Download presentation

Presentation is loading. Please wait.

Published byValerie Rundell Modified over 3 years ago

1
K c and Equilibrium Problems (4 types)

2
Additional KEY Terms ICE table Solve 4 types of problems involving equilibrium constants.

3
1. Plugand solve *Always make sure the equation is balanced, FIRST*

4
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

5
3. Substitute the concentrations and calculate K. (Note: no units for K)

6
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

7
and solve 2. Rearrange

8
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)

9
3. Substitute concentrations then solve.

10
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 ]

11
[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

12
3. Simple I.C.E. Tables

13
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

14
[Eq]1.320.34 H 2(g) + F 2(g) 2 HF (g)

15
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

16
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.

17
4. HARD I.C.E. Tables

18
[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

19
Get rid of the square by taking the square root of both sides. Substitute known values.

20
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

21
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 ]

22
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

23
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.

24
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

25
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

26
Additional KEY Terms ICE table CAN YOU / HAVE YOU? Solve 4 types of problems involving equilibrium constants.

Similar presentations

Presentation is loading. Please wait....

OK

Solving Equilibrium problems using the RICE method.

Solving Equilibrium problems using the RICE method.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Download ppt on coordinate geometry for class 9th model Ppt on water our lifeline program Think pair share ppt online Raster scan and random scan display ppt online Ppt online examination project Ppt on non ferrous metals Ppt on hard gelatin capsule composition Ppt on cost leadership strategy By appt only movie pro Ppt on types of business communication