Thermochemistry

An Introduction to Thermochemistry The Laws of Conservation of Energy states that: the total energy of the universe is a constant; energy cannot be created or destroyed.

First Law of Thermodynamics: Universe = System + Surroundings ∆Euniverse = ∆Esystem + ∆Esurroundings = 0

Open system  open to the surroundings. Closed system  closed to the surroundings.

Types of Energy: 1. Kinetic Energy: the energy of motion.

2. Potential energy: stored energy.

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Calculating Heat Transfer The formula for calculating heat transfer is as follows: q = m ∙ c ∙ ∆T

Where q is the amount of heat in J (joules), m is the mass in g (grams), c is the specific heat capacity in J/g°C (the values for this variable are usually found in a table), and ∆T is the change in temperature in °C (degrees Celsius). Moreover, ∆, which is called Delta, means “change in,” and in this case: “change in” temperature.

Things to Remember:

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Energy and heat are the same.

•

If mass doubles, so does heat!

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Questions for Calculating Heat Transfer 1. 100 g of Ethanol at 25°C is heated until it reaches 50°C. How much heat does the ethanol gain?

q = m ∙ c ∙ ∆T q = (100 g)(2.46 J/ g°C)(25°C) q = 6150 J

m = 100 g ∆T = 25°C c = 2.46 J/ g°C q = ?  6150 J

2. Beaker A contains 50 g of liquid at room temperature. The beaker is heated until the liquid gains 10°C. Beaker B contains 100 g of the same liquid at room temperature. This beaker is also heated until the liquid gains 10°C. In which beaker does the liquid absorb the most heat?

Beaker A q = m ∙ c ∙ ∆T

m = 50 g

q = (50 g)(4.184 J/ g°C)(10°C) q = 2092 J

∆T = 10°C c = 4.184 J/ g°C q = ?  2092 J

Beaker B q = m ∙ c ∙ ∆T

m = 100 g

q = (100 g)(4.184 J/ g°C)(10°C) q = 4184 J

∆T = 10°C c = 4.184 J/ g°C q = ?  4184 J

* When your mass (g) doubles, then so does your heat!

3. How much heat is required to raise the temperature of 789 g of liquid ammonia from 25.0°C to 82.7°C? Remember to subtract 82.7°C from 25.0°C to determine ∆T, as we usually do not use negative temperatures, yet.

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q = m ∙ c ∙ ∆T q = (789 g)(4.70 J/ g°C)(57.7°C)

m = 789 g ∆T = 57.7°C

q = 213968.91 J

c = 4.184 J/ g°C

q = 214 Kj

q = ?  214 Kj

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* Remember if the final number for “q” is large, you can convert it to Kj (kilojoules) by using the very elementary “KHD[]dcm”. If you are going left, then divide by 1000, if you are going right, then multiply by 1000.

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Enthalpy The change in heat, specific to phase change – changing state from solid to liquid, etc. – that does not involve temperature change.

---------- SUBLIMINATION ---------melting

Solid

----->

boiling/vaporizing

Liquid

<----solidifying/freezing

----->

Gas

<----condensing

---------- SUBLIMINATION ----------

The formula for calculating enthalpy is as follows: H = n ∙ ∆H

Where H is the enthalpy change in Kj (kilojoules), n is the number of moles in mols, and ∆H (vaporization, condensation, melting, freezing, etc.) is the molar enthalpy in Kj/mol. Moreover, the value for ∆H is usually found in a table that is already provided, or easily found.

Molar Enthalpy of Vaporization/Boiling, ∆Hvap: The enthalpy change for the state change of one mole from liquid to gas.

Molar Enthalpy of Condensation, ∆Hcond: The enthalpy change for the state change of one mole of a substance from gas to a liquid.

Molar Enthalpy of Melting, ∆Hmelt: The enthalpy change for the state change of one mole of a substance from solid to liquid.

Molar Enthalpy of Freezing, ∆Hfre: The enthalpy change for the state change of one mole of a substance from liquid to solid.

