Solving Equilibrium Problems 7.5

We then compare the value to the equilibrium constant How do we know if a system has reached equilibrium? The Reaction Quotient Q = [Products]coefficient [Reactants] coefficient Tells you the direction the reaction will go to reach equilibrium Calculated the same as the equilibrium constant, but for a system not at equilibrium We then compare the value to the equilibrium constant

What Q tells us If Q<K If Q>K If Q=K system is at equilibrium Not enough products Shift to right If Q>K Too many products Shift to left If Q=K system is at equilibrium

Example for the reaction 2NOCl(g) 2NO(g) + Cl2(g) K = 1.55 x 10-5 M at 35ºC In an experiment 0.10 mol NOCl, 0.0010 mol NO(g) and 0.00010 mol Cl2 are mixed in 2.0 L flask. Which direction will the reaction proceed to reach equilibrium?

Q= Q= 5x10-9 1.55 x 10-5 M = [Cl2][NO]2 [NOCl ]2 [NOCl] = 0.05 mol/L [NO] = 0.0005 mol/L [Cl2] = 0.00005 mol/L [0.00005 mol/L ][0.0005 mol/L]2 [0.05 mol/L ]2 Q= Q= 5x10-9 Q<K reaction proceeds to the right

Solving Equilibrium Problems Given the starting concentrations and one equilibrium concentration. Use stoichiometry to figure out other concentrations and K. Create a table of initial and final conditions. You may include changes I C E Chart

1. Consider the following reaction at 600ºC 2SO2(g) + O2(g) 2SO3(g) In a certain experiment 2.00 mol of SO2, 1.50 mol of O2 and 3.00 mol of SO3 were placed in a 1.00 L flask. At equilibrium 3.50 mol SO3 were found to be present. Calculate the equilibrium concentrations of O2 and SO2 and K

2SO2(g) + O2(g) 2SO3 (g) I 2.00 mol/L 1.50 mol/L 3.00mol/L C -2x -x +2x E 2.00-2x 1.5-x 3.00+2x [SO3] = 3.00+2x 3.50=3.00 +2x X=0.25 mol/L [SO2]=2.00-2x = 1.50 mol/L [O2] = 1.5-x = 1.25 mol/L

K = [3.5mol/L]2 [1.50mol/L]2 [1.25mol/L] = 4.36 = 4.4

2. Consider the same reaction at 600ºC In a different experiment 0 2. Consider the same reaction at 600ºC In a different experiment 0.500 mol SO2 and 0.350 mol SO3 were placed in a 1.000 L container. When the system reaches equilibrium 0.045 mol of O2 are present. Calculate the final concentrations of SO2 and SO3 and K

2SO2(g) + O2(g) 2SO3 (g) I 0.500 mol/L 0 mol/L 0.350 mol/L C +2x +x -2x E 0.500+x x 0.350-2x [O2] =0.045 mol/L = x [SO3] = 0.350-2x =0.260 mol/L [SO2] =0.500+x =0.590 mol/L

K = [0. 260 mol/L]2 [0.590mol/L ]2 [0.045mol/L ] = 4.32

Calculating equilibrium concentration using K and initial concentrations Write the balanced equation Write the equilibrium law expression for this reaction Let x represent the amount of reactant used up or created. Set up an ICE chart Solve…possibly using the quadratic formula, unless you can simplify if the initial concentration divided by k is greater than 100

Example [H2]i = (15.7g/2.02)/5.00 L = 1.56 mol/L H2(g) + I2(g) 2HI(g) K = 7.1 x 102 at 25ºC Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.9 g of H2 294 g I2 . [H2]i = (15.7g/2.02)/5.00 L = 1.56 mol/L [I2]i = (294g/253.8)/5.00L = 0.2316 mol/L [HI]i = 0

H2(g) I2(g) HI(g) I 1.56 M 0.232 M 0 M C -x -x +2x Let x(mol/L) represent the amount of H2 used up H2(g) I2(g) HI(g) I 1.56 M 0.232 M 0 M C -x -x +2x E 1.56-x 0.232 –x 2x K = [HI]2 = 7.1 x 102 Solve for x x=0.231 [H2] = 1.33mol/L [I2] almost 0 [HI] = 0.464 mol/L [H2] [I2] = [2x]2 [1.56-x] [0.232 –x ]

Practice For the reaction Cl2 + O2 2ClO(g) K = 156 In an experiment 0.100 mol ClO, 1.00 mol O2 and 0.0100 mol Cl2 are mixed in a 4.00 L flask. If the reaction is not at equilibrium, which way will it shift? Calculate the equilibrium concentrations.

Let x(mol/L) represent the amount of H2 used up Cl2(g) O2(g) ClO(g) I 0.0025 M 0.250 M 0.025 M Q=1 which is < Ke; forward reaction results C -x -x +2x E 0.0025-x 0.250 –x 0.025 + 2x

For example For the reaction 2NOCl 2NO +Cl2 K= 1.6 x 10-5 If 1.20 mol NOCl, and 0.45 mol of NO are mixed in a 1 L container What are the equilibrium concentrations

k = [NO]2[Cl2] [NOCl]2 x =1.14 x 10 -4 .

What are the equilibrium concentrations. Practice Problem For the reaction 2ClO(g) Cl2 (g) + O2 (g) K = 6.4 x 10-3 In an experiment 1.00 mol ClO(g), 0.100 mol O2 and 1.00 x 10-2 mol Cl2 are mixed in a 4.00 L container. What are the equilibrium concentrations.

[ClO] [O2 ] [Cl2] 6.4 x 10-3 = [O2 ][Cl2] [ClO]2 I 0.250 0.025 0.0025 Q= .001; reaction proceeds in the forward direction. Let x(M) be the amount of Cl2 made. C -2x +x +x E 0.250 - 2x 0.025+x 0.0025+x

6.4 x 10-3 = [0.025+x ][0.0025+x] [0.25 - 2x ]2 X = .00808M [Cl2] = 0.0105 M [O2] = 0.1058 M [ClO]= 0.0884M

Calculating new concentrations when the equilibrium has been disturbed. Consider the effect of introducing more product after this equilibrium has been established. SO2 (g) + NO2(g) SO3 (g) + NO(g) K = [NO][SO3] [NO2 ][SO2]

Analysis of the equilibrium mixture show the following: [SO2] = 4.0 M [SO3] = 3.0 M [NO2] = 0.50 M [NO] = 2.0 M What is the new equilibrium concentration for NO when 1.5 mol of NO2 is added?

K = [NO][SO3] [NO2 ][SO2] find K K = [2.0 M][3.0 M] [0.50 M ][4.0 M] = 3.0 Let x(M) be the amount of NO2 used up in the forward rxn [NO2 ] [SO2] [NO ] [SO3] I 0.50 +1.5 4.0 2.0 3.0 C -x -x +x +x E 1.5 –x 4.0-x 2.0+x 3.0+x

3.0 = [2.0+x ][ 3.0+x] [1.5 –x ][4.0-x] x = 0.59 M [SO2] = 3.6 M [SO3] = 3.0 M [NO2] = 0.90 M [NO] = 2.6 M