15 ENERGETICS AHL YR 11 DP CHEMISTRY ROB

15 ENERGETICS (AHL) YR 11 DP CHEMISTRY ROB SLIDER

STANDARD ENTHALPY CHANGES Let’s define some terms: Standard State This refers to reactants and products being at 298 K and 101. 3 k. Pa Standard enthalpy change of formation ΔHfθ This is the enthalpy change associated with the formation of 1 mole of a compound from its elements under standard conditions Standard enthalpy change of combustion ΔHcθ This is the enthalpy change associated with the combustion of a compound under standard conditions

STANDARD ENTHALPIES OF FORMATION Standard molar enthalpies of formation can be looked up in external resources. The table on the right is one such resource and your IB Chemistry Data Booklet contains many more. These can be used to determine standard enthalpies of reaction – see next slide. Notice that standard enthalpies of formation for stable elements is zero (0).

USING HESS’S LAW We can use the standard enthalpies of formation to determine standard enthalpies of reaction by using Hess’s Law Reactants ΔH Products ΣΔHf (products) ΣΔHf (reactants) Elements Construct a formula for ΔH using this the above enthalpy cycle (answer on next slide)

STANDARD ENTHALPIES OF REACTION FROM ENTHALPIES OF FORMATION Systematic way to compare energy changes of reactions. Can be written in terms of the standard enthalpies of formation of products minus reactants (Hess's Law). ΔHθrxn = Σ ni ΔHθf (products) – Σ nj ΔHθf (reactants) Calculate from tables of reference data, such as the IB Chemistry Data Booklet. Example: Ca. CO 3(s) + heat –––> Ca. O(s) + CO 2(g) ΔHθrxn = ΔHθf (Ca. O, s) + ΔHθf (CO 2, g) – ΔHθf (Ca. CO 3, s) = [(1 mol)(-635)] +[(1 mol)(-393. 51)]- [(1 mol)(-1207. 6)] = +179 k. J/mol (note: this should be a +ve value as heat is absorbed in the rxn)

STANDARD ENTHALPIES OF REACTION FROM ENTHALPIES OF COMBUSTION Standard enthalpies of combustion ΔHcθ can also be used to solve enthalpy problems using Hess’s Law. Example: The enthalpy of combustion for H 2, C(graphite) and CH 4 are -285. 8, -393. 5, and -890. 4 k. J/mol respectively. Calculate the standard enthalpy of formation ΔHfθ for CH 4. First, write out the equations: 1) H 2(g) + 0. 5 O 2(g) -> H 2 O(l) 2) C(graphite) + O 2(g) -> CO 2(g) 3) CH 4(g) + 2 O 2(g) -> CO 2(g) + 2 H 2 O(l) From the above equations, derive C + 2 H 2 -> CH 4 Answer: C + 2 H 2 -> CH 4 -74. 7 Hint: 2*(1) + (2) - (3), Thus, ΔHfθ = [2 * (-285. 8)] + (-393. 5) - (-890. 4) = -74. 7 k. J/mol ΔHcθ /(k. J/mol) -285. 8 -393. 5 -890. 4

ANOTHER EXAMPLE ΔHfθ for ethanol can be determined using the enthalpy cycle below: 2 C(s) 2 X ΔHfθ (CO 2) 2 O 2(g) + 3 H 2(g) 3 X ΔHfθ (H 2 O) + 1/2 O 2(g) ΔHfθ (C 2 H 5 OH) C 2 H 5 OH 1½O 2(g) 2 O 2(g) ΔHcθ (C 2 H 5 OH) 2 CO 2(g) + 3 H 2 O(l) First, add the appropriate reactants to the cycle with correct coefficients Next, add the correct enthalpy change values multiplied by the appropriate coefficients. Finally, set up and solve the equation using Hess’s Law ΔHfθ (C 2 H 5 OH) = [2 X ΔHfθ(CO 2)] + [3 X ΔHfθ(H 2 O)] - ΔHcθ (C 2 H 5 OH) ΔHfθ (C 2 H 5 OH) = [2 X -393. 5] + [3 X -285. 8] – (-1371) = -273. 4 k. J mol-1

