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# K c and Equilibrium Problems (4 types). Additional KEY Terms ICE table Solve 4 types of problems involving equilibrium constants.

## Presentation on theme: "K c and Equilibrium Problems (4 types). Additional KEY Terms ICE table Solve 4 types of problems involving equilibrium constants."— Presentation transcript:

K c and Equilibrium Problems (4 types)

Additional KEY Terms ICE table Solve 4 types of problems involving equilibrium constants.

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

At 225°C, a 2.0 L container holds 0.040 moles of N 2, 0.15 moles of H 2 and 0.50 moles of NH 3. If the system is at equilibrium, calculate K C. 1. Change all into concentrations - mol/L 2. Write the equilibrium law for the reaction N 2(g) + 3 H 2(g) 2 NH 3(g) 0.040 mol N 2 2.0 L = 0.020 M N 2 = 0.075 M H 2 = 0.25 M NH 3

3. Substitute the concentrations and calculate K. (Note: no units for K)

The equilibrium concentrations of N 2 and NO are 1.40 mol/L and 5.20 mol/L respectively. Calculate the K C. According to Eq stoichiometry – the same amount of O 2 will be produced as N 2. 2 NO (g) N 2(g) + O 2(g) 1.40 M 5.20 M 1.40 M K c = [N 2 ][O 2 ] [NO] 2 K c = [1.40][1.40] [5.20] 2 =0.0725

and solve 2. Rearrange

At 210°C, the K c is 64.0 The equilibrium concentrations of N 2 and O 2 are 0.40 mol/L and 0.60 mol/L, respectively. Calculate the equilibrium concentration of NO. 1. Write out the Eq Law. 2. Rearrange for [NO]. N 2(g) + O 2(g) 2 NO (g)

3. Substitute concentrations then solve.

At 10.0°C, the K c is 215.0 The equilibrium concentrations of SO 2 is 9.40 mol/L. Calculate the [S] and [O 2 ] at equilibrium. SO 2(g) S (g) + O 2(g) Values should be the same amount for S and O 2 – assign the unknown “x” x M 9.40 M x M K c = [S][O 2 ] [SO 2 ] = K c [S][O 2 ] [SO 2 ]

[x][x] = 215 [9.40] = K c [S][O 2 ] [SO 2 ] [x] 2 = 2021 √√ x = 45.0 M [S] eq and [O 2 ] eq = 45.0 M

3. Simple I.C.E. Tables

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 K C. H 2(g) + F 2(g) 2 HF (g) [Eqlbm]1.320.34 [Initial] 1.00 M 0 [Change] - 0.66 + 1.32 Tip #1 – the changes in concentration match the stoichiometry of the reaction. 1 1

[Eq]1.320.34 H 2(g) + F 2(g) 2 HF (g)

Initially 2.0 mol of SO 2, 1.0 mol of O 2 are mixed in a 3.0 L reaction container. At equilibrium, 0.20 mol of O 2 are found to remain. Calculate the K c. 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) [E][E] 0.150.0670.52 [I][I] 00.67 mol/L0.33 mol/L [C][C] - 0.52+ 0.52- 0.26 Tip #2 – make sure the +/- reflect the appropriate side being made or used up

K c = 0.00557 K c = 5.6 x 10 -3 [E][E] 0.150.0670.52 K c = [SO 2 ] 2 [O 2 ] [SO 3 ] 2 2 SO 3 (g) 2 SO 2 (g) + O 2 (g) K c = [0.15] 2 [0.067] [0.52] 2 Tip #3 – Write Eq Law for equation as written.

4. HARD I.C.E. Tables

[E][E] 6.0 - x 2x 6.0 - x 6.0 moles of N 2 and O 2 gases are placed in a 1.0 L container, what are all the concentrations at equilibrium? The K c is 6.76. N 2(g) + O 2(g) 2 NO (g) [I][I] 6.0 mol/L 0 [C][C]- x + 2x Tip #4 – use x values to match the stoichiometry changes of the reaction. 1 1

Get rid of the square by taking the square root of both sides. Substitute known values.

Isolate, and solve for x. [I][I] [C][C] [E][E] 6.0 mol/L 0 - 3.4 2(3.4) 6.8 2.6 6.0 - x2x6.0 - x - x + 2x

H 2 (g) + I 2 (g) 2 HI (g) 2.0 moles of HI were placed in a 1.0 L flask at 430ºC. (K c = 54.3) Calculate the equilibrium concentrations. [I][I] [C][C] [E][E] 002.0 + x - 2x 2.0 - 2x+ x K c = [HI] 2 [H 2 ][I 2 ]

54.3 = [2.0 - 2x] 2 [+ x][+ x] K c = [HI] 2 [H 2 ][I 2 ] 7.37 = 2.0 – 2x x 7.37 x = 2.0 – 2x √√ 7.37 x + 2x = 2.0 9.37 x = 2.0 x = 0.21 [I][I] [C][C] [E][E] 002.0 + x - 2x 2.0 - 2x+ x 0.21 - 0.42 1.580.21

The K c for the reaction 5.3 x 10 -2 at 0ºC. Initially, 2.5 mol each particle was injected into a 1 L reaction vessel. Find the Eq concentrations. 2.5 + x 2.5 - 2x -2x+ x 2.5 mol/L [E][E] [C][C] [I][I] H 2 (g) + Cl 2 (g) 2 HCl (g) Tip #5 – use the Kc value to predict the changes in the reaction.

5.3 x 10 -2 = [2.5 - 2x] 2 [2.5 + x][2.5 + x] K c = [HCl] 2 [H 2 ][Cl 2 ] √ 5.3 x 10 -2 = [2.5 - 2x] 2 [2.5+x] 2 √ √ 0.23 = 2.5 – 2x 2.5+ x

x = 0.86 mol/L x = 1.92 2.23 0.58 + 0.23x = 2.5 - 2x 2.5 mol/L [E][E] [C][C] [I][I] + 0.86 2(0.86) 0.8 mol/L 3.4 mol/L 2.23 x = 1.92

Additional KEY Terms ICE table CAN YOU / HAVE YOU? Solve 4 types of problems involving equilibrium constants.

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