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Thermochemistry

Enthalpy

Hess’s Law: The enthalpy change (ΔH) of an overall or net process is the sum of the enthalpy changes for the

individual steps in that process.

Types of reactions:

Endothermic Reactions Exothermic Reactions Take in heat (products have less potential energy than

reactants) Release heat (products have less potential energy than

reactants) A + B + heat C

A + B C + heat

A + B C ΔH = +ve (enthalpy is positive) A + B C ΔH = -ve (enthalpy is negative)

Example:

The following heats of reaction are known:

ClF(g) + ½ O2(g) ½ Cl2O(g) + ½ F2O ΔH0= 83.7 kJ/mol

2F2(g) + O2(g) 2F2O(g) ΔH0= -43.4 kJ/mol

Calculate the ΔH for the following reaction:

4ClF(g) + O2(g) 2Cl2O(g) + 2F2(g) ΔH0= ?

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SOLUTION:

Step 1: The objects in blue are those we want to have in the final equation, the objects in red are those we

want to cancel:

ClF(g) + ½ O2(g) ½ Cl2O(g) + ½ F2O ΔH0= 83.7 kJ/mol

2F2(g) + O2(g) 2F2O(g) ΔH0= -43.4 kJ/mol

---------------------------------------------------------------------------------

4ClF(g) + O2(g) 2Cl2O(g) + 2F2(g) ΔH0= ?

Step 2: Your goal in these questions is to add the reactions given until they equal the one asked for.

- When you multiply a reaction, the ΔH is multiplied by the same factor (e.g. 3× rxn1 => new ΔH = 3×ΔH). When you divide a reaction, the ΔH is similarly divided

- When you flip a reaction, the ΔH changes sign (e.g. flip a reaction with ΔH = 59 kJ, ΔH is now –59 kJ)

- When you add reactions, you add ΔHs (e.g. ΔH1 + ΔH2 = ΔHnet). Reagents on the same side add, reagents on opposite sides cancel/substract.

4(ClF(g) + ½ O2(g) ½ Cl2O(g) + ½ F2O ΔH0= 83.7 kJ/mol) +

-(2F2(g) + O2(g) 2F2O(g) ΔH0= -43.4 kJ/mol)

---------------------------------------------------------------------------------

4ClF(g) + O2(g) 2Cl2O(g) + 2F2(g) ΔH0= 83.7×4 + -43.4×-1=378.2 kJ/mol

Uses of Hess’s Law:

ΔH of any reaction (rxn) can be calculated using ΔHrxn = ∑ΔHf0products – ∑ΔH f

0reactants

(the heat of reaction is equal to the sum (∑) of the heats of formation of the products minus the sum (∑) of the

heats of formation of the reactants). Here is why:

Let’s start with a combustion reaction:

C6H12O6(s) + O2(g) CO2(g) + H2O(g), first we must balance the reaction:

1) Balance C/H using CO2 and H2O:

C6H12O6(s) + O2(g) 6CO2(g) + 6H2O(g)

2) Balance O using O2:

6Oxy. xOxy. 12 Oxy. 6 Oxy.

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g)

Always balance elements last - changing their coefficient does not affect any other reagent in the reaction.

3) If the question asks for the coefficients of the equation and you have a fraction/decimal (e.g. ½ or 0.5, etc.),

multiply the entire equation (all coefficients) by 2 to get whole numbers (the diploma numerical response only

accepts whole number coefficients for balancing reactions).

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Now the fun part, ΔHrxn = ∑ΔHproducts – ∑ΔHreactants:

6H2(g)+ 3O2(g) 6H2O(g) ΔH= -1450.8 kJ

6C(s) + 6O2(g) 6CO2(g) ΔH= -2361.0 kJ

C6H12O6(s) 3O2(g) + 6C(s) + 6H2(g) ΔH= +1273.3 kJ

-----------------------------------------------------------------------------------------------------------------------

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g)

ΔHrxn = 6×(ΔHf0 CO2(g)) + 6×(ΔHf

0 H2O(g) ) – (ΔHf0 C6H12O6(s)) = ∑ΔHproducts – ∑ΔHreactants

ΔH is written as kJ but can be expressed as kJ/mol

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ΔHrxn = -2538.5 kJ

This reaction has a heat of -2538.5 kJ per 1 mole C6H12O6(s) OR

-2538.5 kJ per 6 moles of H2O = -423.1 kJ/mol water created)

The ΔH symbol explained:

delta (Δ) stands for final - initial.

ΔHf0

the zero stands for standard conditions (SATP) Standard Ambient Temperature and Pressure: 25 ºC = 298 K and 1 atm = 101.3 kPa the subscript “f” stands for the type of reaction: f = formation, c = combustion rxn = reaction decomp. = decomposition (opposite of “f”), vap = vaporization/condensation (liquid gas) fus = fusion/melting (solid liquid)

Note that STP (Standard Temperature and Pressure) is a different set of conditions:

0 ºC = 273 K, 1 atm = 101.3 kPa

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There are multiple ways to represent Hess’s law:

Notice that in the above reaction, the more V-Cl bonds that are formed, the more exothermic the reaction is!