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Questions for Enthalpy Change 1. An ice cube with a mass of 8.2 g is placed in some lemonade. The ice cube melts. How much heat does the ice cube absorb from the lemonade as it melts?

8.2 g of H20

1 mol

=

18.02 g

=

0.46 mols of H20

H = n ∙ ∆H H = (0.46 mols)(6.02 Kj/mol)

H = ?  2.77 Kj n = 0.46 mols

H = 2.77 Kj

∆H = 6.02 Kj/mol

2. What mass of Diethly Ether, C4H10O, can be vaporized by adding 80.7 Kj of heat?

H = n ∙ ∆H

H = 80.7 Kj

80.7 Kj = n (29 Kj/mol) 80.7 Kj ÷ 29 Kj/mol = n

n = ?  2.78 mols ∆H = 29 Kj/mol

2.78 mols = n

2.78 mols of C4H10O

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=

74.1 g

=

205.998 g of C4H10O

1 mol

=

206 g of C4H10O

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Total Energy Changes

Things to Remember: • •

The values of “c” changes based on its physical state (“q”) on the curve. Remember that, when you are drawing/determining Total Energy Changes (Curves), use the following formulas: q = m ∙ c ∙ ∆T & H = n ∙ ∆H.

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Questions for Total Energy Changes 1. A metal bucket containing 532.1 g of ice at 0°C is placed on a wood-burning stove. The ice is melted and then heated to 45.21°C. (a) Draw a heating curve. (b) How much heat is required to melt the ice and warm the water to 45.21°C?

q = m ∙ c ∙ ∆T

(a)

m = 532.1 g

q = (532.1 g)(4.184 J/g°C)(45.21 °C)

∆T = 45.21°C

q = 100651.312 J

c = 4.184 J/g°C

q = 100.7 Kj

q = ?  100.7 Kj

H = n ∙ ∆H

H = ?  177.7 Kj

H = (29.52 mol)(6.02 Kj/mol)

n = 29.52 mol

H = 177.7 Kj

∆H = 6.02 Kj/mol

532.1 g of H2O =

1 mol 18.02 g

=

29.52 mols of H2O

(b) Etotal = q + H Etotal = 100.7 Kj + 177.7 Kj Etotal = 278.4 Kj

2. Calculate the total heat released when 200 g of water vapor at 300.0°C is cooled until it reaches -20.0°C. The specific heat capacity of ice is 2.01 J/g°C. The specific heat capacity of water vapor is 2.01 J/g°C. (a) Draw a heating curve.

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Questions for Total Energy Changes Continued… (a)

q1 = m ∙ c ∙ ∆T

m = 200 g

q1 = (200g)(2.01 J/g°C)(200°C)

∆T = 200°C

q1 = 80400 J

c = 2.01 J/g°C

q1 = 80.4 Kj

q1 = ?  80.4 Kj

H1 = n ∙ ∆H

H1 = ?  451.77 Kj

H1 = (11.1 mols)(40.7 Kj/mol)

n = 11.1 mols

H1 = 451.77 Kj

∆H = 40.7 Kj/mol

q2 = m ∙ c ∙ ∆T

m = 200 g

q2 = (200g)(2.01 J/g°C)(200

∆T = 100°C

q2 = 40200 J

c = 2.01 J/g°C

q2 = 40.2 Kj

q1 = ?  40.2 Kj

H2 = n ∙ ∆H

H2 = ?  66.82 Kj

H2 = (11.1 mols)(6.02 Kj/mol)

n = 11.1 mols

H2 = 66.82 Kj

∆H = 6.02 Kj/mol

q3 = m ∙ c ∙ ∆T

m = 200 g

q3 = (200g)(2.01 J/g°C)(-20.0°C) ∆T = 45.21°C q3 = -8040 J

c = 4.184 J/g°C

q3 = -8.04 Kj

q3 = ?  -8.04 Kj

q3 = + 8.04 (we are only using + right now)

200 g of H2O =

1 mol 18.02 g

=

11.1 mols of H2O

Etotal = q1 + h1 + q2 + h2 + q3 Etotal = 80.4 Kj + 451.77 Kj + 40.2 Kj + 66.82 Kj + 8.04 Kj Etotal = 647.23 Kj

Things to Remember: •

For questions such as the one featured above, repeat “q = m ∙ c ∙ ∆T & H = n ∙ ∆H” until there are no more “q#” or “H#” to solve for. Moreover, remember that the values of “c” changes based on its physical state (“q”) on the curve.

