Equilibrium law
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Equilibrium Law. Mass action expression. Equilibrium constant. Kc =. Introduction to the Equilibrium law. [H 2 O]. [H 2 O] 2. [H 2 ] 2 [O 2 ]. [H 2 ] 2 [O 2 ]. Read 14.3 to PE1. 2H 2 (g) + O 2 (g)  2 H 2 O (g). 2 H 2 (g) + O 2 (g)  2H 2 O (g). 2H 2 (g) + O 2 (g)  2H 2 O (g).

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Equilibrium Law

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Equilibrium law

Equilibrium Law


Introduction to the equilibrium law

Mass action expression

Equilibrium constant

Kc =

Introduction to the Equilibrium law

[H2O]

[H2O]2

[H2]2[O2]

[H2]2[O2]

  • Read 14.3 to PE1

2H2(g) + O2(g)  2H2O (g)

2H2(g) + O2(g)  2H2O (g)

2H2(g) + O2(g)  2H2O (g)

2H2(g) + O2(g)  2H2O (g)

2H2(g) + O2(g)  2H2O (g)

Step 1:Set up the “equilibrium law” equation

Step 2:Product concentrations go in numerator

Step 3:Concentration in mass action expression is raised to the coefficient of the product

Step 4:Reactant concentrations go in denominator

Step 5:Concentrations in mass action expression are raised to the coefficients of reactants


Equilibrium law important points

Equilibrium law: important points

  • State (g, l, s, aq) may or may not be added at this point since we will only be dealing with gasses for this section. Later it will matter.

  • The equilibrium law includes concentrations of products and reactants in mol/L (M)

  • The value of Kc will depend on temperature, thus this is listed along with the Kc value

  • Tabulated values of Kc are unitless

  • By substituting equilibrium concentrations into equilibrium law, we can calculate Kc …

  • Do RE 14.31, 35, 36, 37 (pg. 589)


Equilibrium law re 14 31 14 35

Kc =

Kc =

[0.00261]

,Kc=

[0.105][0.250]2

[0.150]

,Kc=

[0.0222][0.0225]

Equilibrium law: RE 14.31, 14.35

[CH3OH]

CO(g) + 2H2(g)  CH3OH(g)

=0.398

[CO][H2]2

C2H4(g) + H2O(g)  C2H5OH(g)

[C2H5OH]

= 300

[C2H4][H2O]


Equilibrium law re 14 36 14 37

Kc =

Kc =

[x]

,Kc=

[0.210][0.100]2

[0.280]2

,Kc=

[0.00840][x]3

Equilibrium law: RE 14.36, 14.37

[CH3OH]

= 0.500

CO(g) + 2H2(g)  CH3OH(g)

[CO][H2]2

x = [0.210][0.100]2 (0.500) =

0.00105 M

N2(g) + 3H2(g)  2NH3(g)

[NH3]2

= 64

[N2][H2]3

x3 = [0.280]2/ [0.00840] (64) =

0.146

x3 = 0.146

x = 0.53 M


When kc mass action expression

Kc =

[0.175]

,Kc=

[0.0197][0.0200]

When Kc  mass action expression

  • We can use the equilibrium law to determine if an equation is at equilibrium or not

  • If mass action expression equals equilibrium constant then equilibrium exists

    Q - consider: C2H4(g) + H2O(g)  C2H5OH(g)

    If Kc = 300, []s = 0.0197 M, 0.0200 M, 0.175 M which direction will the reaction need to shift?

[C2H5OH]

= 300

[C2H4][H2O]

444  300

444 must be reduced to 300. Thus, the top must decrease and the bottom must increase. A shift to left is required to establish equilibrium.


More equilibrium law problems

PCl3 (g) + Cl2(g)  PCl5(g)

Kc =

Kc =

[0.00600]

,Kc=

[0.0520][0.0140]

[0.0100][0.0400]

,Kc=

[0.00150][0.00300]

More equilibrium law problems

  • Do RE 14.32, 33 (pg. 589). For each, state in which direction the reaction needs to shift

[PCl5]

= 0.18

[PCl3][Cl2]

8.24  0.18

Top must , bottom must -shift to left is needed

SO2(g) + NO2(g)  NO(g) +SO3(g)

[NO] [SO3]

= 85

[SO2][NO2]

88.9  85

Top must , bottom must -shift to left is needed

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