CM4106 Chemical Equilibria & ThermodynamicsLesson 1Introduction to Chemical EquilibriaA Chemistry Education Blog by Mr Tanhttp://chemistry-mr-tan-yong-yao.blogspot.sg/
Equilibrium Constant, K- describes the extent to which reaction proceeds (position of chemical equilibrium) aA + bB ⇌ cC + dD [C]c[D]d (pC)c(pD)dKc = Kp = [A]a[B]b (pA)a(pB)b equilibrium equilibrium concentrations pressures
(I) Writing K expressions1. Write K as for reaction proceeding from left to right2. Do not include solids and liquids in K expressions3. Be careful not to leave out reacting ratios
(II) Manipulating KKp = Kc (RT) n n = (moles of gaseous product) - (moles of gaseous reactant)Examples:SO2(g) + Cl2(g) ⇌ SO2Cl2(g) n = 1 – 2 = -1CaCO3(s) ⇌ CaO(s) + CO2(g): n=1–0=1
(II) Manipulating K1. The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. [NO2]2N2O4(g) ⇌ 2 NO2 (g) Kc = = 0.212 at 100 C [N2O4] [N2O4]2 NO2 (g) ⇌ N2O4(g) Kc’ = 2 = 4.72 at 100 C [NO2] 1 Kc = Kc’
(II) Manipulating K2. The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. [NO2]2N2O4(g) ⇌ 2 NO2 (g) Kc = = 0.212 at 100 C [N2O4] [NO2]42 N2O4(g) ⇌ 4 NO2 (g) Kc = [N O ]2 = (0.212)2 at 100 C 2 4 Kc’ = (Kc)n
(II) Manipulating K3. The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps. [B]2 2 A (g) ⇌ 2 B (g) Kc’ = [A]2 = at 100 C 2 B(g) ⇌ 3 C(g) [C]3 Kc’’ = = at 100 C [B]2 [C]3 = Kc’ KC’’ 2 A(g) ⇌ 3 C(g) Kc = x [A]2 [B]2 [C]3 = x [A]2 [B]2 = at 100 C
(III) Equilibrium CalculationsSulfur trioxide decomposes at high temperature in a sealed container: 2SO3(g) ⇌ 2SO2(g) + O2(g)Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm.At equilibrium, the SO3 partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K. Step 1: Find I-C-E 2SO3(g) ⇌ 2 SO2(g) + O2(g) Initial (atm) 0.500 atm 0 0 Change (atm) -0.300 atm +0.300 atm +0.150 atm Equilibrium (atm) 0.200 atm 0.300 atm 0.150 atm Step 2: Substitute equilibrium values into Kp (pSO2)2(pO2)1 Kp = (pSO3)2 (0.300)2(0.150)1 Kp = (0.200)2 Kp = 0.338 Page 6
(III) Equilibrium Calculations For the equilibrium PCl5(g) ⇌ PCl3(g) + Cl2(g), the equilibrium constant Kp has the numerical value 0.497 at 500 K. A gas cylinder at 500 K is charged with PCl5(g) at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl5, PCl3 and Cl2 at this temperature? PCl5(g) ⇌ PCl3(g) + Cl2(g) Initial (atm) 1.66 0 0 Change (atm) -y +y +y Equilibrium (atm) 1.66 - y y y [PPCl3] [PCl2]Kp = = 0.497 [PPCl5] (y)(y) P (Cl2) = 0.693 atm0.497 = (1.66 – y ) Solve for y, P (PCl3) = 0.693 atm P (PCl5) = 0.967 atm y = 0.693 or -1.19 (rejected)
(III) Equilibrium Calculations At a certain temperature a 2.00 L flask initially contained 0.298 mol PCl3(g) and 8.70 x 10-3 mol PCl5(g). After the system had reached equilibrium, 2.00 x 10-3 mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes according to the reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g) Calculate the equilibrium concentrations of all species and the value of Kc. PCl5(g) ⇌ PCl3(g) + Cl2(g) Initial [ ] 4.35 x 10-3 M 0.149 M 0 Change [ ] - 1.00 x 10-3 M + 1.00 x 10-3 M + 1.00 x 10-3 M Equilibrium  3.35 x 10-3 M 0.150 M 1.00 x 10-3 M [PCl3] [Cl2] Take note of volume of system especiallyKc = for Kc calculation [PCl5] [1.00 x 10-3] [0.150]Kc = Kc = 0.0449 (to 3.s.f.) [3.34 x 10-3]
(III) Equilibrium Calculations Q<K Q=K Q>K Q<K Q>K The [product] is The [product] too small and is too large [reactant] is System is at and [reactant] too large. equilibrium is too small. there is no net movement Reaction will Reaction will proceed from proceed from left to right right to left forming more forming more products. reactants Page 9
(III) Equilibrium CalculationsAt 1000 K the value of Kp for the reaction 2SO3(g) ⇌ 2SO2(g) + O2(g)is 0.338.Predict the direction in which the reaction will proceed towardequilibrium if the initial partial pressures are PSO3 = 0.16 atm,PSO2 = 0.41 atm and PO2 = 2.5 atm.Compare Q and Kp to predict direction reaction will proceedtowards eqm (PSO2)2(PO2)Q= (PSO3)2 (0.41)2(2.5) Since Q > K,Q= (0.16)2 Reaction will proceed from right to left to achieve equilibrium by forming Q = 16.4 > Kp more reactants
(IV) Le Châtelier’s Principle Factor Rate of Rate Position of Equilibrium constant, reaction constant, k equilibrium Kc (or Kp)Increase in reactant No change Shifts to reduce No effectconcentration ↑ reactant conc.Increase in pressure No change Shifts in direction No effect(by decreasing ↑ w ↓ no. of molesvolume)Increase in Shifts in direction endothermic reaction ↑ ↑temperature of endothermic rxn ↑ exothermic reaction ↓Adding a catalyst No change No effect ↑ ↑ Catalysts increase rate of forward and reverse reaction equally.
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