This means forming bonds releases energy. The more energy released, the more stable the products (they are

less energetic now!). It takes energy to break bonds (endothermic) however making bonds releases energy

(exothermic).

In some diagrams we do not show the activation energy and instead focus on the differences in energy

between the different products generated:

This allows us to measure the stability of the different products and calculate the differences in their heats of

reaction. For example, what is the heat of the following reaction?

VCl2 + Cl2 VCl4

From the above diagram, this is simply -300 kJ – (-200kJ) = -100 kJ!

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Catalysts Explained

Consider the following reaction:

CH3Cl + I− → CH3I + Cl−

This reaction is a substitution where Cl is substituted with I. The reaction has a high activation energy and

proceeds through an activated complex. If you look below, this complex is a carbon with what looks to be five

(5!!!) bonds – this is an extremely unstable species and requires extra energy beyond the ∆H to form, hence

we have an “activation energy” (Ea).

The activation energy is the minimum energy needed for the reaction to proceed. A catalyst lowers this energy

by creating a slightly different activated complex that is easier to reach. During the reaction, the catalyst is

regenerated (it makes a different activated complex and then goes back into solution to work again).

Another way to make sure more molecules have the energy needed to form products and overcome the

activation energy is to raise the temperature – this way the molecules have more kinetic energy and therefore

more collisions that successfully result in a product.

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Reactions you have to know/distinguish:

Combustion: C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ΔHrxn = -2538.5 kJ

Respiration: C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l) ΔHrxn = -2802.5 kJ

Photosynthesis: 6CO2(g) + 6H2O(g) sunlight C6H12O6(s) + 6O2(g) ΔHrxn = +2538.5 kJ

Note that respiration is more efficient (releases more heat per mole of sugar) than combustion, this is because

extra energy does not have to be expended in order to vaporize water.

As in the diagram below, energy must be input to melt and vaporize water (an endothermic process)

Relative Potential Energy and Phase Changes of Water

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Heat

The law of conservations of energy states that energy is neither created or destroyed. Therefore:

Heat lost + Heat gained = 0

Heat lost = -Heat gained

Qlost + Qgained = 0

Qlost = - Qgained

*The negative is important as it makes sure you have the right sign for ΔH, heat capacity, etc.

Q is described by one of three terms in Chemistry 30:

Q = mcΔT (used for a single substance heating/cooling) (kinetic energy change – Ek)

CΔT (used for a calorimeter composed of multiple substances) (kinetic energy change – Ek)

nΔH (used for a reaction or phase change) (potential energy change – Ep)

m = mass c = specific heat capacity (J/g°C) ΔT = final temperature - initial

C = heat capacity (J/°C) ΔT = final temperature - initial

n = moles of reactant ΔH= change in enthalpy

Example:

When 25.0 g of potassium hydroxide are added to a calorimeter containing 100 mL of water and composed of

100 g of copper and 50.0 g of aluminum at an initial temperature of 25.0 °C, the temperature of the system

rose to 77.0 °C. What is the heat of solution of potassium hydroxide?

Start by stating

Heat lost = -Heat gained and listing the variables.

Then verbalize the problem:

Heat lost by KOH = - Heat gained by copper + aluminum + water

Then, substitute in terms for each heat transfer:

nΔH = -( mcΔTcopper + mcΔT aluminum + mcΔTwater)

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Now that we have an equation, we can put in the variables we know:

(25 g KOH (1 𝑚𝑜𝑙

56.11 𝑔)) ∆H = − ((100𝑔𝐶𝑢)(

0.385𝐽

𝑔°𝐶)((77 − 25)°𝐶) + mcΔTaluminum + mcΔTwater)

*for space reasons Al and water were not expanded, cAl = 0.897 J/g°C, cwater = 4.19 J/g°C.

This gives:

(0.446 mol KOH )∆H = −(2009.7 𝐽 + 2341.17 J + 21,871.8 J), ∆H = −58,797 J

mol KOH= −

58,8 kJ

mol KOH

Which is very close to the literature value of -57.61 kJ/mol (Wikipedia)

Notice that ΔH uses kJ/mol whereas c uses J/g˚C. Make sure you use the right units when solving!

When heating a single substance it undergoes phase changes at specific temperatures:

Heating Curve for 100 g of water:

Notice that the slopes of the solid, liquid and gas phase of water are different – indicating different specific heat

capacities (c).

There are two plateaus for the phase changes – this is because they are potential energy changes.

Temperature changes on the other hand are kinetic energy changes.

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Example:

Build an equation that gives the amount of heat is required to warm 100 g of water from -20 °C to + 120 °C.

Start with

Q = heat gained by water

Then, we put in all five sections of the graph as terms in the equation:

Q = mcΔTsolid + nΔHs→l + mcΔTliquid + nΔHl→g + mcΔTgas

The overall heat, Q is positive if the heat is entering water and negative if it is entering the environment