CHM 2045C Module Seven Sample Exam Answers

Module Seven Part G: Hess Law of Constant Heat Summation    10 points

1. Using the following equations:

S (s)        +  3/2 O2 (g)               SO3 (g)      H1 =   -395.2 kJ

2 SO2 (g)     +       O2 (g)             2 SO3 (g)   ∆ H2 =    -198.2 kJ

calculate the ∆ H for the reaction:

S (s)          +       O2 (g)                     SO2 (g)

From p 261 section 6.7 Hess’s Law states that if a reaction is the sum of two or more reactions,

H  for the overall process is the sum of the ∆ H values of those reactions

The solution to the above is

(1)     to leave Equation 1 alone

(2)     reverse Equation 2 and multiple the reaction by ½

so that the sum of Eq1 & Eq2 = the resultant reaction.

S (s)  +  3/2 O2 (g)     à     SO3 (g)                       H1 =   -395.2 kJ

½[2 SO3 (g)                à  2 SO2 (g) +  O2 (g) ]    ∆ H2 =    ½(+ 198.2 kJ)

Simplify:

S (s)  +  3/2 O2 (g)     à     SO3 (g)                       H1 =   -395.2 kJ

SO3 (g)                      à    SO2 (g) +  ½O2 (g)     ∆ H2 =    + 99.1 kJ)

----------------------------------------------------------------------------------------------------

Add the two equations and apply Hess’s Law:

S (s)          +       O2 (g)                   SO2 (g)           H  =  - 296.1 kJ

Read section 6.7 and Example 6.8 on page 263. Try Exercise 6.11 page 264 as well as end of chapter exercises #43-46 pages 274-275.

2. Given the following equations:

B2O3 (s)   +  3H2O (g)         B2 H6 (g)   +  3 O2 (g)          ∆ H1 =  +2035 kJ

H2O (l)           H2O (g)                               ∆ H2 =    +44 kJ

2 B (s)     +    3 H2 (g)         B2 H6 (g)                             ∆ H3 =    +36 kJ

H2 (g)         + 1/2  O2 (g)        H2O (l)                                ∆ H4 =   -286 kJ

Calculate the ∆ H for the reaction:

2 B (s)     +    3/2  O2 (g)     B2 O3 (s)

Solution:

(1)   In order to get the B2O3 on the right, you need to reverse Equation #1. This puts B2H6 on the left so it will cancel the B2H6 in Equation #3

B2 H6 (g)   +  3 O2 (g) à B2O3 (s)   +  3H2O (g)      ∆ H1 =  -2035 kJ

(2) Reverse equation 2 and multiply it by 3 so the H2O (g) will cancel with the reversed Equation #1

3 [ H2O (g)           H2O (l) ]                              ∆ H2 =   3 [- 44 kJ ]

3 H2O (g)          3 H2O (l)                                ∆ H2 =   - 132 kJ

(3)   Reverse equation #4 and multiply through by 3:

3[ H2O (l)    à   H2 (g)         + 1/2  O2 (g)]             ∆ H4 = 3 [+286 kJ]

3 H2O (l)    à    3H2 (g)         + 3/2  O2 (g)]             ∆ H4 = +858 kJ

(4)     The revised reactions to sum to give Equation 5:

B2 H6 (g)   + 6/2 O2 (g) à   B2O3 (s)   +  3H2O (g)       ∆ H1 =  -2035 kJ

3 H2O (g)                      à 3 H2O (l)                              ∆ H2 =   - 132 kJ

2 B (s)      +  3 H2 (g)   à   B2 H6 (g)                            ∆ H3 =   +  36 kJ

3 H2O (l)                      à   3H2 (g)        + 3/2 O2 (g)]     ∆ H4 =   +858 kJ

-------------------------------------------------------------------------------------------

2 B (s)    +  3/2 O2 (g)   à  B2 O3 (s)                       ∆ H5 =   - 1273 kJ