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Solutions to Practice questions (Chapter 6)

Q1. Use LeChateliers principle to predict the sense of the shift in equilibrium for the system

2NO2 N2O4

when it is cooled to -75C and use the vant Hoff equation to confirm your prediction with a quantitative

calculation of K at -75C, given H = -13.67 kcal/mol (-57.2kJ/mol) and K=6.97 at 25 C.

Solution

Cooling an exothermic reaction causes it to produce more heat by running forward hence more

dimerization will occur, as observed.

Ln(K2/K1) = +[13.670 cal/mol/1.9872 cal/Kmol)](1/198K -1/298K)

=11.6

K2 = (1.16 X 105) (6.97) = 8.06 X 10

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This large value of K implies nearly complete reaction.

Q2. The system 2NO2 N2O4 is brought to equilibrium at a total pressure of 1.00 atm at 25C. K=6.97

at 25 C. What are the partial pressures of NO2and N2O4

Solution

Let P0is the initial pressure of NO2

2NO2 N2O4 Total pressure

Initial P0 0 P0

Change -2x x

Equilbm P0-2x x p0-x

P0-x = 1 atm

P0=1-x

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K=x/(1-x)2

x may vary from 0-1

The resulting quadratic equation has a solution of x=0.686, 1.457.

Only 0.686 is acceptable, yielding

PN2O4 = 0.686 atm and PNO2 = 0.314 atm. Position of equilibrium lies to the right as expected.

Q3. Assuming the rapid attainment of equilibrium, which of the following conditions would be most

favourable for the production of ethanol from the reaction of ethylene with steam

CH2=CH2(g) + H2O(g) CH3CH2OH(g)

a.

P=5 atm, T=600K

b.

P=50 atm, T=400K

c.

P=50 atm, T=600K

d.

P=5 atm, T=400K

Q4. For the synthesis of ammonia at 500C, the equilibrium constant is 6.0 X 10-2

Predict the direction in

which the system will shift to reach equilibrium in each of the following cases.

a.

[NH3]0= 1.0 X 10-3

M; [N2]0 = 1.0X10-5

M; [H2]0= 2.0X10-3

M

b.

[NH3]0= 2.0 X 10-4 M; [N2]0 = 1.5X10

-5M; [H2]0= 3.54X10-1M

c.

[NH3]0= 1.0 X 10-4M; [N2]0 = 5.0 M; [H2]0= 1.0X10

-2M

Solution:

a.

Q = 1.3 X 107

Q>K The system shift to the left.

b.

Q = 6.01 X 10-2

Q=K No shift will occur in the system.

c.

Q = 2.0 X 10-3

Q

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[HF]0= [H2]0= [F2]0= 3.000 mol/1.5 L = 2.000 M

Q = 1.000, Q