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Chapter 15: Chemical Equilibrium. N 2 O 4 (g) ⇋ 2 NO 2 (g). A reaction is in equilibrium when the rate of the forward reaction:. rate for = k for [N 2 O 4 ]. equals the rate of the reverse reaction:. rate rev = k rev [NO 2 ] 2. rate for = rate rev. k for [N 2 O 4 ] = k rev [NO 2 ] 2.

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slide1

Chapter 15: Chemical Equilibrium

N2O4(g)⇋ 2 NO2(g)

A reaction is in equilibrium when the rate of the forward reaction:

ratefor = kfor[N2O4]

equals the rate of the reverse reaction:

raterev = krev[NO2]2

ratefor = raterev

kfor[N2O4] = krev[NO2]2

Kc = the equilibrium constant for the reaction when Molar concentrations are used to calculate K.

slide2

Note that it is only possible to establish an equilibrium for a reaction that proceeds via a single step mechanism. Hence the concentration of each reactant and product is raised to its corresponding coefficient.

It is important to remember that even though the [reactants] and [products] are constant at equilibrium the reaction hasn’t stopped Chemical equilbrium is a dynamic process.

N2O4

2 NO2

slide3

What is Kc for the following reactions?

N2(g) + 3 H2(g)⇋ 2 NH3(g)

H2O(l)⇋ H2O(g)

CO2(g) + H2O(l)⇋ H+(aq) + HCO3(aq)

2 Ag+(aq) + CO32-(aq) ⇋ Ag2CO3(s)

Note that reactants and products in the solid or liquid state do not appear in the equilibrium constant expression because their molarity does not change as a result of the position of the equilibrium.

slide4

Sometimes, it is more convenient to determine the equilibrium constant for gas phase reactions in terms of the partial pressures of the gasses.

N2O4(g)⇋ 2 NO2(g)

Kp and Kc are related through the ideal gas law as follows:

PNO2V = nNO2RT, so PNO2 = (nNO2/V)RT = [NO2]RT

Similarly: PN2O4 = [N2O4]RT

In general: Kp = KcRTn

slide5

2 NOBr(g)⇋ 2 NO(g) + Br2(g)

(A)

What is Kc for 2 NO(g) + Br2(g) ⇋ 2 NOBr(g)

(B)

Which reaction, A or B, favors product formation?

Rxn B because Kc > 1

Q: What is the relative value of Kc if reactants are favored?

Kc < 1 for reactant favored rxns

Q: If neither reactants nor products are favored?

Kc = 1

slide6

Combining equilibria a la Hess’s Law:

If you know that:

2 NOBr(g)⇋ 2 NO(g) + Br2(g)

And that Br2(g) + Cl2(g)⇋ 2 BrCl(g)

What is Kc for the combined reaction?

2 NOBr(g) + Cl2(g) ⇋ 2 NO(g) + 2BrCl(g)

Kc = (Kc1)(Kc2) = 0.10

slide7

Calculations with Equilibrium Constants: ICE Tables

15.24 A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00 L vessel at 700K. These substances react as follows:

At equilibrium, the vessel is found to contain 0.566 g of H2. What are the equilibrium partial pressures of each species?

H2(g) + Br2(g) ⇋ 2 HBr(g)

Initial mol H2 = 1.374 g/(2.01588 g/mol) = 0.68159 mol

Initial mol Br2 = 70.31 g/(159.808 g/mol) = 0.43997 mol

Eq mol H2 = 0.566 g/(2.01588 g/mol) = 0.2808 mol

H2(g) + Br2(g) ⇋ 2 HBr(g)

Initial

0.68159 mol

0.43997 mol

0 mol

Change

-0.4008 mol

-0.4008 mol

+0.8016 mol

Equilibrium

0.2808 mol

0.0392 mol

0.8016 mol

slide8

Use the number of moles of each gas present at equilibrium to determine their partial pressures.

PH2 = nRT/V = (0.2808 mol)(0.0821Latm)(700K)/(2.00 L)

PH2 = 8.07 atm

PBr2 = nRT/V = (0.0392 mol)(0.0821Latm)(700K)/(2.00 L)

PBr2 = 1.1 atm

PHBr = nRT/V = (0.8016 mol)(0.0821Latm)(700K)/(2.00 L)

PHBr = 23.0 atm

d) What is Kp?

= 58

e) What is Kc?

Kp = KcRTn

n = 0, so Kp = Kc

slide9

Stoichiometry and Equilibrium ICE Tables

Ex: A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the equilibrium constant Kc for the reaction shown below at 448°C is 50.5. What are the equilibrium concentrations of H2, I2, and HI in moles per liter?

H2(g) + I2(g) ⇋ 2 HI(g)

I

1.000 M

2.000 M

0

C

- x

+2x

- x

E

1.000 - x

2x

2.000 - x

E

0.444 M

1.11 M

1.444 M

Solve quadratic and get that x = 0.556 M

slide10

Using the reaction quotient to predict equilibrium shifts

H2(g) + I2(g) ⇋ 2 HI(g)

Kc = 50.5

Ex: Predict in which direction the reaction will proceed to reach equilibrium if we start with 2.0 × 10−2 mol of HI, 1.0 × 10−2 of H2, and 3.0 × 10−2 of I2 in a 2.00-L container.

Initial concentrations:

[HI] = (0.020 mol HI)/2.00 L = 0.010 M

[H2] = (0.010 mol H2)/2.00 L = 5.0 x 10−3 M

[I2] = (0.030 mol H2)/2.00 L = 0.015 M

Qc < Kc so not enough products

Right shift

slide11

Le Chatelier’s Principle: When a stress is applied to an equilibrium, the equilibrium will shift to alleviate the stress.

Fe+3(aq) + SCN-1(aq) ⇋ FeSCN+2(aq)

Colorless ⇋ Dark red

Reactants

Products

Left shift = lighter color

Right shift = darker color

Initial color

slide12

Other ways to cause a Le Châtelier Shift:

N2(g) + 2 O2(g)⇋ 2 NO2(g)

H = 67.7 kJ

What kind of shift would you see if:

Pressure increased?

Right shift 

Volume increased?

 Left shift

Right shift 

Heating temperature increased?

slide13

CH4(g) + 2 Cl2(g)⇋ CCl4(g) + 2 H2(g)

H = -32 kJ

What kind of shift would you see if:

Pressure increased?

No Change

Heating temperature increased?

 Left shift

H2 removed?

Right shift 

CH4 removed?

 Left shift

Catalyst added?

No Change