15. Energetics HL

Define and apply the terms standard state, standard enthalpy change of formation, and standard enthalpy change of combustion AND

Determine the enthalpy change of a reaction using standard enthalpy changes of combustion and formation

Standard State- state in which matter has an enthalpy of zero and is under standard thermochemical conditions

(i.e. is it a solid, liquid, or gas at room temperature?)

Standard enthalpy of formation (ΔH_f^°)- the amount of energy required to form one mole of a compound in its standard state from its elements in their standard states.

(i.e. Like using Hess’ Law, except working with individual compounds and not the intermediate reactions)

            ΔH_f^°= ΣΔH_products - ΣΔH_reactants

EXAMPLE: Find the ΔH_f^° for the reaction 2NaHCO₃ à NaCO₃ + CO₂ + H­₂O using the following data: NaHCO₃= -948, NaCO₃ = -1131, CO₂ = -395, H₂O= -286

ΔH_f^°= ΣΔH_products - ΣΔH_reactants

ΔH_f^°= (-1131 + -395 + -286) – (2 x -948)

ΔH_f^°= -1815 + 1896

ΔH_f^°= +81 kJ mol^-1

Standard enthalpy of combustion (ΔH_comb^°)- the amount of energy required for one mole of a substance to undergo complete combustion in excess oxygen under standard conditions

            (i.e. Like finding ΔH_f^°, except something gets burned)

            ΔH_c^°= ΣΔH_products - ΣΔH_reactants

Define and apply the terms lattice enthalpy and electron affinity

Lattice Enthalpy (ΔH_lat^°)- The enthalpy change when one mole of a solid ionic compound is separated into gaseous ions

            (i.e. The energy gained/lost when a lattice breaks down into component ions)

            Because it is being broken down, it will always be ENDOTHERMIC

Electron Affinity- The amount of energy needed or released when a mole of a substance to gain an additional electron (each)

Construct a Born-Haber cycle for group 1 and 2 oxides and chlorides and use it to calculate an enthalpy change

Born-Haber Cycle- Hess cycle used to measure lattice enthalpies

Steps to a Born Haber Cycle- Atomization, Ionization, Determining bonds (AID)

1. Determine the enthalpy of atomization for both substances.

Enthalpy of Atomization (ΔH_lat^°)- Energy change when one mole of gaseous atoms are formed from the element in its standard states

* This can be calculated using info from your data booklet

2. Determine the ionization/electron affinity enthalpy change

Ionization enthalpy change (ΔH_i^°)- Enthalpy change for one mole of gaseous atoms or cations to LOSE an elec. to form a mole of gaseous positive ions.

Electron affinity enthalpy change (ΔH_e^°)- Enthalpy change for one mole of gaseous atoms or anions to GAIN an elec to form a mole of gaseous negative ions

* This can be calculated using info from the periodic table

**If you need an ion with a multiple charge (ex. Mg²⁺), you need to use the sum of two values (ΔH_e^° when Mg à Mg⁺ AND ΔH_e^° when Mg⁺ à Mg²⁺)

3. Determine the lattice enthalpy of the bond

ΔH_lat^° =  (ΔH_atom^°[ A] + ΔH_atom^°[B] + ΔH_i^°[A] + ΔH_e^°[B] ) - ΔH_f^° [Compound AB]

ΔH_lat^°= (Everything) – (Standard Enthalpy of Formation)

EXAMPLE!!! Construct a Born-Haber Cycle for the reaction: Na (s) + ½Cl­₂ (g) à NaCl

Given that ΔH_f[NaCl] = -411kJ and ΔH_atom[Na] = +109 kJ

1. Atomization

ΔH_a [Na] = +109        *This was stated above

ΔH_a [Cl] = + 121         *This was found by taking the Cl-Cl bond enthalpy and dividing it by two (1/2 mole of Cl₂)

2. Ionization

ΔH_i [Na] = + 494        *These were found using the data booklet

ΔH_e [Cl] = -364

3. Determining Lattice

ΔH_lat^° =  (ΔH_atom^°[ Na] + ΔH_atom^°[Cl] + ΔH_i^°[Na] + ΔH_e^°[Cl] ) - ΔH_f^° [NaCl]

ΔH_lat^° = (109 + 121 + 494 -364) – 411

ΔH_lat^° = 360 - (- 411)

ΔH_lat^° = + 711 kJ mol^-1

***Sometimes, you may be asked to Draw or Construct a Born Haber Cycle. You will still need to do the mathematics behind it, but you will need to draw one as well. A sketch is included on the following page. The 3-steps listed above are written in red.

Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds AND

Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds in terms of their covalent character

A theoretical method for predicting lattice enthalpy- the Ionic Model

Ionic model- assumes that ions are perfect spheres and the only interaction is due to attraction between ions

A decrease in ionic radius means an increase in the force of attraction between ions

An increase in ionic radius means a decrease in the force of attraction between ions.

Ionic model is more accurate for smaller ions than larger ions. Larger ions have a larger percent difference as compared to experimental (Born-Haber) models

***Works best for small, highly charged, atoms

State and explain the factors that increase the entropy of a system.

Entropy- A measure of the degree of disorder or randomness in a system. A.k.a.

ENTROPY IS DISORDER!!!

If ΔS^° is positive- entropy increases

If ΔS^° is negative- entropy decreases

Predict whether the entropy change (ΔS) for a given process is positive or negative

Less moles à More moles= increase in entropy

More moles à Less moles= decrease in entropy

If those are the same, use the following tricks:

• From solid à liquid à gas, there is more entropy
• Production of gas = increase in entropy
• Using up gas = decrease in entropy

Calculate the standard entropy change for a reaction (ΔS) using standard entropy values.

Change in entropy (ΔS^°)= ΔS^°_products - ΔS^°_reactants

I won’t go through an example. It’s the same general idea as ΔH

Units of entropy- J kg^-1 mol^-1

***WATCH THE UNITS!!!

Predict whether a reaction or a process will be spontaneous using the sign for ΔG

Spontaneous- A change that tends to happen

            Ex. Salt dissolves in water, gas expands to fill its container

Gibbs Free Energy (ΔG^°)- The change in the amount of energy available to do usable work

If ΔG^° is positive- the process IS NOT spontaneous

If ΔG^° is negative- the process IS spontaneous

Calculate the Δhmm.

if it was great sex, excusable. if he was otherwise normal average regular all around

nope

he's getting kicked out of bed.

why, is the guy you've been screwing a smoker? :P

G for a reaction using the equation ΔG^° = ΔH^° - ΔS^°T

EXAMPLE!!!

For the reaction of PCl₃ and Chlorine gas to form PCl₅ at 25^oC, the entropy change is -85J mol^-1 K^-1 and the enthalpy change is -124 kJ mol^-1. What is the approximate value of the Gibbs free energy? Is this reaction spontaneous?

 ΔG^° = ΔH^° - ΔS^°T

ΔG^° = -124 – (85/1000) x (25 + 273)

REMEMBER to change J mol^-1 K^-1 à kJ mol^-1 K^-1 by dividing the value by 1000

REMEMBER ALSO to change Celcius à Kelvin by adding 273

REMEMBER REMEMBER the 5^th of November the gun…nevermind. Back to chem.

ΔG^° = -124 – (298 x 0.085)

ΔG^° = -124 – 25.33

ΔG^° = -98.67

It’s negative, then it’s spontaneous!!!!

Predict the effect of a change in temperature on the spontinety of a reaction using standard entropy and enthalpy changes

Favored reactions are exothermic ones that increase entropy, though neither is entirely sufficient

Exothermic reactions with a negative entropy change will always be spontaneous

Exothermic reactions with a positive entropy change will be spontaneous at high temperatures because only there will ΔH > ΔST

Endothermic reactions with a negative entropy change will be spontaneous at low temperatures because only there will ΔH <ΔST

Endothermic reactions with a positive entropy change will never be spontaneous

ΔG^° = ΔH^° - ΔS^°T

            If you’ve seen the HP series, remember this

Gryffindor = Harry – Slytherin Tendencies

EXAMPLE!!!

For the reaction of PCl₃ and Chlorine gas to form PCl₅ at 25^oC, the entropy change is -85J mol^-1 K^-1 and the enthalpy change is -124 kJ mol^-1. What is the approximate value of the Gibbs free energy? Is this reaction spontaneous?

 ΔG^° = ΔH^° - ΔS^°T

ΔG^° = -124 – (85/1000) x (25 + 273)

REMEMBER to change J mol^-1 K^-1 à kJ mol^-1 K^-1 by dividing the value by 1000

REMEMBER ALSO to change Celcius à Kelvin by adding 273

REMEMBER REMEMBER the 5^th of November the gun…nevermind. Back to chem.

ΔG^° = -124 – (298 x 0.085)

ΔG^° = -124 – 25.33

ΔG^° = -98.67

It’s negative, then it’s spontanteous!!!!

Follow this link for some good practice problems (E-H)
http://coffman.dublin.k12.oh.us/teachers/teacherpages/brown/mr._browns_chemistry_pages/Topic_5_files/practiceenergetics.pdf