Calculating enthalpy changes (when a phase, chemical or nuclear change occurs.)
The enthalpy change or
H for each reaction is unique. What relationship exists between the heat produced or absorbed by a reaction and the amount of reacting substance? Complete the graphing lab below to examine this relationship.
| | Graphing Lab
What relationship exists between the heat released or absorbed in a reaction and the amount of substance reacted?
Click on the pencil and complete a dry lab to examine this question |
As can be seen by the assignment above , the energy change in these reactions varies directly as the number of moles of substance reacted or formed. The factor that determines the amount of heat absorbed or released is the
H of the reaction.
| Moles of compound reacted or formed | Varies directly with | Heat absorbed or released in the reaction |
| <---------------------> |
H reaction (kJ) |
Two methods may then be used to calculate heat changes during phase, chemical or nuclear changes
Calculating heat changes given an equation using an formula.
To calculate the heat absorbed or released we can use the following formula.
| Q | = | n | * | H substance |
Heat released or absorbed during reaction = Q
Number of moles = n
Molar enthalpy (enthalpy change per mole) for the substance in the reaction = H substance
The molar enthalpy for many substances in certain reactions have been determined by chemists and recorded in tables like the one below. (Such tables can be found in many handbooks or textbooks)
For example:
1) Vaporization (phase change from liquid to gas) of the substance or
2) Fusion (phase change from solid to liquid) of the substance.
| Molar Enthalpy's of Vaporization and Fusion ( under standard conditions) |
| Substance | Formula | H vap (kJ / mol ) | H fus (kJ/ mol ) |
| Ammonia | NH3 | + 23.3 | + 5.66 |
| Ethanol | C2H5OH | + 38.6 | + 4.94 |
| Methanol | CH3OH | + 35.2 | + 3.22 |
| Water | H2O | + 40.7 | + 6.01 |
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3) Combustion (burning of substance in oxygen to produce carbon dioxide and water) of the substance.
| Molar enthalpies of combustion for selected substances |
| Substance | Formula | H combustion |
| Carbon ( graphite ) | C | - 394 |
| Ethanol | C2H5OH | - 1367 |
| Glucose | C6H12O6 | - 2800 |
| Hydrogen | H2 | - 286 |
| Methane | CH4 | - 891 |
| Methanol | CH3OH | - 726 |
| Propane | C3H8 | - 2219 |
4) For other substances in reactions where molar enthalpies are not recorded we can use the following procedure to calculate them.
To determine the H substance (kJ /mol) when given an equation we need only divide the
H given by the equation by the balance of the substance.
| | Question:
What is the molar enthalpy of CO2 (g) in the reaction for the burning of butane below?
| 2 C4H10 (l) | + | 13 O2 (g) | | 8 CO2 (g) | + | 10 H2O (g) | H | = | -5315 kJ |
Answer:
Molar enthalpy is the enthalpy change in equation divided by the balance of CO2 (g)
Molar enthalpy, H substance = 5315 kJ ÷ 8 mol = 664 kJ / mol.
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Important procedure | To solve heat calculations of enthalpy using the equation method we need to do the following:
1) Determine the information given.
2) Calculate the # of moles substance and the molar enthalpy from the equation, or take it from a table of values, if needed.
3) Plug values into the equation and calculate missing value.
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| | Question:
How much heat will be released if 65 grams of butane is burned in a lighter according the equation in the example above.
Answer:
1) Given
65 grams of butane
H reaction = 5315 kJ
2) a) Moles of butane
mass ÷ gram molecular weight ( butane )
65 grams ÷ 58.14 g/mol = 1.12 moles
b) Molar enthalpy
5315 kJ ÷ 2 mol C4H10 = 2657.5 kJ/mol
3) Plug values into equation.
Q = n * H substance
Q = 1.12 mol * 2657.5 kJ/mol = 2976.4 kJ
Considering significant figures = 3.0 MJ
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Check your understanding | Answer the following question using the equation method above
Calculate the heat released when 120 grams of Iron (III) oxide is formed by the following equation | 
click to view answer |
| 2 Fe2O3 (s) | | 4 Fe (s) | + | 3 O2 (g) | | H | = | 1625 kJ |
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Assignment: Calculating Heat using the formula method.
Click on notepad and complete assignment using the equation method
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Ratio Method of calculating heat changes during enthalpy change
Since the heat released or absorbed in any phase, chemical and nuclear change is directly proportional to the H in the equation, ratios can be used to solve problems involving heat and the amount of substance.
Examine the following reaction:
| 2 C4H10 (l) | + | 13 O2 (g) | | 8 CO2 (g) | + | 10 H2O (g) | H | = | -5315 kJ |
The equation tells us that 8 moles of CO2 (g) releases 5317 kJ , since the relationship is a direct one then 2 moles would release one quarter as much heat or:
2 moles would release 1,329 kJ.
| 8 moles | | 5315 kJ | |
| = |
| x = 1,329 kJ |
| 2 moles | | x kJ | |
important
procedure | To calculate heat changes using equations we will perform the following steps
1) calculate the number of moles of substance reacted or formed.
2) create a proportion using the balances and heat in the chemical equation.
3) solve for missing quantity.
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| | Calculate the amount of heat released when 25 grams of C4H10 (l) is burned in oxygen using the equation above.
Solution:
1) Calculate the number of moles of C4H10 (l)
mole = mass / molar mass ( gram molecular weight )
mole = 25 g / 58.14 g/mol
mole = 0.4299 mol.
2) Proportion
| 2 mol | | 5315 kJ |
| = |
|
| 0.4299 mol | | x kJ |
3) Solving for x, by cross multiplying
x = 1,142 kJ
4) 1,100 kJ of heat would be released when 25 grams is burned.
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| | Check your understanding Question:
Calculate the amount of heat absorbed in kJ when 120 grams of Iron (III) oxide is decomposed by the reaction below.
|
click to view answer |
| 2 Fe2O3 (s) | | 4 Fe (s) | + | 3 O2 (g) | | H | = | 1625 kJ |
| |
Assignment : Calculating Heats using Ratio Method
click on the notepad and complete assignment
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