WATCH YOUR STATES! States of matter are very important when looking up enthalpy change values. You must make sure you are choosing the correct value. For example: CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(l) ΔHcθ = -890. 4 k. J/mol CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2 O(g) ΔHcθ = -802. 4 k. J/mol Notice that the production of water vapour instead of liquid results in a lower enthalpy change of combustion. This is because liquid water must absorb energy (endothermic) to become vapour: 2 H 2 O(l) 2 H 2 O(g) ΔHcθ = +88 k. J/mol

EXERCISES 1. Given the standard enthalpy change of formation values for ammonia (g), hydrogen bromide (g) and ammonium bromide (s) are -46, -36 and -271 k. J/mol respectively, calculate ΔHrxnθ for the reaction between ammonia and hydrogen bromide. 2. Given the standard enthalpy of formation of water vapour and carbon dioxide are -286 and -394 k. J/mol respectively and the standard enthalpy of combustion of methanol (l) (CH 3 OH) is -715 k. J/mol, construct an enthalpy cycle and calculate the ΔHfθ for methanol.

SOLUTIONS 1. NH 3(g) + HBr (g) NH 4 Br(s) ΔHθrxn = Σ ni ΔHθf (products) – Σ nj ΔHθf (reactants) = -271 – (-46+ -36) = -189 k. J mol-1 2. 2 C(s) 2 X ΔHfθ (CO 2) 2 O 2(g) 2 CO 2(g) + 4 H 2(g) 4 X ΔHfθ (H 2 O) + + 2 x ΔHfθ (CH 3 OH) O 2(g) 2 CH 3 OH(l) 2 O 2(g) 4 H 2 O(l) 3 O 2(g) 2 x ΔHcθ (CH 3 OH) [2 x ΔHfθ (CH 3 OH)] + [2 x ΔHcθ (CH 3 OH)] = [2 X ΔHfθ (CO 2)] + [4 X ΔHfθ (H 2 O)] [2 x ΔHfθ (CH 3 OH)] = - [2 x ΔHcθ (CH 3 OH)] + [2 X ΔHfθ (CO 2)] + [4 X ΔHfθ (H 2 O)] 2 x ΔHfθ (CH 3 OH) = -(2 X -715) + (2 X -394) + (4 X -286) ΔHfθ (CH 3 OH) = [1430 – 788 – 1144]/2 = -502/2 = -251 k. J mol-1

LATTICE ENTHALPY Recall that ionic compounds are made up of repeating positive and negative ions in a lattice. Different ionic compounds have different stabilities based on the strength of the interactions between the ions in the lattice. The relative stabilities are measured by the lattice enthalpy values Lattice enthalpy – the energy required to split 1 mole of a solid ionic substance into its gaseous ions Note: Lattice enthalpies cannot be measured directly. However, we can calculate them using other values such as ionisation energies and electron affinities

ELECTRON AFFINITY Recall electron affinity is a measure of how strongly an atom attracts electrons to itself Electron affinity ΔHEAθ – the energy change that occurs when 1 mole of electrons is accepted by one mole of atoms in the gaseous state forming one mole of negative ions. Recall the pattern of electron affinity values on the Periodic Table

