Chem 2 IB Unit 4 – Energy and Thermochemistry 

Energy – the capacity to do workTwo major types

Kinetic – E of motionPotential – E stored-up (can do work, but not doing it yet) 

W=F x d

First Law of ThermodynamicsThree Parts:

E of universe is constantE cannot be created or destroyedE can be transformed

Average KE of particles temperatureSo how are temperature and heat related?

•Heat – how energy is transferred from one substance to another

•Higher kinetic energy lower kinetic energy•Hot cold •When there are energy changes in chemical reactions there are two parts to the universe. The SYSTEM (part of the rxn i.e. reactants/products) and the SURROUNDINGS (everything else). •The E w/in a system is called the INTERNAL ENERGY and is = to the sum of KE and PE of all particles in the system. •It’s impossible to truly isolate system from surroundings during rxn; more important to measure Δ in E of a system when rxn happens. The ΔE will occur in form of work (W) or heat (q), so… ΔE=q+w

W = F x d = F x Δh Since P=F/A and F=PA, then…

W = P x A x Δh = PΔV

Remember: the sign for Δs in E always represents the system’s point of view. A positive ΔV means system has done work on surroundings (lost E), so we represent W as –PΔV

ΔE = q-PΔVThis value for heat (q) is important, b/c rxns are usually in open containers where V can Δ; so, this is given a special symbol, ΔH, and called “standard enthalpy change” for the rxn

•The most common type of work done by chemical reaction is expansion (ex – combustion in a piston cylinder)

•Equation in a piston (Δh = Δ in height of piston)

Exothermic - Energy is given off (lost) by the system

Endothermic - Energy is absorbed (gained) by the system

Type of Reaction

Heat energy Δ

Temperature Δ Relative enthalpies

Sign of ΔH

Exothermic Heat energy evolved

Becomes hotter Hp < Hr Negative ( – )

Endothermic Heat energy absorbed

Becomes colder Hp > Hr Positive ( + )

Here’s a table to help with endothermic/exothermic ideas:

IB ♥’s to ask questions about…Endothermic process Take in E → DECREASE in temperature Exothermic process give out E temperature INCREASE

What about relative stabilities? Show with an enthalpy level diagram.

Math with energetics

Specific and molar heatThe heating or cooling of a substance

1)Heat Capacity- the amount of energy a substance can absorb before its temperature is increased. General equation is: C = heat absorbed/increase in tempa) Molar heat capacity: the energy required to raise the temp of 1 mole of a substance by 1◦C. (units =J/mol·C)b) Specific Heat: the energy required to raise the temperature of 1 gram of a substance by 1◦C

symbol= cp units= J/g°C

examples: 4.18J/gC for H2O, 0.45J/gC for Fe, 0.71J/gC for carbon.

Bomb calorimeter

Equation for Specific heat

Equation using specific heat q= cp x m x ∆T where:

cp is specific heat

m=mass in grams∆T= change in temp.q= energy in joules

Temperature Changes

Measuring the temperature change in a calorimetry experiment can be difficult since the system is losing heat to the surroundings even as it is generating heat.

By plotting a graph of time vs. temperature it is possible to extrapolate back to what the maximum temperature would have been had the system not been losing heat to the surroundings.

9A time v temperature graph

•We use process of calorimetry to measure energy changes•We use water in our calorimeter because it can absorb/give off a lot of heat w/o Δ’ing temp no extremes (i.e. state changes, etc.)•In a calorimeter, Δ in enthalpy measured by Δ in temperature

Q=CΔT•Q=heat•C=heat capacity (not Cp – specific heat capacity)

•In Normal Lab conditions: (can Δ volume ) (not in a bomb calorimeter), q≈ΔH ΔH=CΔT

C = m x Cp•If Δ temp easily (ex – metals ) lower Cp

•If d/n/Δ temp easily (ex – plastic) higher Cp Q=m x Cp x ΔT (Same equation from last slide)

Some values to know…Cp H2O = 4.18 J/gCCm H2O = 75.3 J/molC (molar heat capacity J/molK)

ΔH=CΔT

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Ex: The Cp of Al is 0.900 J/g 0 C. How much energy is needed to raise the temperature of an 851 g block of Al from 22.8 to 94.6 0 C? What is the molar heat capacity of Al?

