Equilibrium

A Getting-It-On Review and Self-Test

Chemical Kinetics and Molecular Equilibrium



The study of the factors that affect the speed of a reaction is called

. How fast a reaction occurs does not depend on the stability of

the . Some of the factors involved in kinetics are:

, , , and . Increasing the temperature

by 10°C approximately the rate. Increasing the concentration of

reactants also the rate as does the addition of a .A

reaction coordinate diagram demonstrates the relationship between the

potential energies of the reactants and products as a function of the

progressive change from the reactants to . The energy barrier

between the reactants and products is called the energy. The

larger the barrier is, the the rate. Addition of a catalyst

effectively the activation energy by changing the mechanism.

The reaction coordinate diagram suggests that reactions are . The

degree to which a reaction is reversible is related to the difference in

potential energies between the and . The more

exothermic the reaction is, the more the are favored and the less

reversible the reaction becomes.

When the rates of converting reactants is equal to the rate of the reverse

reaction, the reaction has reached . Le Chatelier’s principle

states that an equilibrium will adjust to offset any applied stress.









Apply Le Chatelier’s principle indicating an equilibrium shift towards the

products as an “ → ”, a shift towards the reactants as an “ ← ”, and no

effect as “ Ø ”.





1. 2 SO2 (g) + O2 (g)  2 SO3 (g) + Heat







a. decreasing the temperature



b. adding O2



c. removing SO3



d. increasing the pressure



e. adding a catalyst

2





2. CH4 (g) + Br2 (g)  CH3Br (g) + HBr (g) + Heat







a. increasing the temperature



b. adding NaOH (s)





c. removing Br2



d. increasing the pressure



e. adding CH3Br







3. Draw a reaction coordinate diagram for the combination of H2 (g)

and I2 (g) to form HI (g) if the activation energy is 10 kcal/mole and

the heat produced is 5 kcal/mole. Label the diagram indicating the

energy axis, reaction progress axis, the reactants, the products,

the activation energy, and the transition state.





4. Write the equilibrium expression and calculate the equilibrium

constant if 1.0 M H2O (g) at high temperature is 25% dissociated

into H2 (g) and O2 (g).





5. Calculate the equilibrium constant for the combination of N2 (g) and

H2 (g) to form NH3 (g) if the equilibrium concentrations are: 0.85 M

N2, 0.17 M H2, and 0.63 M NH3.





6. At a certain temperature, Kc for the dissociation of H2CO is

2.5×10–5. Calculate the concentration of H2CO, CO, and H2 at

equilibrium if the initial [H2CO] = 2.0 M.





7. Calculate the percent dissociation of 5.0 M NH3 if Kc = 4.0×10–6.

3





ANSWERS



chemical kinetics products reactant reactivity



temperature reactant concentration catalyst



doubles increases catalyst products



activation slower lowers reversible reactants



products products equilibrium







1.



a. decreasing the temperature →

b. adding O2 →

c. removing SO3 →

d. increasing the pressure →

e. adding a catalyst Ø





2.



a. increasing the temperature ←

b. adding NaOH (s) →

c. removing Br2 ←

d. increasing the pressure Ø



e. adding CH3Br ←

4





3.

Transition State





Activation Energy

(10 kcal/mole)

Energy



H2, I2

Heat Produced

HI (5 kcal/mole)





Reaction Progress →









4. The reactions is: 2 H2O (g)  2 H2 (g) + O2 (g)



H  O2 

2



The equilibrium expression is: Kc  2

H2O

2









Calculate the equilibrium concentration:



75 mole H2O

H2O  1.0 M   0.75 M

100 mole H2O



0.25 mole H2O react 2 mole H2

  0.25 M H2

1 liter 2 mole H2O



0.25 mole H2O react 1 mole O2

  0.13 M O2

1 liter 2 mole H2O





0.25 0.13  0.014

2



Calculate Kc: Kc 

0.75

2

5





5. N2 (g) + 3 H2 (g)  2 NH3 (g)



NH3   0.63

2 2



Kc   95

N2 H2  0.850.17

3 3









6. H2CO  H2 + CO Kc 

H2  CO  2.5  105

H2CO

let x  H2   CO at equilibrium,then H2CO  2.0 M  x





Substitute into Kc:

 x  x   2.5  105

2.0  x 

Assume that x is so much smaller than 2.0 that Kc becomes:



 x  x   2.5  105

2.0

Now solve for x :



x 2  (2.5  10 5 )(2.0)

x  (2.5  10 5 )(2.0)  7.1  10 3 M



Equilibrium concentrations are: [H2CO] = 2.0 M, [H2] = [CO] =

7.1×10–3.

6





7. 2 NH3 (g)  N2 (g) + 3 H2 (g)



N H2 

3



Kc  2

NH3 

2









x mole NH3 1 mole N2 x

let x  NH3  that reacts, then N2    

liter 2 mole NH3 2

x mole NH3 3 mole H2 3x

and H2    

liter 2 mole NH3 2





 x  32x 

3



Substitute into Kc: 2

 4.0  10 6

 5.0  x 

2









However since x is much smaller than 5.0 M, the Kc becomes:



 x  32x 

3

2

 4.0  10 6

 5.0 

2









Now solve for x :



27 x 4

6 27x 4

16

 4.0  10 or  4.0  106

25 400



x 4   400   4.0  106   5.9  105

27





so x  4 5.9  105  0.088 M



The equilibrium concentrations are: [NH3] = 5.0 M, [H2] = 0.13 M,

[N2] = 0.044 M.



0.088

% dissociation   100  1.8%

5.0