PROCESSES INVOLVED IN FORMING AN IONIC LATTICE Electron Affinity ΔHEAθ – the enthalpy change when 1 mole of gaseous ions is formed from one mole of atoms (these are generally exothermic, at least for the first electron) Ionisation Energy ΔHIEθ – the energy required to remove electrons from atoms/ions in the gaseous state (this process is endothermic as the process requires energy) Enthalpy of Atomisation ΔHatθ – the enthalpy change when 1 mole of gaseous atoms is formed from its elements in the standard state. For a diatomic molecule (e. g. Cl 2), this is ½ the bond enthalpy (this process occurs prior to ionisation and is endothermic) Enthalpy of Dissociation ΔHDθ (aka bond enthalpy) – the enthalpy change when 1 mole of gaseous atoms is formed from the dissociation of a covalent bond in the standard state. (this process occurs prior to ionisation and is endothermic) Lattice Enthalpy ΔHlattθ – the enthalpy change when 1 mole of an ionic lattice is formed (exothermic) or gaseous ions are formed from the lattice (endothermic) THE GREATER THE LATTICE ENTHALPY, THE MORE STABLE THE COMPOUND

Enthalpy change of formation for Na. Cl using a Born-Haber cycle If we want to find the ΔHfθ of sodium chloride, we can use a Born-Haber cycle and consider all of the steps that we would need to get from the reactants to the products -349 Step 1 – convert Na from a solid to a gas (atomisation) Step 2 – convert diatomic Cl 2 to the atom Cl (½ dissociation or atomisation) Step 3 – convert Na to Na+ (ionisation) Step 4 – convert Cl to Cl- (electron affinity) Step 5 – formation of the Na. Cl lattice (lattice energy – exothermic) Step 6 – solve for ΔHfθ = ΔHatθ + ½ΔHDθ + ΔHIθ + ΔHEAθ + ΔHlattθ = 109 + 121 + 494 + (-349) + (-787) = -412 k. J mol-1 -787

Another Example: Construct a Born-Haber cycle for the formation of Mg. O Note: Mg forms Mg 2+, so ΔHIθ will be the sum of the first two ionisation enthalpy changes Mg 2+(s) + O 2 -(g) ΔHIθ = ΔHIθ(1)+ ΔHIθ(2) ΔHEAθ ΔHlattθ Mg(g) + O(g) ΔHatθ ½ΔHDθ Mg(s) + ½O 2(g) ΔHfθ Mg. O(s)

EXERCISE Construct a Born-Haber cycle and find the enthalpy change for the following reaction: Mg(s) + Cl 2(g) Mg. Cl 2(s) Use the following data: ΔHatθ = +148 k. J/mol ΔHIEθ (Mg 1)= +736 k. J/mol ΔHIEθ (Mg 2)= + 1451 k. J/mol ΔHDθ (Cl 2)= +244 k. J/mol ΔHEAθ (Cl)= -349 k. J/mol ΔHlattθ = -2542 k. J/mol Mg 2+(s) + 2 Cl-(g) ΔHIEθ = ΔHIEθ(1)+ ΔHIEθ(2) = +2187 k. J/mol 2ΔHEAθ= -698 k. J/mol ΔHlattθ= -2542 k. J/mol Mg(g) + 2 Cl(g) ΔHfθ =+148+244+2187 -698 -2542 = -661 k. J mol-1 ΔHatθ= +148 k. J/mol ΔHDθ= +244 k. J/mol Mg(s) + Cl 2(g) ΔHfθ Mg. Cl 2(s)

FACTORS AFFECTING LATTICE ENTHALPIES Lattice enthalpy depends on two factors: 1. Ion Size Small ions are close to each other and exhibit stronger attraction. When charges are same, smaller ions have higher L. E. Li. Cl (845 k. J/mol) versus Na. Cl (787 k. J/mol) The charges are the same (+1 & -1) Since Li+1 is smaller than Na+1, Li. Cl has greater L. E. 2. Ion Charge Ion with higher charges exhibit stronger attraction. When size is similiar, higher charges lead to higher L. E. KCl (709 k. J/mol) versus Ca. Cl 2 (2258 k. J/mol) K+1 and Ca+2 are similiar in size (same period next to each other). Since Ca+2 has higher charge, Ca. Cl 2 has greater L. E.

