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Chemical Equilibrium Lesson 1 Chapter 13.

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Presentation on theme: "Chemical Equilibrium Lesson 1 Chapter 13."— Presentation transcript:

1 Chemical Equilibrium Lesson 1 Chapter 13

2 Reactions are reversible
A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the forward reaction is possible As C and D build up, the reverse reaction speeds up while the forward reaction slows down. Eventually the rates are equal

3 Forward Reaction Reaction Rate Equilibrium Reverse reaction Time

4 What is equal at Equilibrium?
Rates are equal Concentrations are not. Rates are determined by concentrations and activation energy. The concentrations do not change at equilibrium. or if the reaction is verrrry slooooow.

5 Law of Mass Action For any reaction jA + kB lC + mD
K = [C]l[D]m PRODUCTSpower [A]j[B]k REACTANTSpower K is called the equilibrium constant. is how we indicate a reversible reaction

6 Playing with K If we write the reaction in reverse. lC + mD jA + kB
Then the new equilibrium constant is K’ = [A]j[B]k = 1/K [C]l[D]m

7 K’ = [A]nj[B]nk = ([A] j[B]k)n = Kn [C]nl[D]nm ([C]l[D]m)n
Playing with K If we multiply the equation by a constant njA + nkB nlC + nmD Then the equilibrium constant is K’ = [A]nj[B]nk = ([A] j[B]k)n = Kn [C]nl[D]nm ([C]l[D]m)n

8 The units for K Are determined by the various powers and units of concentrations. They depend on the reaction.

9 K is CONSTANT At any temperature. Temperature affects rate.
The equilibrium concentrations don’t have to be the same only K. Equilibrium position is a set of concentrations at equilibrium. There are an unlimited number.

10 One for each Temperature
Equilibrium Constant One for each Temperature

11 Calculate K (Experiment #1)
N2 + 3H NH3 Initial At Equilibrium [N2]0 =1.000 M [N2] = 0.921M [H2]0 =1.000 M [H2] = 0.763M [NH3]0 =0 M [NH3] = 0.157M

12 Calculate K (Experiment #2)
N2 + 3H NH3 Initial At Equilibrium [N2]0 = 0 M [N2] = M [H2]0 = 0 M [H2] = M [NH3]0 = M [NH3] = 0.203M K is the same no matter what the amount of starting materials

13 Equilibrium and Pressure
Some reactions are gaseous PV = nRT P = (n/V)RT P = CRT C is a concentration in moles/Liter C = P/RT

14 Equilibrium and Pressure
2SO2(g) + O2(g) SO3(g) Kp = (PSO3) (PSO2)2 (PO2) K = [SO3] [SO2]2 [O2]

15 Equilibrium and Pressure
K = (PSO3/RT) (PSO2/RT)2(PO2/RT) K = (PSO3)2 (1/RT) (PSO2)2(PO2) (1/RT)3 K = Kp (1/RT)2 = Kp RT (1/RT)3

16 General Equation jA + kB lC + mD
Kp= (PC)l (PD)m= (CCxRT)l (CDxRT)m (PA)j (PB)k (CAxRT)j(CBxRT)k Kp= (CC)l (CD)mx(RT)l+m (CA)j(CB)kx(RT)j+k Kp = K (RT)(l+m)-(j+k) = K (RT)Dn Dn=(l+m)-(j+k)=Change in moles of gas

17 Homogeneous Equilibria
So far every example dealt with reactants and products where all were in the same phase. We can use K in terms of either concentration or pressure. Units depend on reaction.

18 Heterogeneous Equilibria
If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change. As long as they are not used up we can leave them out of the equilibrium expression. For example

19 For Example H2(g) + I2(s) 2HI(g) K = [HI]2 [H2][I2]
But the concentration of I2 does not change. K = [HI]2 = K’ [H2]


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