Ex: 15.0 g of copper is heated to 75.0 0C and added to 50.0g of water at 25.0 0C. If the final temperature of both is 26.4 0C, what is the specific heat capacity of the metal?

Average Bond Enthalpies(Same as Bond Dissociation Energy in Chem 1)

•Break substance up into individual gaseous diatomic molecules•Then, measure energy required to break them apart•Luckily for us, values already established listed in IBDB and on Handout•These values can be used to determine ΔH for a reaction

•Let’s try…(drumroll)…combustion of methane!

CH4 + 2O2 CO2 + 2H2O

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Pseudonyms (other names) for H Heat of Reaction: Hrxn heat produced in a chemical reaction

Heat of Combustion: Hcomb heat produced by a combustion reaction

Heat of Neutralization: heat produced in a neutralization reaction (when an acid and base are mixed to get water, pH = 7)

Heat of solution: Hsol heat produced by when something dissolves

Heat of Fusion: Hfus heat produced when something melts

Heat of Vaporization: Hvap heat produced when something evaporates

Heat of Sublimation: Hsub heat produced when something sublimes

Heat of formation: Hf change in enthalpy that accompanies the formation of 1 mole of compound from it’s elements (this has special uses in chemistry…)

5.1.1 Define the terms exothermic reaction, endothermic reaction and standard enthalpy change of reaction (ΔHo).

Heats of Formation

∆H for a reaction can be determined for a chemical change when you know the change in enthalpy for the formation (∆Hf) of each reactant and product.

B/C different conditions will result in different values for ∆Hf a Thermodynamic Standard State has been defined as 298.15 K (250C), 1 atm of pressure for each gas, and 1M concentration for each solution (n.b. – this is not STP).

Standard Heats of Formation - ∆Hfө:

The enthalpy change for the hypothetical formation of 1 mol of a substance in its standard state from the most stable forms of its constituent elements in their standard states.

To calculate the change in enthalpy for a reaction (∆Hrxn):

∆Hrxn = Σnp∆Hfө(products) - Σnr∆Hf

ө(reactants)

Ex: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

[1 mol(-394kj/mol) + 2 mol(-286kj/mol)] – [1 mol(-75 kj/mol) + 2(0 kj/mol)]

= (-394 kJ – 572 kJ) – (-75 kJ) = -966 kJ + 75 kJ = - 891 kJ

Now rewrite the equation as a Thermodynamic Equation for the formation of 1 mol of water:

½ CH4(g) + O2(g) → ½ CO2(g) + H2O(g) + 446 kJ

What about for other amounts reacted or formed? For example:

If only 5.83 g of water is produced, how much energy is released?

5.83 g x 1 mol x - 891 kJ = - 144 kJ 18.02g 2 mol H2O

n. b. - ∆Hfө for elemental substances Al, O2, S8, P4, etc. is always 0.

For reversible reactions – The sign for ∆H in the reverse rxn is opposite that for the forward rxn. So if the the forward rxn is exo, the reverse rxn is? Endothermic.

Hess’ Law – Indirect Enthalpy Calculations by Rearranging Reactions

Hess’ Law provides a way to calculate enthalpy changes even when the reaction cannot be performed directly.

If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy change for the individual steps

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This is because the value for ΔH is the same no matter what path is taken. Thus we can sum enthalpy changes for a set of reactions which together give us the overall reaction for which we are trying to determine the value of ΔH.

Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction.

Techniques Equations may be multiplied, divided, or reversed

and then added together to form a new equation If an equation is multiplied or divided the enthalpy

of the reaction is multiplied or divided by the same factor

If the direction an equation is reversed the sign of the enthalpy is the opposite as well

When adding equations together the enthalpies are added together as well

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Hess’ Law: Example 1

N2 (g) + O2 (g) 2 NO (g) H1 = +181 kJ

2 NO (g) + O2 (g) 2 NO2 (g) H2 = -113 kJ

Find the enthalpy change for

N2 (g) + 2 O2 (g) 2 NO2 (g)

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Hess’ Law: Example 1The required equation is really the sum of the

two given equationsSolution:

