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 Hesss 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

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

 

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

 

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

 

Simplify:

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

 

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

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Add the two equations and apply Hesss 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) 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) H2 (g) + 1/2 O2 (g)] ∆ H4 = 3 [+286 kJ]

 

3 H2O (l) 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) B2O3 (s) + 3H2O (g) ∆ H1 = -2035 kJ

3 H2O (g) 3 H2O (l) ∆ H2 = - 132 kJ

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

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

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

 

Answer: -1273 kJ