1. Equilibrium Calculations Lesson 8

2. 1.When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. Implies initial and not equilibrium concentrations- ICE

3. 1.When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. Implies initial and not equilibrium concentrations- ICE 2SO 2 (g) + O 2 (g) ⇋ 2SO 3 (g) I C E

4. 1.When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. Implies initial and not equilibrium concentrations- ICE 2SO 2 (g) + O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C E

5. 1.When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. Implies initial and not equilibrium concentrations- ICE 2SO 2 (g) + O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C E0.300 M

6. 1.When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. Implies initial and not equilibrium concentrations- ICE 2SO 2 (g) + O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C+0.300 M E0.300 M X 2/2

7. 1.When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. Implies initial and not equilibrium concentrations- ICE 2SO 2 (g) + O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C-0.300 M+0.300 M E0.300 M X 1/2

8. 1.When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. Implies initial and not equilibrium concentrations- ICE 2SO 2 (g) + O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C -0.300 M -0.150 M +0.300 M E0.300 M

9. 1.When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. Implies initial and not equilibrium concentrations- ICE 2SO 2 (g) + O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C -0.300 M -0.150 M +0.300 M E0.100 M0.250 M0.300 M Equilibrium concentrations go in the equilibrium equation!

10. 1.When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. Implies initial and not equilibrium concentrations- ICE 2SO 2 (g) + O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C -0.300 M -0.150 M +0.300 M E0.100 M0.250 M0.300 M Equilibrium concentrations go in the equilibrium equation! Keq= [SO 3 ] 2 [SO 2 ] 2 [O 2 ]

11. 1.When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. Implies initial and not equilibrium concentrations- ICE 2SO 2 (g) + 1O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C -0.300 M -0.150 M +0.300 M E0.100 M0.250 M0.300 M Equilibrium concentrations go in the equilibrium equation! Keq= [SO 3 ] 2 (0.3) 2 = [SO 2 ] 2 [O 2 ](0.1) 2 (0.25)

12. 1.When 0.800 moles of SO 2 and 0.800 moles of O 2 are placed into a 2.00 litre container and allowed to reach equilibrium, the equilibrium [SO 3 ] is to be 0.300 M. Calculate the Keq value. Implies initial and not equilibrium concentrations- ICE 2SO 2 (g) + O 2 (g) ⇋ 2SO 3 (g) I0.400 M0.400M0 C -0.300 M -0.150 M +0.300 M E0.100 M0.250 M0.300 M Equilibrium concentrations go in the equilibrium equation! Keq= [SO 3 ] 2 (0.3) 2 ==36.0 [SO 2 ] 2 [O 2 ](0.1) 2 (0.25)

13. 2.0.80 moles of H 2 and 1.4 moles of S are initially in a 4.0 L flask and allowed to reach equilibrium. Calculate the [H 2 ] at equilibrium. H 2(g) + S (s) ⇄ H 2 S (g) Keq= 14 I0.20 M0 C x x E0.20 - xx Equilibrium concentrations go in the equilibrium equation! Keq=[H 2 S]x ==14 [H 2 ].2 - x

14. x=14.2 - x1 cross multiply 1x=14(.2 - x) 1x=2.8 - 14x 1x + 14x = 2.8 15x =2.8 x=0.19 M [H 2 ]=0.20 - x [H 2 ]=0.20 - 0.19 [H 2 ]=0.01 M

15. 3.If 6.0 moles of HI are initially in a 3.00L vessel and allowed to reach equilibrium, what is the equilibrium concentration of H 2 ? 2HI (g) ⇄ H 2(g) + I 2(g) Keq = 0.0183 I2.0 M00 C 2x xx E2.0 - 2xxx Keq= [H 2 ][I 2 ]x 2 ==0.0183 [HI] 2 (2 - 2x) 2

16. x 2 =0.0183 (2 - 2x) 2 take the square root of both sides x=0.13528 (2 - 2x) 1 cross multiple 1x=0.270555-0.27055x 1.27055x=0.270555 x=[H 2 ]=0.21 M

17. 4. The same number of moles of I 2 and Cl 2 are placed in a 1.0L flask and allowed to reach equilibrium. If the equilibrium concentration of ICl is 0.40 M, calculate the initial number of moles of I 2 and Cl 2. I 2 (g) + Cl 2 (g) ⇄ 2ICl (g) Keq= 10.0 Ixx0Ixx0 C 0.20 M 0.20 M0.40 M Ex - 0.20x - 0.200.40 M Keq= [ICl] 2 =10.0 [I 2 ][Cl 2 ] 0.40 2 =10.0 (x -.20) 2

18. 0.40 2 =10.0 (x -.20) 2 square root both sides 0.40=3.16227 x -.20 1 0.4 = 3.16227x - 0.63246 1.03246 = 3.16227x x=[I 2 ]=[Cl 2 ] =0.33 M 1 Lx0.33 mole = 0.33 moles 1L

19. 5.Sketch the changes in concentrations of [O 2 ] and [N 2 ] as equilibrium is obtained. Calculate the value of the Keq. N 2 O 4(g) ⇋ 2O 2(g) +N 2(g) I4.0000 C1.201.202.40 E2.801.202.40 4.00 M 2.80 M [N 2 O 4 ] Time (min) 2040 1.20 M 2.40 M

20. Keq=[N 2 ][O 2 ] 2 [N 2 O 4 ] Keq=[1.20][2.40] 2 [2.80] =2.47