Thermochemistry Review answer

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Thermochemistry Review Session 30–A1.1k recall the application of Q = mc∆t to the analysis of heat transfer

Kinetic energy (Ek)

 Kinetic energy (Ek) is the energy of motion of ions, atoms, molecules (Translational, rotational, vibrational motion). Ek is temperature dependant.

 Different substances will heat up and cool down at different rates, depending on their molecular makeup. Metals tend to heat up quickly, and will cool down quickly.

 Water tends to take more energy to heat up, but once it gets hot, it tends to take longer to cool down. To make this concept of heating and cooling more quantitative, we use heat capacity.  Specific heat capacity shows us how a substance changes temperature. Substances like metals

have low specific heat capacities. Substances like water have high specific heat capacities.

Examples of Problems You Will Be Expected To Do For Kinetic Energy Changes Ek = mc∆t

Where:

Ek = kinetic energy (J)

m = mass of substance that is heating or cooling (g) c = specific heat capacity of the substance (J/g·°C) ∆t = change in temperature (°C) where ∆t = tfinal - tinitial

If you are given any 3 of the 4 values, you will be expected to find the 4th Ek has a negative value when something loses heat (temperature decreases)

Ek has a positive value when something gains heat (temperature increases)

Remember that Ek , m , c, t values are all for the same substance

Example 1:

Find the energy needed to heat 150.0 g of water from 25.3 °C to 75.0 °C. Ek = mc∆t

Ek = 150.0 g x 4.19 J/g°C x (75.0°C – 25.3 °C)

Ek = 31200 J

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Potential Energy (Ep)

30–A1.2k explain, in a general way, how stored energy in the chemical bonds of hydrocarbons originated from the sun

Potential energy (Ep) is stored energy. Solar energy from the sun is converted to chemical Ep by plants and stored as carbohydrates (photosynthesis). Plants and animals convert this Ep to other forms of energy for living using cellular respiration.

30–A1.3k define enthalpy and molar enthalpy for chemical reactions

The change in potential energy is dependent upon the amount of substance (AKA moles), and the type of substance (not only what it is, but also what state it is in). The type of substance is called the molar enthalpy (∆H).

We can have changes in potential energy for:

 Phase changes – melting/freezing and vaporization/condensation – here intermolecular bonds break or are formed. ∆Hfus/∆Hvap

 Dissociation – when ionic compounds dissolve. ∆Hsoln

 Chemical reactions – where intramolecular bonds break and reform. ∆Hreaction

So for any calculation involving potential energy, we can use the formula: Ep = n∆H

Where Ep = Potential Energy

n = number of moles of a substance ∆H = molar enthalpy for that substance

Remember, for an equation, if you change the number of moles, then you must change the ∆H by the same factor.

Enthalpy of reaction will change directly with the coefficients in a chemical reaction.

H2(g) + ½ O2(g)  H2O(l) ∆H°f = -285.8 kJ

2H2(g) + O2(g)  2H2O(l) ∆H°f = -571.6 kJ

Chemical reactions are reversible, and if you reverse the reaction, then you need to reverse the sign on the ∆H as well.

H2(g) + ½ O2(g)  H2O(l) ∆H°formation = -285.8 kJ

H2O(l)  H2(g) + ½ O2(g) ∆H°decomposition = +285.8 kJ

You may also see ∆H = n∆H, because Alberta Learning does not separate enthalpy from molar enthalpy. We use Ep because we think it makes

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Progress of Reaction Ep C2H12O6(s) + 6O2(g)

6CO2(g) + 6H2O(l)

∆H = -2802.5kJ

Exothermic Reaction

Progress of Reaction Ep

6CO2(g) + 6H2O(l)

C2H12O6(s) + 6O2(g)

∆H = +2802.5kJ

Endothermic Reaction

30–A1.4k write balanced equations for chemical reactions that include energy changes

30–A1.5k use and interpret ∆H notation to communicate and calculate energy changes in chemical reactions

30–A2.2k explain the energy changes that occur during chemical reactions, referring to bonds breaking and forming and changes in potential and kinetic energy

30–A2.3k analyze and label energy diagrams of a chemical reaction, including reactants, products, enthalpy change and activation energy

You can be expected to communicate the enthalpy in a reaction three ways:

1 Using ∆H notation and written to the right of an equation.

2 Using ∆H notation and written within the equation.

3 Visually using a potential energy diagram.

In endothermic reactions overall, energy is added (photosynthesis):

2802.5 kJ + 6CO2(g) + 6H2O(l)  C2H12O6(s) + 6O2(g) ∆H is positive

In exothermic reactions overall, energy is released (cellular respiration):

C2H12O6(s) + 6O2(g)  2802.5 kJ + 6CO2(g) + 6H2O(l) ∆H is negative

∆Hrxn can also be shown visually with potential energy diagrams that show chemical potential energy change over the progress of a reaction

Cellular Respiration Photosynthesis

.

