48.(D) All intermediate mechanisms must add up to yield the original, overall balanced

equation.

() () ()

() () ()

() () ()

NO g O g NO g

NO g O g NO g

NO g NO g NO g

22

2

22

2

23

3

"

"

"

+

+

+

49.(C) Remember that Ka×Kb= 10–14. Therefore,

b

.

K Lac.

15 10

(^10) 67 10
5
(^1410)

• ==- #
  
  
  • 
    
    
    
    • 
      
    
    
    
    
  
  
  
  

50.(D) Use the relationship gram-equivalents acid = gram-equivalents base. Solve for the

gram-equivalents of acid.

. ..
.

liter acid

liter

gram equiv acid

1 gram equivalents of acid

0 700

1

30

# 21

• =-

At neutralization, the gram-equivalents of acid = the gram-equivalents of base. Therefore,

..

.

gram equiv NaOH.

gram equiv NaOH

g NaOH

1 g NaOH

21

1

40 00

# = 84

51. (D) ∆G°represents the free energy at standard conditions: 25°C and 1 atm pressure. ∆G
    represents the free energy at nonstandard conditions. In this problem, we have the non-
    standard condition of 100°C. In order to solve for the free energy of this reaction, you
    must use the following equation:
    ∆G= ∆G°+ 2.303 RT log Qp

where the constant R= 8.314 J ⋅K–1⋅mole–1and Qpis called the reaction quotient. The

reaction quotient has the same form as the equilibrium constant Kp.

Step 1:Write a balanced equation.

CaCO 3 (s) →CaO(s) + CO 2 (g)

Step 2: Determine the value of Qp, the reaction quotient.

Qp= [CO 2 (g)] = 1.00

Step 3: Substitute into the equation.

∆G= ∆G°+ 2.303 RT log Qp

= 130,240 J/mole + 2.303(8.314 J ⋅K–1⋅mole–1)(373 K)(log 1.00)

= 130,240 J/mole

= 130.240 kJ/mole

Part IV: AP Chemistry Practice Test