<p> Equilibrium</p><p>For most chemical reactions, as soon as the forward reaction begins to produce product the reverse reaction begin to break the product apart to reform reactants.</p><p>+1 -1 HC2H3O2(aq) + H2O(l)  H3O (aq) + C2H3O2 (aq)</p><p>+1 -1 H3O (aq) + C2H3O2 (aq)  HC2H3O2(aq) + H2O(l)</p><p>When the rate of the forward reaction is equal to the rate of the reverse reaction in a closed system, the system is said to be in a state of dynamic equilibrium. A closed system is one where nothing is allowed to enter or leave the system.</p><p>H2O(l)   H2O(g)</p><p>The volume of water inside a closed bottle is constant because the rate of evaporation is exactly the same as the rate of condensation.</p><p>As a reaction proceeds the concentration of the reactants decreases and the concentration of the products increases. Equilibrium is reached when the concentrations of both reactant and product remains constant. THE CONCENTRATIONS DO NOT NEED TO BE THE SAME – JUST CONSTANT.</p><p>The ratio of products to reactants will always be the same under the same experimental conditions. Exactly where and when equilibrium will be reached can be calculated by solving for Keq.</p><p>Keq = [products] [reactants]</p><p>If Keq = 1, when equilibrium is reached, there will be approximately equal amounts of reactants and products.</p><p>If Keq > 1, when equilibrium is reached, there will be more products than reactants.</p><p>If Keq < 1, when equilibrium is reached, there will be more reactants than products.</p><p>A(g) + 2 B(g)   3 C(g) + 4 D(g)</p><p>3 4 Keq = [C(g) ] [D(g) ] - each of the concentrations is raised to the same 2 [A(g)] [B(g)] number that appears as a coefficient in the balanced equation</p><p>Rules …</p><p>1) include all gases and aqueous materials 2) never include solids 3) never include single liquids (only 1 liquid is found in the equation) 4) include liquids when there are 2 or more within the equation Calculations</p><p> calculating Keq from equilibrium concentrations</p><p>At a certain temperature, a mixture of H2 and I2 was prepared by placing 0.100 mole of H2 and 0.100 mole of I2 into a 1.00 L flask. After a period of time equilibrium was established and it was found that the I2 concentration had -1 dropped to 0.020 mole L . Calculate Keq.</p><p>H2(g) + I2(g)   2 HI(g)</p><p>H2(g) + I2(g)   2 HI(g) i 0.100 0.100 0.00  -x -x + 2x 0.100 - x 0.100 – x = 0.0200 2x</p><p> e x = 0.080 0.0200 0.0200 0.160</p><p>2 Keq = [HI(g) ] </p><p>[H2(g)] [I2(g)] </p><p>= [0.160] 2 [0.0200] [0.0200] </p><p>= 64</p><p> calculating equilibrium concentrations using Keq</p><p>CH4(g) + H2O(g)   CO(g) + 3 H2(g)</p><p>At 1500 oC, an equilibrium mixture of these gases were found to contain the following concentrations; 0.300 M CO(g), 0.800 M H2(g) and 0.400 M CH4(g). If Keq </p><p>= 5.67, calculate the equilibrium concentration of H2O(l).</p><p>3 Keq = [CO(g)] [H2(g) ] </p><p>[CH4(g)] [H2O(g)] 5.67 = [0.300] [0.800] 3</p><p>[0.400] [H2O(g)] 3 [H2O(g)] = [0.300] [0.800] [0.400] (5.67)</p><p>[H2O(g)] = 0.0678 M</p><p> calculating equilibrium concentrations using Keq and initial concentrations</p><p>CO(g) + H2O(g)   CO2(g) + H2(g)</p><p>If Keq = 4.06 and 0.100 mole of CO is mixed with 0.100 mole of H2O in a 1.0 L flask, calculate all of the equilibrium concentrations.</p><p>CO(g) + H2O(g)   CO2(g) + H2(g) i 0.100 0.100 0.000 0.000  - x - x + x + x e 0.100 - x 0.100 – x x x</p><p>Keq = [COg)] [H2(g)]</p><p>[CO(g)] [H2Og)] 4.06 = [x] [x] [0.100 - x] [0.100 - x] 4.06 = [x] 2 [0.100 - x]2 2.01 = [x] [0.100 - x] 2.01[0.100 - x] = x 0.201 – 2.01x = x 0.201 = 3.01 x x = 0.201 3.01 x = 0.0668 M</p><p>0.100 – x 0.100 – x x x</p><p> e 0.100 – 0.0668 0.100 – 0.0668</p><p>0.0330 0.0330 0.0668 0.0668</p>