Less Ep More Ep

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Progress of Reaction Ep C12H22O11(aq) + 12O2(g)

12CO2(g) + 11H2O(l)

∆H = -5640.3 kJ

Combustion of Sugar

Example 1:

When sucrose is burned in our bodies with excess oxygen, carbon dioxide, water and 5640.3 kJ of energy are produced according to the following equation:

C12H22O11(aq) + 12O2(g)  12CO2(g) + 11H2O(l)

Rewrite the equation by (a) using ∆H notation, (b) placing the energy term within the equation, and (c) drawing a potential energy diagram.

a) C12H22O11(aq) + 12O2(g)  12CO2(g) + 11H2O(l) ∆H = -5640.3 kJ

b) C12H22O11(aq) + 12O2(g)  12CO2(g) + 11H2O(l) + 5640.3 kJ

c)

Diploma Examination Practice

Use the following information to answer the next question.

1. If this equation is rewritten to show the production of one mole of CH4(g) and the energy is expressed

as a term in the equation, then the energy will be A. 587.7 kJ on the reactant side

B. 1763.0 kJ on the reactant side C. 587.7 kJ on the product side D. 1763.0 kJ on the product side

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Calorimetry:

30–A1.8k use calorimetry data to determine the enthalpy changes in chemical reactions

30–A1.9k identify that liquid water and carbon dioxide gas are reactants in photosynthesis and products of cellular respiration and that gaseous water and carbon dioxide gas are the products of hydrocarbon combustion in an open system

Use the principle of Heat Transfer to do calorimetry calculations:

1. Find the energy gained by the calorimeter (Ek = mc∆T)

2. Use the Principle of Heat Transfer to flip the sign for energy change by the reaction.

3. Use Energy term, and moles of substance reacted to calculate ( )

Experimental method to find heat (enthalpy) absorbed or released in:

Dissolving Chemical reactions

Calorimetry is based on the principle of heat transfer where energy gained by one substance = energy lost by another

Three types:

Styrofoam cup – assumed to be perfectly insulated. Good for aqueous solutions, and reactions.

1. Find E absorbed (released) by the calorimeter (Ek = mc∆t)

2. Use the Principle of Heat Transfer Ep reaction = - Ek calorimeter

3. Find E released (absorbed) by the reaction (Ep = n∆H)

Metal Can – finding heat of a flame, or combustion reactions. You need to include mc∆T of the can as well as the water.

1. Find E absorbed (released) by the calorimeter (Ek = = mc∆Twater + mc∆Tmetal)

3. Find E released (absorbed) by the reaction (Ep = n∆H)

Bomb calorimetry – Reaction occurs in oxygen gas, and the reaction centre is surrounded by liquid water. Used for explosive reactions all heat needs to be transferred to the water, and so final products are CO2(g) and H2O(l).

1. Find E absorbed (released) by the calorimeter (Ek = C∆t)

(C is the heat capacity of the entire bomb calorimeter)

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On diploma examinations, the general principle that will be followed is that if combustion reactions are performed empirically in a bomb calorimeter, liquid water will be the product, and if the combustion occurs in an ambient

environment and a theoretical heat of combustion is to be determined, the product will be water vapour.

Use the following information to answer the next two questions.

3. The amount of energy involved in this change is ___________ kJ.

4. The molar heat of solution for NaOH(s) is –44.6 kJ/mol. If 25.0 g of NaOH(s) is dissolved in water in

a calorimeter, the heat released inside the calorimeter is A. 27.9 kJ

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Enthalpy of Formation

30–A1.6k predict the enthalpy change for chemical equations using standard enthalpies of formation

30–A2.1k define activation energy as the energy barrier that must be overcome for a chemical reaction to occur

30–A2.4k explain that catalysts increase reaction rates by providing alternate pathways for changes, without affecting the net amount of energy involved; e.g., enzymes in living systems.

