CHAPTER 16

1. (3 pts.)  Circle T (true) or F (false):

  T  F  a. An equilibrium constant of 10 would 
           indicate a large amount of product is formed.

  T  F  b. The Kc value for an equation is dependent 
           on pressure.

  T  F  c. If the equilibrium constant for a reaction 
           increases as the temperature decreases, the 
           reaction is exothermic.

2. (4 pts.)  H2O <---> H2(g) + ½O2(g)	 H  = +242 kJ


Answer increase, decrease, or no change for the effect on 
[H2] in the system when:

   a. the temperature is increased.______________________

   b. the pressure is decreased._________________________

   c. one mole of argon is added.________________________

   d.a catalyst is added.________________________________

3. (4 pts.)  At 520 C, Kc = 0.016 for the reaction

	2HI(g) <---> H2(g) + I2(g)

Calculate the concentrations of all species at equilibrium 
in a 6.0 liter container starting with 0.40 mole of H2 and 
0.40 mole of I2.












4.	(2 pts.)  Write the equilibrium expression for

	3Fe(s) + 4H2O(g) <---> Fe3O4(s) + 4H2(g)
5.	(3 pts.)  N2(g) + 3H2(g) <--->   2NH3(g)

One mole of N2, three moles of H2, and two moles of NH3 are 
placed in a one-liter flask at 150 C.  If Kc at 150 C is 0.90, 
the system will:

a. shift right   b. shift left   c. not shift (at equilibrium)

















6. (4 pts.) The following reaction is carried out at 500 C:

	2N2H4(g) + N2O4(g) <---> 3N2(g) + 4H2O(g)

When 0.60 moles of N2H4 are mixed with 0.50 moles of N2O4 in 
a 3.0 liter flask, 0.30 moles of H2O is found at equilibrium.  
What is the equilibrium concentration of N2H4?










			KEY

1. (3 pts.)  Circle T (true) or F (false):

  T  F* a. An equilibrium constant of 10-3 would indicate 
           a large amount of product is formed.
  T  F* b. The Kc value for an equation is dependent on pressure.
  T* F  c. If the equilibrium constant for a reaction increases 
           as the temperature decreases, the reaction is exothermic.

2. (4 pts.)  H2O <---> H2(g) + ½O2(g)  H  = +242 kJ   endothermic;
                                                     absorbs heat.

Answer increase, decrease, or no change for the effect on [H2] in the 
system when:

  a. the temperature is increased.		increase
  b. the pressure is decreased.			increase
  c. one mole of argon is added.		no change
  d. a catalyst is added.      no change; both forward and reverse
	                       reactions are speeded up equally.  

3. (4 pts.) At 520 C, Kc = 0.016 for the reaction

	2HI(g) <---> H2(g) + I2(g)

Calculate the concentrations of all species at equilibrium in a 
6.0 liter container starting with 0.40 mole of H2 and 0.40 mole of I2.
                 2HI <---------------->  H2    +       I2
Init.			                0	0.40 mole = 0.067
     	0.40 mole/6.0 L = 0.067
Change		+2x		        -x	       -x
Equil.		 2x		      0.067-x	     0.067-x


Kc = 0.016 = [H2][I2]				0.26x = 0.067 - x
             ___________
               [HI]2				1.26x = 0.067

0.016 = (0.067 - x)(0.067 - x)		    x = 0.053
               (2x)2

Take square root of both sides:

0.13 = (0.067 - x)
            2x


[H2] = [I2] = 0.067 - 0.053 = 0.014 M

[HI] = 2x = 2(0.053) = 0.11 M

4.(2 pts.)  Write the equilibrium expression for

3Fe(s) + 4H2O(g) <---> Fe3O4(s) + 4H2(g)

Kc = [H2]4  	only gases appear in the Kc expression.
          [H2O]4

5. (3 pts.)  N2(g) + 3H2(g) <---> 2NH3(g)

One mole of N2, three moles of H2, and two moles of NH3 are placed 
in a one-liter flask at 150 C.  If Kc at 150 C is 0.90, the system 
will:

a. shift right*	 b. shift left	c. not shift (at equilibrium)

	  [NH3]2   = 0.90 	Put initial conc. in Kc equation to 
	_________			calculate Q.
	[N2][H2]3

         
	Q =  (2)2  =  4 = 0.148 initially.
         (1)(3)   27

Q < KcEquilibrium will shift to raise 0.148 to 0.90, 
so system moves toward more product; inc. size of numerator.

6. (4 pts.) The following reaction is carried out at 500 C:

	2N2H4(g) + N2O4(g) <---> 3N2(g) + 4H2O(g)

When 0.60 moles of N2H4 are mixed with 0.50 moles of N2O4 in a 
3.0 liter flask, 0.30 moles of H2O is found at equilibrium.  
What is the equilibrium concentration of N2H4?

           2N2H4            +    N2O4 <--->              3N2 + H2O
Init. 0.60 mole/3.0 L = 0.20 M  0.50 mole/3.0 L = 0.17 M  0	0
Change      -2x		          -x		         +3x   +4x
Equil.   0.20 - 2x		0.17 - x		  3x	4x


at equilibrium there is

0.30 mole H2O or 0.10 M H2O
  3.0 L

  4x		=	0.10 M
   x		=	0.025 M
[N2H4]	        =	0.20 - 2x
		=	0.20 - 0.05
[N2H4]	        =       0.15 M