SO2, O2 and SO3 is allowed to reach equilibrium at 852 K. The equilibrium concentration areSO2 = 3.61x 10-3 mol/L, O2 = 6.11x10-4, and SO3 = 1.01x10-2 mol/L. Calculate the equilibrium constant (KC) for the reaction, 2SO2(g) + O2(g) 2SO3(g) KC = SO32 / O2 SO22 = (1.01 x 10-2)2/ (3.61x10-3)2 x (6.11x10-4) = 1281.5 One of the primary components in the aroma of rotten eggs is H2S. At a certain temperature, it will decompose via the following reaction. 2H2S(g)  2H2(g) + S2(g) If an equilibrium mixture of the gases contained the following pressures of the components, what would be the value of Kp? PH2S = 1.19 atm; PH2 = 0.25 atm; PS2 = 0.25 atm A. 0.011 B. 91 C. 0.052 D. 0.013 The liquid metal mercury can be obtained from its ore cinnabar via the following reaction: HgS(s) + O2(g)  Hg(l) + SO2(g) Which of the following shows the proper expression for Kc? A. Kc = [Hg][SO2]/[HgS][O2] B. Kc = [SO2]/[O2] C. Kc = [Hg][SO2]/[O2] D. Kc = [O2]/[SO2] At a certain temperature, FeO can react with CO to form Fe and CO2. If the Kp value at that temperature was 0.242, what would you calculate as the pressure of CO2 at equilibrium if a sample of FeO was initially in a container with CO at a pressure of 0.95 atm? FeO(s) + CO(g)  Fe(s) + CO2(g) A. 0.24 atm B. 0.48 atm C. 0.19 atm D. 0.95 atm