These reactions are balanced, and record values based on one mole of the compound.

e.g. ½ H2(g) + ½ I2(s)  HI(g) ∆H°f = +26.5 kJ

e.g. 6C(s) + 6H2(g) + 3O2(g)  C6H12O6(s) ∆H°f = -1273.1kJ

The ∆H means the energy absorbed or released (on a per mole basis) The ° means it is standard conditions (25°C and 101.3 kPa)

The “f” means formation.

It makes a difference what state the substances are in, so please record the state!!!

e.g. H2(g) + ½ O2(g)  H2O(l) ∆H°f =-285.8 kJ

versus

H2(g) + ½ O2(g)  H2O(g) ∆H°f =-241.8 kJ

 Molar enthalpies of formation are given for compounds only, the molar enthalpy of formation for

elements are assumed to be zero for this course.



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Progress of Reaction Ep (kJ)

½ H2(g) + ½ I2(s)

HI(g)

∆H = +26.5 kJ ½ H2(g) + ½ I2(s) HI(g) ∆H°f = +26.5 kJ

0 265

H2(g) + ½ O2(g)

H2O(l)

∆H =-285.8 kJ H2(g) + ½ O2(g) H2O(l) ∆H°f =-285.8 kJ

0

285.8

 Most values are negative for molar enthalpy of formation.

 In reality, all exothermic reactions have an activation energy. This energy requirement prevents a

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6C(s) + 6H2(g)+3O2(g)

C6H12O6(s)

∆H =-1273.1kJ Uncatalyzed Reaction (No Enzyme)

0

Activation Energy

6C(s) + 6H2(g)+3O2(g)

C6H12O6(s)

∆H =-1273.1kJ Catalyzed Reaction (Enzyme)

0 Activation Energy

Catalysts simply reduce the activation energy required to move the reaction forward, and they get us to equilibrium (completion) more quickly, but do not release or absorb any more energy.

Visually an uncatalyzed reaction and a catalyzed reaction look like this:

Diploma Practice Questions:

5. The energy released when 1.00 mol of AgI(s) is formed from its elements is __________ kJ. Use the following information to answer the next two questions.

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6. When 10.0 g of ethane gas was originally formed from its elements, the decrease in enthalpy was A. 3.92 kJ

B. 28.2 kJ C. 84.7 kJ D. 255 kJ

7. Liquid ethane, at a pressure of approximately 5 000 kPa and a temperature of –35°C, enters the plant via a pipeline. It passes through a series of heat exchangers and eventually emerges as a gas at 900 kPa and 35°C. This change in the ethane is primarily due to

A. an increase in potential energy and kinetic energy B. an increase in potential energy only

C. a decrease in density and kinetic energy

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Hess’s Law

30–A1.7k explain and use Hess’ law to calculate energy changes for a net reaction from a series of reactions

There is actually a reason why we make you do all that work with those Heats of Formation Reactions and tables!

We use those Heats of Formations to apply Hess’s Law.

Hess’s Law lets us calculate the energy released or required by a reaction, simply by adding up the energy of formation for each reactant and compound. We don’t have to perform a calorimetry experiment to find the energy absorbed or released by a reaction.

More Formally, Hess’s Law states:

If a chemical reaction can be expressed as the algebraic sum of two or more reactions, then its enthalpy of reaction (∆H rxn)is also the algebraic sum of the separate enthalpy of reactions (∆Hf).

If you can add the equations, you can add the ∆H’s We need to follow some basic rules to make Hess’s law work:

 Write a balanced equation for the reaction for which ∆H is needed.  Add all the H°f values of the compounds on the product side.  Add all of the H°f values on the reactant side.

 Remember to multiply any H°f on either the product side by the coefficient found in the original equation!

 Subtract the sum of the reactants from the sum of the products and you have the ∆H of the reaction.

It looks like this:

EXAMPLES FOR HESS’ LAW USING THE MATH METHOD: Example 1: You could be asked to find molar enthalpy:

Find the molar enthalpy of combustion for ethane, C2H6(g). Reaction is:

C2H6(g) + O2(g)  2CO2(g) + 3H2O(g) ∆H = ??

∆H = (3H2O(g) + 2CO2(g)) - (C2H6(g))

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Example 2: You could be asked to find the enthalpy of formation for a reaction, given the enthalpy of reaction:

Establish the standard enthalpy of formation for carbon disulphide, CS2(l), given the enthalpy of

combustion for CS2(l) is -1075.1 kJ/mol. Carbon disulphide burns in oxygen to produce carbon dioxide gas and sulphur dioxide gas.