THEORETICAL VS. EXPERIMENTAL Lattice enthalpies can be determined experimentally as we have seen with the Born-Haber cycles. We can also determine lattice enthalpies theoretically using Coulomb’s Law: F=k q 1 q 2 r 2 Where: • F is the force of attraction on one ion on another, • q 1&q 2 are the charges on the ions, • r is the distance between ions and • k is a constant. This equation shows mathematically what we stated in the previous slide – increasing charge and decreasing radii increases lattice enthalpy. See next slide for comparative data

Comparing ΔHlattθ for some ionic compounds Compound Lattice Enthalpy (k. J mol-1) Experimental Theoretical Difference (%) Na. F 902 891 1. 2 Na. Cl 771 766 0. 6 Na. Br 733 732 0. 1 Na. I 684 686 0. 3 Ag. F 955 870 8. 9 Ag. Cl 905 770 14. 9 Ag. Br 890 758 14. 8 Ag. I 876 736 16. 0 What do the values of %difference tell us about the ionic/covalent character of the bonds in these substances? Firstly, notice that the experimental values are always higher than theoretical values. Secondly, notice that the silver halides have a much greater % difference than sodium halides. Since theoretical values are based on charge interactions (ionic bonding), this implies that there is some other bonding influencing the lattice enthalpy values. The experimental values that are closest to theoretical value fit the ionic model more closely. For example, the ionic model for Na. Cl fits quite well, so we say it has a large ionic character. Ag. Cl on the other hand, has a greater % difference which is due to more covalent character, which we would expect due to the smaller electronegativity difference.

ENTROPY When a tyre is punctured with a nail, why does the air escape spontaneously? When perfume is sprayed on one side of the room, why does it spread across the room where you can smell it?

ENTROPY (S) Answers: Air escapes from the tyre spontaneously as energy tends to disperse naturally unless it is hindered from doing so (in the unpunctured tyre). Similarly, the perfume droplets will also diffuse throughout the room to your nose, again due to natural dispersion of the particles. This natural dispersion of energy is known as entropy, symbol S and is often described as a measure of the degree of randomness or disorder in a system. “Entropy: A measure of the amount of energy in a physical system not available to do work. As a physical system becomes more disordered, and its energy becomes more evenly distributed, that energy becomes less able to do work. For example, a car rolling along a road has kinetic energy that could do work (by carrying or colliding with something, for example); as friction slows it down and its energy is distributed to its surroundings as heat, it loses this ability. The amount of entropy is often thought of as the amount of disorder in a system. ” Source: The American Heritage® Science Dictionary Copyright © 2005 by Houghton Mifflin Company.

ENTROPY (S) http: //art. bradley. edu/bug/? p=1675 http: //contrarianinconsistent. wordpress. com/2011/05/06/entropy-a-dish-best-served-hot / http: //www. cartoonstock. com/directory/e/entropy. asp

ENTROPY CHANGE (ΔS) Because energy naturally tends to become more dispersed (i. e. more random). This would represent a positive change in entropy over time (ΔS = +). The opposite is true for a negative change in entropy (ΔS = -). The standard units are J K-1 or J K-1 mol-1 Identify whether the examples below are positive or negative entropy changes: Process Ethanol evaporating Water freezing A salt dissolving in water Decomposition: Cu. CO 3(s) Cu. O(s) + CO 2(g) Heating water from 150 C to 450 C 2 Al(s) + 3 S(s) Al 2 S 3(s) CH 4(g) + H 2 O(g) 3 H 2(g) + CO 2(g) ΔS (+/-)

ENTROPY CHANGE (ΔS) Because energy naturally tends to become more dispersed (i. e. more random). This would represent a positive change in entropy over time (ΔS = +). The opposite is true for a negative change in entropy (ΔS = -). The standard units are J K-1 or J K-1 mol-1 Identify whether the examples below are positive or negative entropy changes: Process ΔS (+/-) Ethanol evaporating + Water freezing - A salt dissolving in water + Decomposition: Cu. CO 3(s) Cu. O(s) + CO 2(g) + Heating water from 150 C to 450 C + 2 Al(s) + 3 S(s) Al 2 S 3(s) - CH 4(g) + H 2 O(g) 3 H 2(g) + CO 2(g) +