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Calorimetry Calorimetry is a way of measuring changes in kinetic energy using an isolated system.

Simple Calorimeter (Coffee-Cup): Hsystem = qwater n ∙ ∆H = m ∙ c ∙ ∆T

Where n is the number of moles in mols, and ∆H (vaporization, condensation, melting, freezing, etc.) is the molar enthalpy in Kj/mol, m is the mass in g (grams), c is 4.184 J/g°C, and ∆T is the change in temperature in °C (degrees Celsius).

Things to Remember: •

• •

Simple Calorimeters always involves water. Any talk of water (warming or cooling) in your question means you are using a Simple Calorimeter. You are usually always solving for “n” or “∆H”. The value of “c” will always be 4.184 J/g°C in a Simple Calorimeter.

•

Isolated/dissolving = warming water.

•

Always include Hsystem = qwater above your formula.

Bomb Calorimeter: Hsystem = qcal n ∙ ∆H = C ∙ ∆T

Where n is the number of moles in mols, and ∆H (vaporization, condensation, melting, freezing, etc.) is the molar enthalpy in Kj/mol, C is the specific heat capacity of the calorimeter in Kj/°C, and ∆T is the change in temperature in °C (degrees Celsius).

Things to Remember:

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Bomb Calorimeters never involve water.

•

You are usually always solving for “n” or “∆H”.

•

Always include Hsystem = qcal above your formula.

•

Combustion is usually involved with Bomb Calorimeters.

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Questions for Calorimetry 1. Suppose 4.6 of LiCl is dissolved in 32 ml of water. The water had an initial temperature of 22°C, which increased to 87.2°C when the reaction completed. Calculate the molar enthalpy of the LiCl.

Hsystem = qwater

n = 0.11 mols

n ∙ ∆H = m ∙ c ∙ ∆T (0.11 mols) ∆H = (32 g)( 4.184 J/g°C)( 65.2°C)

∆H = ? m = 32 ml (32 g)

∆H = 8729.50 J ÷ 0.11 mols ∆H = 79359.07 J/mol

c = 4.184 J/g°C ∆T = 65.2°C

∆H = -79.36 Kj/mol

4.6 g of LiCl

1 mol

=

42.4 g

=

0.11 mols of LiCl

2. The molar enthalpy of combustion for a gasoline assumed to be octane (C8H18) is -1.3 Mj/mol. A particular engine has a heat capacity of 105 Kj/°C. What is the minimum mass of octane that must be burned to change the temperature of an engine from 18°C to 120°C?

Hsystem = qcal

n=?

n ∙ ∆H = C ∙ ∆T n (-1300 Kj/mol) = (105 Kj/°C)( 102°C)

∆H = -1.3 Mj/mol (-1300 Kj/mol) C = 105 Kj/°C

n = 10710 Kj ÷ -1300 Kj/mol n = -8.24 mols

∆T = 102°C

n = 8.24 mols

8.24 mols of C8H18

=

114.26 g 1 mol

=

941.50 g of C8H18

Things to Remember: • •

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Exothermic  Gives off heat (Increase Temp)(Increase T) – ∆H is negative Endothermic  Takes in heat (Decrease Temp)(Decrease T) – ∆H is positive

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Hess’s Law of Heat Summation Chemists can determine the enthalpy change of any reaction using an important law, known as Hess’s law of heat summation, which states: the enthalpy change of a physical or chemical process depends only on the beginning conditions (reactants) and the end conditions (products). The enthalpy change is independent of the pathway of the process and the number of intermediate steps in the process. It is the sum of the enthalpy changes of all the individual steps that make up the process.