N2 (g) + O2 (g) 2 NO (g) H1 = +181 kJ

2 NO (g) + O2 (g) 2 NO2 (g) H2 = -113 kJ

-------------------------------------------------------------N2 (g) +2O2 (g)+ 2 NO (g) 2 NO (g) + 2 NO2 (g)

N2 (g) +2O2 (g) + 2 NO2 (g)

H = H1 + H2 = +181 kJ +(-113) = + 68 kJ

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Hess Law: Example 2 From the following reactions and enthalpy changes: 2 SO2 (g) + O2 (g) 2 SO3 (g) H = -196 kJ

2 S (s) +3 O2 (g) 2 SO3 (g) H = -790 kJ

Find the enthalpy change for the following reaction: S (s) + O2 (g) SO2 (g)

Solution:2 SO3 (g) 2 SO2 (g) + O2 (g) H = +196 kJ

2 S (s) +3 O2 (g) 2 SO3 (g) H = -790 kJ

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

Reversing the order of the first equation reverses the sign of H

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Hess Law Example 2 From the following reactions and enthalpy changes: 2 SO2 (g) + O2 (g) 2 SO3 (g) H = -196 kJ 2 S (s) +3 O2 (g) 2 SO3 (g) H = -790 kJ Find the enthalpy change for the following reaction: S (s) + O2 (g) SO2 (g)

2 SO3 (g) 2 SO2 (g) + O2 (g) H = +196 kJ 2 S (s) +3 O2 (g) 2 SO3 (g) H = -790

kJ--------------------------------------------------------------------------------------------------------------

2 SO3(g) +2 S(s) + 2 3 O2 (g) 2 SO3 (g)+2 SO2 (g) + O2 (g) H = -594 kJ

2 S(s) + 2 O2 (g) 2 SO2 (g) H = -594 kJ21

Hess Law: Example 2 From the following reactions and enthalpy changes: 2 SO2 (g) + O2 (g) 2 SO3 (g) H = -196 kJ 2 S (s) +3 O2 (g) 2 SO3 (g) H = -790 kJ Find the enthalpy change for the following reaction: S (s) + O2 (g) SO2 (g)

2 SO3 (g) 2 SO2 (g) + O2 (g) H = +196 kJ

2 S (s) +3 O2 (g) 2 SO3 (g) H = -790 kJ--------------------------------------------------------------------------------------------------------------

2 SO3(g) +2 S(s) + 2 3 O2 (g) 2 SO3 (g)+2 SO2 (g) + O2 (g) H = -594 kJ

2 S(s) + 2 O2 (g) 2 SO2 (g) H = -594 kJ S(s) + O2 (g) SO2 (g) H = -297 kJ

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Example: Use the following information to calculate ΔHcombθ for the

combustion of methane.

CH4(g) + O2(g) → CH2O(g) + H2O(g) ΔHθ = -284 kJ

CH2O(g) + O2(g) → CO2(g) + H2O(g) ΔHθ = -518 kJ

H2O(l) → H2O(g) ΔHθ = -44 kJ

Enthalpy Cycles

An enthalpy cycle is a way of representing energy changes that take place in a process or reaction. It is similar to the idea of Hess’s Law since ΔH values in parts of the cycle can be used to calculate an unknown ΔH value in another part of the cycle.

Consider the enthalpy cycle below:

3 C(s) + 4 H2(g) 5 O2(g) C3H8(g) + 5 O2(g)

3 CO2(g) + 4 H2O(g)

ΔH1

ΔH3 ΔH2

Notice that the reactants in equation 1 (ΔH1 above the arrow) could form the products in equation 3 (ΔH3). The products from equation 1 could also react to form the same products (ΔH2).

We can write reactions 2 and 3 as we would if we were solving a Hess’s Law problem:

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔH2

3 C(s) + 4 H2(g) 5 O2(g) + 3 CO2(g) + 4 H2O(g) ΔH3

Notice that if equation 2 is reversed then the CO2 and H2O are cancelled out and we get equation 1. (Work it out)

So what this tells us is that to determine the value for ΔH1 the value for ΔH3 could be added to the opposite (negative) of ΔH2 (Since the reverse of reaction 2 would produce the products of equation 1).

That is ΔH1 = ΔH3 - ΔH2 (just as we would expect using Hess’s Law)