STANDARD ENTROPY CHANGE The standard entropy change, ΔSrxnθ is calculated just as the standard enthalpy change using standard entropy values (all positive) at 101. 3 k. Pa and 298 K. Many of these values can be found in the IB Chemistry Data Booklet. ΔSrxnθ = Σ ni Sθ (products) – Σ nj Sθ (reactants) Example: Evaluate the entropy change for the reaction: CO + 3 H 2 -> CH 4 + H 2 O in which all reactants and products are gaseous. Entropy values are respectively 198, 131, 186, 189 J (K mol)-1 Solution Standard entropies of reaction, ΔSrxnθ , equals the entropy of products minus the entropy of reactants. The standard entropies of the reactants and products have been given above: CO + 3 H 2 -> CH 4 + H 2 O ΔSrxnθ = ((186 + 189) - (198 + 3*131)) J (K mol)-1 = -216 J (K mol)-1

ENTROPY AND PROBABILITY Entropy is often equated with probability: If the boxes below represent gases in a closed system, which of the two configurations do you think is more probable? Which has more entropy? There is a higher probability that a system will be in disorder rather than ordered. (2 nd Law of Thermodynamics) The greater the probability a state exists, the higher it’s entropy. So, we can conclude that a positive entropy change is more probable ΔS = + (more favourable)

SPONTANEITY We have now looked at two values that describe whether a reaction is spontaneous or not. The following values predict spontaneity: ΔHrxnθ = - (negative enthalpy) Consider the reaction: ΔSrxnθ = + (positive entropy) N 2(g) + 3 H 2(g) 2 NH 3(g) ΔHrxnθ ΔSrxnθ -92. 6 k. J mol-1 -198. 5 J K-1 Predicted spontaneous Predicted non-spontaneous Which is correct? Is it spontaneous or not? ?

GIBBS FREE ENERGY In reality, spontaneity is determined by Gibbs Free Energy (ΔG) and depends on: • Enthalpy (k. J mol-1) ΔG>0 non-spontaneous • Entropy (k. J K-1 mol-1) ΔG<0 spontaneous • Temperature (K) ΔG = ΔH - T ΔS Predict spontaneity: ΔH ΔS + + + - - ΔG Spontaneous

GIBBS FREE ENERGY In reality, spontaneity is determined by Gibbs Free Energy (ΔG) and depends on: • Enthalpy (k. J mol-1) ΔG>0 non-spontaneous • Entropy (k. J K-1 mol-1) ΔG<0 spontaneous • Temperature (K) ΔG = ΔH - T ΔS Predict spontaneity: ΔH ΔS ΔG Spontaneous + + Depends on T With high T + - + No - + - Yes - - Depends on T With low T

GIBBS FREE ENERGY EXAMPLE Calculate ΔG for the following reaction at 25°C. Will the reaction occur (be spontaneous)? How do you know? Also given for this reaction: NH 3(g) + HCl(g) → NH 4 Cl(s) ΔH = -176. 0 k. J·mol-1 ΔS = -284. 8 J·K-1·mol-1

GIBBS FREE ENERGY EXAMPLE Solution We will calculate ΔG using the formula ΔG = ΔH - TΔS but first we need to convert units for ΔS and temperature to Kelvin: ΔS = -284. 8 J·K-1 mol-1 = -0. 2848 k. J·K-1 mol-1 K = 273 + °C = 273 + 25 = 298 K Now we can solve our equation: Since ΔG < 0 the reaction will be spontaneous. ΔG = -176. 0 - (298)(-0. 2848) ΔG = -176. 0 - (-84. 9) ΔG = -91. 1 k. J mol-1