and Equilibrium Problems Keq and Equilibrium Problems

C12-4-04 EQUILIBRIUM AND Keq OUTCOME QUESTION(S): C12-4-04 EQUILIBRIUM AND Keq Relate the concept of equilibrium to physical and chemical systems. Include: conditions necessary to achieve equilibrium. Write equilibrium law expressions from balanced chemical equations and solve problems involving equilibrium constants. Include: ICE tables Vocabulary & Concepts

*Always make sure the equation is balanced, FIRST* 1. Plug and solve *Always make sure the equation is balanced, FIRST*

Always check the volume At 225°C, a 2.0 L container holds 0.040 moles of N2, 0.15 moles of H2 and 0.50 moles of NH3. If the system is at equilibrium, calculate KC. N2(g) + 3 H2(g) 2 NH3(g) 1. Change all into concentrations - mol/L 0.040 mol N2 = 0.020 M N2 2.0 L Always check the volume = 0.075 M H2 = 0.25 M NH3 2. Write the equilibrium expression Kc = [N2][H2]3 [NH3]2

Don’t worry about the units for the constant 3. Substitute the concentrations and calculate K. 0.020 M N2 Kc = [N2][H2]3 [NH3]2 0.075 M H2 0.25 M NH3 Kc = [0.020][0.075]3 [0.25]2 Don’t worry about the units for the constant Kc = 7.4 x 103

The equilibrium concentrations of N2 and NO are 1. 40 mol/L and 5 The equilibrium concentrations of N2 and NO are 1.40 mol/L and 5.20 mol/L. Calculate the KC. 2 NO(g) N2(g) + O2(g) 5.20 M 1.40 M 1.40 M According to the equation stoichiometry – the same amount of O2 will be produced as N2 (1:1 ratio) Kc = [N2][O2] [NO]2 Kc = [1.40][1.40] [5.20]2 = 0.0725

2. Rearrange and solve

Concentrations are already in mol/L (M) At 210°C, the Kc is 64.0 At equilibrium, [N2] = 0.40 mol/L and [O2] = 0.60 mol/L. Calculate the [NO]eq. Concentrations are already in mol/L (M) N2(g) + O2(g) 2 NO(g) Kc = [N2][O2] [NO]2 Temperature isn’t needed here – but remember the constant changes if temperature changes Write Eqbm expression = Kc[N2][O2] [NO]2 2. Rearrange for missing value = Kc[N2][O2] [NO]

= Kc[N2][O2] [NO] = (64.0)(0.40)(0.60) [NO] = 3.9 mol/L [NO] 3. Substitute concentrations then solve = Kc[N2][O2] [NO] = (64.0)(0.40)(0.60) [NO] = 3.9 mol/L [NO] These question types are not difficult, so to increase difficulty look for: Unbalanced equations Volume values Unmatched units Stoichiometry assumptions

Calculate the [S] and [O2] at equilibrium. At 10.0°C, the Kc is 215.0 The equilibrium concentrations of SO2 is 9.40 mol/L. Calculate the [S] and [O2] at equilibrium. SO2(g) S(g) + O2(g) 9.40 M x M x M Assume values should be the same for sulfur and oxygen – but we will need to use a variable here… Kc = [S][O2] [SO2] Write Eqbm expression 2. Rearrange for missing value = [S][O2] [SO2]Kc

[S]eq and [O2] eq = 45.0 M = [S][O2] [SO2]Kc = [x][x] [9.40](215) = 3. Substitute concentrations then solve 1. Write Eqbm expression = [S][O2] [SO2]Kc 2. Rearrange for missing value = [x][x] [9.40](215) = [x]2 (2021) = [x] (2021) = [x] 45.0 M [S]eq and [O2] eq = 45.0 M

The 4th type – “hard” ICE tables – will appear in another unit 3. Simple I.C.E. Tables The 4th type – “hard” ICE tables – will appear in another unit

1.00 M of hydrogen and 1.00 M of fluorine are allowed to react at 150.0°C. At equilibrium, the [HF] is 1.32 M. Calculate KC H2(g) + F2(g) 2 HF(g) [Initial] ICE table questions involve reaction progress – to identify this type, look for expressions showing the passage of time Tables are used to calculate the equilibrium concentrations from known progress values to use in the equilibrium expression [Change] [Eqlbm]

1.00 M of hydrogen and 1.00 M of fluorine are allowed to react at 150.0°C. At equilibrium, the [HF] is 1.32 M. Calculate KC H2(g) + F2(g) 2 HF(g) 1 1 [Initial] 1.00 [Change] - 0.66 - 0.66 + 1.32 [Eqlbm] 0.34 0.34 1.32 Enter available data into the ICE table. – if no initial value is given assume a value of zero Find the change for one species and make the appropriate changes in the others. Notice: the changes in concentration reflect the stoichiometry of the reaction.

(What is favoured? How much?) H2(g) + F2(g) 2 HF(g) [Eqlbm] 0.34 0.34 1.32 Use the calculated equilibrium values to solve the equilibrium expression for the constant. Kc = [H2][F2] [HF]2 Kc = 0.1156 1.742 = 15.1 Kc Kc = [0.34][0.34] [1.32]2 Look at the constant value – how does it reflect the concentrations at equilibrium (What is favoured? How much?)

Be mindful of the “tricks” Initially 2.0 mol of SO2, 1.0 mol of O2 are mixed in a 3.0 L reaction container. At equilibrium, 0.20 mol of O2 are found to remain. Calculate the Kc. 2 SO3 (g) 2 SO2 (g) + O2 (g) [I] 0.67 0.33 [C] [E] 0.067 2.0 mol SO2 = 0.67 M SO2 3.0 L = 0.33 M O2 Be mindful of the “tricks” = 0.067 M O2

Initially 2. 0 mol of SO2, 1. 0 mol of O2 are mixed in a 3 Initially 2.0 mol of SO2, 1.0 mol of O2 are mixed in a 3.0 L reaction container. At equilibrium, 0.20 mol of O2 are found to remain. Calculate the Kc. 2 SO3 (g) 2 SO2 (g) + O2 (g) 1 [I] 0.67 0.33 [C] 0.52 + 0.52 – – 0.26 [E] 0.52 0.15 0.067 This reaction is being run “backwards” – this only affects the ICE table and how information is recorded: make sure the appropriate side is being made or used up (+/-) to reflect the progress of the reaction

Kc = 5.6 x 10-3 2 SO3 (g) 2 SO2 (g) + O2 (g) [E] 0.52 0.15 0.067 Kc = A “backwards” reaction only affects the ICE table and how information is recorded: Write the equilibrium expression for reaction as written Kc = [SO2]2[O2] [SO3]2 Kc = 0.00557 Kc = 5.6 x 10-3 Kc = [0.15]2[0.067] [0.52]2 Again…look at the constant value – how does it reflect the concentrations at equilibrium

C12-4-04 EQUILIBRIUM AND Keq CAN YOU / HAVE YOU? C12-4-04 EQUILIBRIUM AND Keq Relate the concept of equilibrium to physical and chemical systems. Include: conditions necessary to achieve equilibrium. Write equilibrium law expressions from balanced chemical equations and solve problems involving equilibrium constants. Include: ICE tables Vocabulary & Concepts