Things to Remember: •

You are trying to obtain an overall formula that is the same as your “Target”.

• •

You may need to multiply or divide one of your “Givens”. You may need to reverse one of your “Givens,” but if you do, be sure to change the sign of the “∆H°r”. For example: if it is +50 Kj, you must change it to -50 Kj, and vice-versa.

•

Be sure to watch what state each compound is in ((g), (l), (s)). The question you are being asked may try to trick you!

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Questions for Hess’s Law of Heat Summation 1. Target: 2NO (g) + O2 (g)  N2O4 (g)

Given: (1) N2O4 (g)  2NO2 (g) (2) NO (g) + ½ O2 (g)  NO2 (g)

(2) 2(NO (g) + ½ O2 (g)  NO2 (g)

∆H°r = ?

∆H°r = +58 Kj ∆H°r = - 56 Kj

∆H°r = - 56 Kj)

(2) 2NO (g) + O2 (g)  2NO2 (g)

∆H°r = - 112 Kj

rev (1) 2NO2 (g)  N2O4 (g)

∆H°r = - 58 Kj

2NO (g) + O2 (g)  2NO2 (g)

∆H°r = - 112 Kj

2NO2 (g)  N2O4 (g)

∆H°r = - 58 Kj

--------------------------------------------------------------------------------------------2NO (g) + O2 (g)  N2O4 (g)

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∆H°r = - 170 Kj

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Hess’s Law of Heat Summation (Formations) Hess’s Law of Heat Summation (Formations) is used to calculate ∆H for a balanced chemical equation.

∆H°rxn = ∑(n ∙ ∆H°f products) - ∑(n ∙ ∆H°f reactants)

Where ∆H°rxn is the reaction and is in Kj, ∑ simply means the “Sum of”, n is the number of mols from the balanced chemical equation, ∆H°f products is the standard heat of formation for each of the products listed in Kj/mol, and ∆H°f reactants is the standard heat of formation for each of the reactants listed in Kj/mol.

Things to Remember: •

The values of both “∆H°f products” and “∆H°f reactants” are found in tables. If “∆H°f products” or “∆H°f reactants” are listed in the Periodic Table, then the value is automatically zero!

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Questions for Hess’s Law of Heat Summation (Formations) 1. C2H4 (g) + H2(g)  C2H6(g)

∆H = ?

∆H°rxn = ∑(n ∙ ∆H°f products) - ∑(n ∙ ∆H°f reactants) ∆H°rxn = [ 1(-84.7 Kj/mol) ] – [ 1(52.3 Kj/mol) + 1(0 Kj/mol) ] ∆H°rxn = [ -84.7 Kj/mol ] – [ 52.3 Kj/mol ] ∆H°rxn = - 84.7 Kj/mol - +52.3 Kj/mol ∆H°rxn = -137 Kj

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Hess’s Law of Heat Summation (Potential Energy Diagrams) Exothermic Reaction Products must have a lower energy (Kj) than reactants (high).

Endothermic Reaction Products must have a higher energy (Kj) than reactants (low).

Things to Remember: •

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If the products are below the reactants, then your value of ∆H°rxn will be negative. If the products are above the reactants, then your value of ∆H°rxn will be positive.

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Questions for Hess’s Law of Heat Summation (Potential Energy Diagrams) 1. C2H4 (g) + H2(g)  C2H6(g)

∆H = ?

∆H°rxn = ∑(n ∙ ∆H°f products) - ∑(n ∙ ∆H°f reactants) ∆H°rxn = [1(-84.7 Kj/mol)] – [1(52.3 Kj/mol) + 1(0 Kj/mol)] ∆H°rxn = [-84.7 Kj/mol] – [52.3 Kj/mol] ∆H°rxn = - 84.7 Kj/mol - +52.3 Kj/mol ∆H°rxn = -137 Kj

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