Reaction is:

CS2(l) + 3O2(g)  CO2(g) + 2SO2(g) ∆H = -1075.3 kJ

-1075.3 kJ/mol = (-393.5 kJ/mol -593.6 kJ/mol) - ∆H°f CS2(l) -88.2 kJ/mol = -∆H°f CS2(l)

+88.2 kJ/mol = ∆H°f CS2(l)

Example 3: You could be asked to find the enthalpy change in a reaction, given the mass of a substance (we need to use stoichiometry here):

Natural gas consists of a mix of ethane and methane. What is the enthalpy released from burning 1.00 kg of ethane in a hot water heater?

Reaction is:

C2H6(g) + 7/2 O2(g) _{2CO2(g) + 3H2O(g)}

∆H°rxn = (2mol(-393.5 kJ/mol) + 3mol(-241.8 kJ/mol)) - ( 1mol(-84.7 kJ/mol)) ∆H°rxn = -1427.7 kJ/mol

This means that 1427.7 kJ of heat energy are released when 1 mole of ethane burns completely to give gaseous products - as it would in a hot water heater. We have 1.00 kg ethane

The energy released when 1.00 kg of ethane is burned is:

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Hess’s Law: Including the Equations

You will only be able to use the math method for Hess’s law when you are asked to find:

 Enthalpy of reaction

 Enthalpy of formation for either reactants or products, BUT there can only be one compound!! Otherwise

you will need to solve the problem using the equation method using heats of formation data, or enthalpy data given to us in a question.

For example:

C12H22O11(aq) + 12O2(g)  12CO2(g) + 11H2O(l)

You could use the math method to find the ∆Hrxn, or you could find the Hf for sugar. It gets to be more

difficult to find the Hf for either water or carbon dioxide, because you would have two unknowns. It

tends to be easier in that case to solve it using the equation method.

We need to follow some basic rules to make Hess’s law work using equations:

1. Write a balanced equation for the reaction for which ∆H is needed (this is the original equation). 2. Using the heats of formation table, write out the balanced equation to form one mole of each product.

3. Using the heats of formation table, write out the balanced equation to decompose one mole of each

reactant (change the sign of the ∆H!).

4. Multiply any of the equations so they match up with the moles of that compound in the original equation (remember to multiply the ∆H as well!).

5. Ignore elements on either side (they will sort themselves out by the end). 6. Add the equations, and they should add up to the original equation.

Example 1: You can find the molar enthalpy of a reaction, just like using the math method:

Find the molar enthalpy of combustion for methanol, CH3OH(l).

Original Equation

CH3OH(l) + O2(g)  CO2(g) + H2O(g) ∆H = ??

C(s) + O2(g)  CO2(g) ∆H = -393.5 kJ

2H2(g) + O2(g)  2H2O(g) ∆H = 2(-241.8 kJ)

CH3OH(l)  C(s) + 2H2(g) + 1/2O2(g) ∆H = +239.0 kJ

___________________________________________________ Net Reaction:

CH3OH(l) + O2(g)  CO2(g) + H2O(g) ∆H = -638.1 kJ

Example 2: You could be expected to find the enthalpy of formation for a reaction, given the molar enthalpy of an overall reaction.

Calculate the molar heat of formation for ammonia, NH3(g), using the reaction given here for the combustion of

ammonia.

Given:

2NH3(g) + 3/2 O2(g)  N2(g) + 3H2O(g) ∆H = -633.2kJ We want to find:

½ N2(g) + 3/2 H2(g)  NH3(g) ∆H = ??

N2(g) + 3H2O(g)  2NH3(g) + 3/2 O2(g) ∆H = +633.2kJ

3H2(g) + 3/2 O2(g)  3H2O(g) ∆H = 3(-241.8 kJ)

___________________________________________________ Net Reaction:

N2(g) + 3 H2(g)  2NH3(g) ∆H = -92.2 kJ

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Diploma Examination Practice

8. The molar enthalpy of reaction for the fermentation of glucose is +/–_________ kJ/mol.

9. The balanced equation and the enthalpy change for the cellular respiration of glucose can be represented as

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10. The amount of energy absorbed when 0.350 mol of VCl4(l) decomposes to form VCl2(s) and Cl2(g) is

_____ kJ.

Use the following equation to answer the next question.

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12. How much heat is produced when 1.00 g of butane in a disposable lighter is completely burned to form gaseous carbon dioxide and water vapour?

A. 45.7 kJ

B. 124.7 kJ

C. 2 656.5 kJ

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