Title: Chemistry 30
Author: P Lawson
Last saved by: Peggy Lawson

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Chemistry 30
Unit 1: Energy Changes in Chemical Reactions
Assignment 3: Sections 2-4 to 2-7

1.	When iron rusts in air, the following reaction occurs:

			4 Fe (s) + 3 O_2 (g)  2 Fe_2O_3 (s)	   H = -1643 kJ

	What is the heat of formation of Fe_2O_3?

	A heat of formation reaction describes the production of one mole of the compound from its elements.	In the reaction given above, two moles of the compound are formed from the elements.  Therefore, 

2

	H_f _= 

2.	The equation for the decomposition of mercury(II) oxide is as follows:

			2 HgO(s) + 182 kJ  2 Hg(l) + O_2 (g)

	Determine H_f for HgO(s)?

2

	The reaction shown in the question is the reverse of a formation reaction. As well, it involves two moles of the compound HgO, whereas heat of formation is concerned with one mole of the compound. 

	The equation needs to be reversed, which will change the sign of H from +182 kJ to -182 kJ.  As well, moles need to be changed to represent one mole, not two.

	Therefore, H_f = 

3.  Using a table of thermochemical data, calculate H for the following reactions. 
	
	Look up heats of formation from the table of thermochemical data. Multiply by coefficients from the balanced equation. Find the total sum of the reactants, and of the products. Then use the formula

			H = H_products -  H_reactants

a. | CO (g) | + | ½ O_2 (g) | | CO_2 (g)
| -110.5 | + | ½ (0) | | -393.5
| -110.5 | | -393.5

2

			H = H_products -  H_reactants

			H =	-393.5 – (-110.5) = - 283 kJ

b. | CH_4 (g) | + | 2 O_2 (g) | | CO_2 (g) | + | 2 H_2O(g)
| -74.8 | + | 2 (0) | | -393.5 | + | 2(-241.8)
| -74.8 | | -877.1

2

			H = H_products -  H_reactants

			H =	-877.1 – (-74.8) = - 802.3 kJ

c. | CS_2(g) | + | 2 H_2O(l) | | CO_2 (g) | + | 2 H_2S(g)
| + 117.4 | + | 2 (-285.8) | | -393.5 | + | 2 (-20.6)
| -454.2 | | -434.7

2

			H = H_products -  H_reactants

			H =	-434.7 – (-454.2) = + 19.5 kJ

4.	Calculate H for the process:

			2 Al(s) + Fe_2O_3 (s)  Al_2O_3 (s) + 2 Fe(s)

	Given that H_f of Fe_2O_3 = -813.0 kJ/mole and H_f of Al_2O_3 is -1,655.0 kJ/mol

| 2 Al(s) | + | Fe_2O_3 (s) | | Al_2O_3 (s) | + | 2 Fe(s)
| 2 (0) | + | -813.0 | | -1655.0 | + | 2 (0)

2

			H = H_products -  H_reactants

			H =	- 1655.0  – (- 813.0) = - 842.0 kJ

5.	The standard heats of formation for CH_4, CHCl_3 and HCl  are -74.8, -132, -92.0 kJ/mole, respectively.  Use this information to calculate the heat of reaction for the following reaction:

			CH_4 + 3 Cl_2  CHCl_3 + 3 HCl

| CH_4 | + | 3 Cl_2 | | CHCl_3 | + | 3 HCl
| - 74.8 | + | 3 (0) | | - 132.0 | + | 3 (-92.0)
| -74.8 | | -408.0

2

			H = H_products -  H_reactants

			H =	- 408.0 – ( - 74.8 ) = - 333.2 kJ

6.  _The _standard _heat _of _formation, _Hf, _for _C2H4 _is _+52.3 _kJ/mol.  _If _C2H4 _(ethylene) _reacts _with _H2 _to _produce _C2H6 _(ethane) _according _to _the _following _equation:

		_C2H4 _+ _H2  _C2H6	_H _= _-137 _kJ

	_what _is _the _heat _of _formation _of _C2H6?	

	(Hint: Solve this question without looking up H_f for C_2H_6 in the Table of Thermochemical Data, although you may want to in order to check your answer.  Solve this using the standard formula, H_reaction = H_products - _reactants. However, this time you know H_reaction  and need to solve for a substance on the product side of the equation.)

| _C2H4 | + | _H2 | | _C2H6
| 52.3 | + | 0 | | x

		H = H_products -  H_reactants

2

		-137 = x – 52.3
	
		-137 + 52.3 = x – 52.3 + 52.3

		-84.7 kJ = x = H_C2H6

7.	Using bond enthalpies, calculate H for the following reaction:

		N_2 + 2 H_2  N_2H_4

	The structural formulas for all reaction participants are shown here:

N  N   +   2  H — H |
		
		You need to refer to the table of bond enthalpies.  Determine how many of each type of chemical bond are broken (on the reactant side of the equation) and how many bonds of each type are formed (on the product side of the equation).

Bonds Broken (energy required):
| 1 N  N | 1  941 | = | 941 kJ
| 2 H – H | 2  436 | = | 872 kJ
| TOTAL | | | 1813 kJ
Bonds Formed (energy released):
| 4 N – H | 4  391 | = | 1564 kJ
| 1 N – N | 1  163 | = | 163 kJ
| TOTAL | | | 1727 kJ

		The net energy change (heat of reaction) is the difference between how much energy is required to break bonds and how much energy is released when bonds form:

2

		H = (energy required) -  (energy released)
	
		H = 1813 – 1727 = +86 kJ

		There is a net requirement of 86 kJ in this endothermic reaction

		Note: Solving for H using bond enthalpies will not always agree completely with values from a Table of Thermochemical Data due to difficulties in accurately measuring bond enthalpies.

8.	The energy from the combustion of hydrazine, N_2H_4, is used to power rockets into space in the reaction:

			N_2H_4 (g) + O_2 (g) N_2 (g) + 2 H_2O (l)		H = -627.6 kJ

		How many kilograms of hydrazine would be necessary to produce 1.0  10^8 kJ of energy?

		Hint: One mole of N_2H_4 produces 627.6 kJ of energy.  How many moles (and then grams) are required to produce 1.0  10^8 kJ of energy?

2

		Use unit analysis to set up and solve this equation.  Since the question concerns mass, you will need to find the molar mass of N_2H_4 as part of the solution:

			2 N = 2  14.0 = 28.0
			4 H = 4    1.0 =  4.0

			molar mass of N_2H_4 = 32.0 g/mol

Unit 1: Energy	Assignment 3