aA + bB  dD + eE that there is an equilibrium constant Calculations With Equilibrium Constants In the last lecture we showed that for a reaction aA + bB  dD + eE that there is an equilibrium constant K 𝑇 = 𝐷 𝑑 𝐸 𝑒 𝐴 𝑎 𝐵 𝑏 And for the reverse reaction dD + eE  aA + bB 𝐾 ′ 𝑇 = 1 𝐾

Calculations With Equilibrium Constants We also showed that for two reactions aA + bB  dD + eE K1 fF + gG  hH + jJ K2 The equilibrium constant for the sum of these reactions aA+bB+fF+gG  dD+eE+hH+jJ K=K1K2

Concentration = 𝑛 𝑉 = 𝑃 𝑅𝑇 Calculations With Equilibrium Constants And finally since concentration can be related to pressure using the ideal gas law Concentration = 𝑛 𝑉 = 𝑃 𝑅𝑇 we showed that Kp=Kc(RT)Dn where n is the difference between the number of moles of gas produced and consumed in a reaction and 𝐾 𝑝 = 𝑃 𝑐 𝑃 𝑑 𝑃 𝑎 𝑃 𝑏 for the gas phase reaction aA +bB = cC +dD

Rules for Equilibrium Constant Problems We learned how to calculate the equilibrium constants given equilibrium concentrations of the products and reactants. Now we want to learn how to calculate equilibrium concentrations given the equilibrium constant and starting concentrations of reactants and products.

Rules for Equilibrium Constant Problems To organize the problem we use a chart with the initial concentrations, the changes needed to reach equilibrium and the final equilibrium concentrations   Initial Change Final

Rules for Equilibrium Constant Problems We write the BALANCED chemical reaction across the top of the chart and the concentrations of reactants and products in the first row aA + bB = cC + dD   [A] [B] [C] [D] Initial Change Final We then fill in the chart with the given information, write out the equilibrium constant expression, formulate the necessary algebraic equations and solve them

Rules for Equilibrium Constant Problems Write the balanced chemical equation Write the equilibrium constant equation Fill out the table using the stoichiometric relations from the balanced equation Use the information in the table to evaluate the equilibrium constant. Solve any remaining algebraic equations.

An Example For H2(g) + I2(g)  2HI(g) What is the equilibrium constant in terms of the concentrations of the product and reactants? K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔

Example (cont.) In an experiment at 490 oC the following concentrations were measured at equilibrium   H2 I2 HI Initial Change Final 0.000862 M 0.00263 M 0.0102 M What is the equilibrium constant?

Example (cont.) What is the equilibrium constant? Substitute into   H2 I2 HI Initial Change Final 0.000862 M 0.00263 M 0.0102 M Substitute into K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔 = ( 0.0102 𝑀) 2 (0.000862 𝑀 (0.00263 𝑀 =45.9

Example (cont.) We can now use the equilibrium constant to describe any equilibrium mixture of these three molecules at 490 C   1.00 moles of H2 and 1.00 mole of I2 are introduced into a 2.00 L box at 490 C. What is the final concentration of all the molecules at equilibrium First we calculate the concentration of all of the reactants and products at the start 𝐻 2 = 𝐼 2 = 1.00 𝑚𝑜𝑙 2.00 𝐿 =0.500 𝑀 𝐻𝐼 =0 And substitute into the ICE table

Example (cont.)   H2 I2 HI Initial  0.500 0.500 0.00 Change Final The change in [H2] and [I2] will be the same, -x, and from the stoichiometry the change in [HI] will be +2x H2(g) + I2(g)  2HI(g)   H2 I2 HI Initial  0.500 0.500 0.00 Change  -x -x +2x Final We can add these to get the algebraic amounts at equilibrium

Example (cont.)   H2 I2 HI Initial  0.500 0.500 0.00 Change  -x -x +2x Final 0.500-x And substitute these expressions into the equilibrium constant expression K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔 = (2𝑥) 2 0.500−𝑥 0.500−𝑥 =45.9 K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔 = (2𝑥) 2 0.500−x 2 =45.9

Example (cont.) (2𝑥) 0.500−x = 45.9 =6.77 2𝑥= 0.500−x 6.77 K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔 = (2𝑥) 2 0.500−𝑥 0.500−𝑥 =45.9 We can simplify this expression K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔 = (2𝑥) 2 0.500−x 2 =45.9 taking the square root of both sides (2𝑥) 0.500−x = 45.9 =6.77 Then multiplying by (0.500-x) to get 2𝑥= 0.500−x 6.77

Example (cont.) 2𝑥= 0.500−x 6.77 8.77𝑥=3.39 𝑥=0.387 Which can be solved for x 8.77𝑥=3.39 𝑥=0.387 So the equilibrium concentrations are 𝐻 2 = 𝐼 2 = 0.500−.387 =0.113𝑀 𝐻𝐼 =2𝑥=0.773 𝑀 To check use these concentrations to calculate the equilibrium constant K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔 = [ 0.773 𝑀 ] 2 (0.113 𝑀 (0.113 𝑀 =46.4 ~ 45.9

Example (cont.) K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔 =45.9 It doesn’t matter whether we start with hydrogen and iodine, or with HI. If two moles of HI are injected into a 1 L box at 490 C what will be the concentration of each species in the box at equilibrium K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔 =45.9   H2 I2 HI Initial Change Final

Example (cont.) K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔 =45.9 Or even if we start with a mixture of all three molecules. One mole of H2, two moles of I2 and three moles of HI are injected into a 1 L box. What will be the concentration of each species at equilibrium at 490 C? K= 𝐻𝐼 𝑔 2 𝐻 2 𝑔 𝐼 2 𝑔 =45.9   H2 I2 HI Initial Change Final

Yet Another Example A 2.500 mol sample of phosphorus pentachloride, PCl5 dissociates at 160 C and 1.00 atm to give 0.338 mol of phosphorus trichloride at equilibrium. What is the composition of the final mixture? What is the equilibrium constant?   H2 I2 HI Initial Change Final

And Yet Another Example Methanol is manufactured commercially by CO(g) + 2H2(g)  CH3OH(g) A 1.500 L vessel was filled with 0.1500 mol CO and 0.3000 mol H2. When this mixture came to equilibrium at 500K the vessel contained 0.1187 mol CO. How many moles of each substance were in the vessel? What is the equilibrium constant?   Initial Change Final