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## Chemical Equilibria

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**Chemical Equilibria**Chapter 16**Chemical reactions**• Can reverse (most of the time) • Though might require a good deal of energy • or double arrows = reversible rxn • It “swings both ways” • Once fwd/rev rxns occur at equal rates • = equilibrium (no net change seen)**Equilibrium constant**• aA + bB cC + dD • Thus,**Give the equilibrium constant for:**H2(g) + I2(s) 2HI(g) Initially, 0.0175 M of reactants Decrease to 0.0037 M reactants What is the concentration of HI formed? Equilibrium constant in action**ICE Table**• Initial Change Equilibrium Table • Great way to explain and show concentration changes • Let’s make an ICE table for the previous reaction on the board**Problems**• Express Keq for: CH3OH(g) CO(g) + 2H2(g) • Express Keq for: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) • Solve Keq for the following: A(g) 2B(g) • Given [A]i = 1.00M, [B]i = 0.00M, and [A]eq =0.75M**Caveats to Keq: solids**• Solids in reversible rxns are excluded from expression since concentration derived from constant densities: • S(s) + O2(g) SO2(g) • Keq = ?**Caveats to Keq: aqueous solns**• Same rule for pure liquids • Ex: NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) • Keq = ?**Caveats to Keq: gases & Kp**• For gases we can use partial pressures • Why? • Hint: Relationship between pressure and concentration • See ideal gas law • Given: aA + bB cC + dD • Kp not necessarily equal to Keq**Problem**• Given 2NO(g) + O2(g) 2NO2(g) • Kp = 2.2 x 1012 @ 25°C • Find Keq**Determining equilibrium constant**• Given 2SO2(g) + O2(g) 2SO3(g) • And: [SO2]i = 1.00 M, [O2]i = 1.00 M & [SO3]f = 0.925 M • What is Keq?**More problem solving**• Sulfuryl chloride (SO2Cl2) dissociates into sulfur dioxide and chlorine in the gas phase: SO2Cl2(g) SO2(g) + Cl2(g) • In an experiment, 3.174 g of SO2Cl2 (MW = 134.96g/mol) is placed in a 1.000 L flask and is at 100.0°C. At equilibrium, the total pressure in the flask is 1.30 atm. Calculate: • a) The partial pressures of each gas at equilibrium. • b) The Kp at 100.0°C for the reaction.**Yet another**• Isopropanol can dissociate into acetone and hydrogen: (CH3)2CHOH(g) (CH3)2CO(g) + H2(g) • At 179°C, the equilibrium constant is 0.444. Calculate the equilibrium partial pressures of all three gases if 10.00 g (MW = 60.10g/mol) of isopropanol are initially placed in a 10.00 L vessel.**Another but with a twist**• At 25°C, the equilibrium constant for the reaction below is 5.9 x 10-13. 2NO2(g) 2NO(g) + O2(g) • Suppose a container is filled with 0.89 atm of NO2 • Calculate the equilibrium partial pressures of each gas • OK to approximate if “x [A]0 < 5%“ • “5% rule”**More on Keq**• Is the rxn product or reactant favored? • I.e., Will it form more product or reactant? • If K 1, then prod concentration higher than reactant concentration prod favored • Makes mathematical sense • See right • If K < 1, then reactant concentration higher than prod concentration reactant favored • If K = 1, neither favored; both equal concentration**Reaction quotient, Q**• When rxn not at equilibrium, use Q • Where aA + bB cC + dD • Then • Remember, Q used for system when system NOT in equilibrium**Q – its benefits**• Answers the question: Is the system at equilibrium (i.e., does Q = K)? • If not, we can predict which way the reaction will continue to proceed • If Q < K, rxn still needs to go to prod side to achieve equilibrium (i.e., where Q = K) • In other words, insufficient product formed for equilibrium conditions • If Q > K, rxn has “overshot” K and needs to go to reactant side to achieve equilibrium (i.e., where Q = K) • In other words, an excess of product is formed for equilibrium conditions**Work on this…**• N2(g) + 3H2(g) 2NH3(g) • The equilibrium constant at 400°C is Keq=0.5. • Suppose we make a mixture with the following concentrations: • [NH3] = 5.0M, [N2] = 3.5M, [H2] = 1.9M • In which direction will the reaction go? • a) products b) reactants**Manipulating Keq**• C(s) + ½ O2(g) CO(g) • Kp’ = ? • Changing it: 2C(s) + O2(g) 2CO(g) • Kp” = ? • How are Kp’ and Kp” related mathematically?**More on Keq**• Given 2NO2(g) N2O4(g) • Keq = ? • What is Keq when rxn is reversed? • Therefore, what can we say about Keq fwd and Keq rev?**Even more on Keq**• Remember Hess’s Law? • 1) A+B C • 2) B+C D • What is the net rxn and it’s Knet using Hess’s Law? • Can one obtain the same values as above not using Hess’s Law? • What can we say about Knet using each rxn’s Keq?**Disturbing chemical equilibria**• Le Châtelier’s Principle • Change one component of the rxn & the rxn will attempt to rectify it • Think of it this way: • If something is changed, how can it be undone or controlled so that equilibrium is achieved once again?**Temperature variation on equilibrium**• 2NO2(g) N2O4(g) + heat • Kp = ? • H° = -57.1 kJ • @ 273 K, Kp = 1300, and @ 298 K, Kp = 170 • Hence, if one raises the temp to 298 K, which way will it swing?**Pressure & volume change on equilibrium**• If volume decreased (pressure increased) favors smaller # of molecules • If volume increased (pressure decreased) favors larger # of molecules • If reversible rxn has = # of molecules on each side, a volume/pressure change will do nothing**Question**• What change in equilibrium will be seen when one adds a solid and why?**How about this?**• 2H2S(g) + O2(g) 2S(s) + 2H2O(g); ΔH = -221.19 kJ/mol • If O2(g) is added to the reaction vessel, what happens to the amount of S(s)? • a) It increases b) It decreases c) Nothing • If the volume of the vessel is cut in half, what happens to the ratio of PH2O/PH2S? • a) It increases b) It decreases c) Nothing • If the temperature is increased, what happens to the equilibrium constant K? • a) It increases b) It decreases c) Nothing • If S(s) is added to the reaction, what happens to PH2O? • a) It increases b) It decreases c) Nothing