Solving Equilibrium Problems Type 1 Determining if Equilibrium has been reached (find Q and Compare to K)

The Reaction Quotient Compare value to equilibrium constant Q = [Products]coefficient [Reactants] coefficient Tells you the directing the reaction will go to reach equilibrium Calculated the same as the equilibrium constant, but for a system not at equilibrium Compare value to 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 Type 2 Using equilibrium concentrations to solve for K

Solving for K Using equilibrium concentrations to solve for K A mixture of H2 and I2 is allowed to react at 448 ºC. When equilibrium is established, the concentrations of the participants are found to be; [H2] = 0.46 M, [I2] = 0.39 M and [HI] = 3.0 M Calculate K.

Solution Write the balanced equation Write the equilibrium law expression for this reaction Substitute the equilibrium concentrations in the K expression and solve.

H2(g) + I2(g) 2HI(g) K = [HI]2 [H2][I2] K = [3.0 M]2 [0.46M][0.39 M ] = 50.

Solving Equilibrium Problems Type 3 Given the starting concentrations and one equilibrium concentration. Calculate K

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, too I C E

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

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+2x =0.590 mol/L

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

Solving Equilibrium Problems Type 4 Calculating equilibrium concentration using K and initial concentrations A- PERFECT SQUARE B- Simplification of Quadratic C- solve a quadratic

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 A) mathematically simplify…perfect square situation B)assumption made…if the initial concentration divided by K is greater than 100 and 5% rule met C) solve the quadratic

Example [H2]i = 0.10M [I2]i = 0.10 M [HI]i = 0 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 0.10M of H2 0.10M I2 . [H2]i = 0.10M [I2]i = 0.10 M [HI]i = 0

H2(g) I2(g) HI(g) I 0.10 M 0.10 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 0.10 M 0.10 M 0 M C -x -x +2x E 0.10M-x 0.10M –x 2x Solve for x 2x/o.1-x= square root of K 2x= 0.1-x (square root K) x= 0.093M [H2] = 0.0070M [I2] = 0.0070M [HI] = 0.186M K = [HI]2 = 7.1 x 102 [H2] [I2] = [2x]2 [0.1-x] [0.1 –x ] (Perfect square!...simplify)

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?

x =1.14 x 10 -4M (x/initialx100 less than 5%) k = [NO]2[Cl2] [NOCl]2 1.6 x 10-5 = (0.45+2x)2(x) (1.20-2x)2 Initial conc./K is GREATER than 100 …simplify 1.6 x 10-5 = (0.45)2(x) (1.20)2 x =1.14 x 10 -4M (x/initialx100 less than 5%) ….calculate equilibrium concentrations [NO]=0.45+2x=0.45M, [Cl2]= 1.14 x 10-4M, [NOCL] = 1.20-2x=1.20M

Example [H2]i = (15.8g/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 15.8 g of H2 294 g I2 . [H2]i = (15.8g/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=0.2318 [H2] = 1.33mol/L [I2] almost 0…0.0002M [HI] = 0.464 mol/L [H2] [I2] = [2x]2 [1.56-x] [0.232 –x ] USE quadratic formula…plug in a,b,c, in calculator

Solving Equilibrium Problems Type 5 Calculating equilibrium concentrations when INItial reactant and product concentrations both given…CALCULATE q FIRST BEFORE STARTING

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

Solving Equilibrium Problems Type 6 Calculating new concentrations when the equilibrium has been disturbed.

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 2.0 –x 4.0-x 2.0+x 3.0+x

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