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YR 11 DP CHEMISTRY ROB SLIDER 15 ENERGETICS (AHL).

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Presentation on theme: "YR 11 DP CHEMISTRY ROB SLIDER 15 ENERGETICS (AHL)."— Presentation transcript:

1 YR 11 DP CHEMISTRY ROB SLIDER 15 ENERGETICS (AHL)

2 STANDARD ENTHALPY CHANGES Standard State This refers to reactants and products being at 298K and 101.3kPa Standard enthalpy change of formation ΔH f θ This is the enthalpy change associated with the formation of 1 mole of a compound from its elements under standard conditions Let’s define some terms: Standard enthalpy change of combustion ΔH c θ This is the enthalpy change associated with the combustion of a compound under standard conditions

3 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).

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

5 STANDARD ENTHALPIES OF REACTION FROM ENTHALPIES OF FORMATION Example: CaCO 3 (s) + heat –––> CaO(s) + CO 2 (g) ΔH θ rxn = ΔH θ f (CaO, s) + ΔH θ f (CO 2, g) – ΔH θ f (CaCO 3, s) = [(1mol)(-635)] +[(1mol)( )]- [(1mol)( )] = +179 kJ/mol (note: this should be a +ve value as heat is absorbed in the rxn) 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). ΔH θ rxn = Σ n i ΔH θ f (products) – Σ n j ΔH θ f (reactants) Calculate from tables of reference data, such as the IB Chemistry Data Booklet.

6 STANDARD ENTHALPIES OF REACTION FROM ENTHALPIES OF COMBUSTION ΔH c θ Standard enthalpies of combustion ΔH c θ can also be used to solve enthalpy problems using Hess’s Law. Example: ΔH f θ The enthalpy of combustion for H 2, C(graphite) and CH 4 are , , and kJ/mol respectively. Calculate the standard enthalpy of formation ΔH f θ for CH 4. First, write out the equations: ΔH c θ ΔH c θ /(kJ/mol) 1)H 2(g) O 2(g) -> H 2 O (l) )C (graphite) + O 2(g) -> CO 2(g) )CH 4(g) + 2O 2(g) -> CO 2(g) + 2H 2 O (l) From the above equations, derive C + 2H 2 -> CH 4 Answer: C + 2H 2 -> CH Hint: 2*(1) + (2) - (3), ΔH f θ Thus, ΔH f θ = [2 * (-285.8)] + (-393.5) - (-890.4) = kJ/mol

7 ANOTHER EXAMPLE ΔH f θ for ethanol can be determined using the enthalpy cycle below: 2C (s) + 3H 2(g) + 1/2O 2(g) C 2 H 5 OH 2CO 2(g) + 3H 2 O (l) 2X ΔH f θ (CO 2 ) 3X ΔH f θ (H 2 O) ΔH c θ (C 2 H 5 OH) 2O 2(g) 1½O 2(g) 2O 2(g) 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 ΔH f θ (C 2 H 5 OH) = [2X ΔH f θ (CO 2 )] + [3X ΔH f θ (H 2 O)] - ΔH c θ (C 2 H 5 OH) ΔH f θ (C 2 H 5 OH) = [2X ] + [3X ] – (-1371) = kJ mol -1 ΔH f θ (C 2 H 5 OH )

8 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: ΔH c θ CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (l) ΔH c θ = kJ/mol ΔH c θ CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g) ΔH c θ = kJ/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: ΔH c θ 2H 2 O (l)  2H 2 O (g) ΔH c θ = +88 kJ/mol

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

10 SOLUTIONS 1.NH 3(g) + HBr (g)  NH 4 Br (s) ΔH θ rxn = Σ n i ΔH θ f (products) – Σ n j ΔH θ f (reactants) = -271 – ( ) = -189 kJ mol C (s) + 4H 2(g) + O 2(g) 2CH 3 OH (l) 2CO 2(g) + 4H 2 O (l) 4X ΔH f θ (H 2 O) 2 x ΔH c θ (C 2 H 5 OH) 2O 2(g) 3O 2(g) 2 x ΔH f θ (CH 3 OH ) 2X ΔH f θ (CO 2 ) [2 x ΔH f θ (CH 3 OH)] + [2 x ΔH c θ (C 2 H 5 OH)] = [2X ΔH f θ (CO 2 )] + [4X ΔH f θ (H 2 O)] [2 x ΔH f θ (CH 3 OH)] = - [2 x ΔH c θ (C 2 H 5 OH)] + [2X ΔH f θ (CO 2 )] + [4X ΔH f θ (H 2 O)] 2 x ΔH f θ (CH 3 OH) = -(2 X -715) + (2 X -394) + (4 X -286) ΔH f θ (CH 3 OH) = [1430 – 788 – 1144]/2 = -502/2 = -251 kJ mol -1

11 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 1mole 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

12 ELECTRON AFFINITY Recall electron affinity is a measure of how strongly an atom attracts electrons to itself ΔH EA θ Electron affinity ΔH EA θ – the energy change that occurs when 1mole 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

13 PROCESSES INVOLVED IN FORMING AN IONIC LATTICE ΔH IE θ Ionisation Energy ΔH IE θ – the energy required to remove electrons from atoms/ions in the gaseous state (this process is endothermic as the process requires energy) ΔH at θ Enthalpy of Atomisation ΔH at θ – the enthalpy change when 1mole 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) ΔH EA θ Electron Affinity ΔH EA θ – the enthalpy change when 1mole of gaseous ions is formed from one mole of atoms (these are generally exothermic, at least for the first electron) ΔH latt θ Lattice Enthalpy ΔH latt θ – the enthalpy change when 1mole of an ionic lattice is formed (exothermic) or gaseous ions are formed from the lattice (endothermic) ΔH D θ Enthalpy of Dissociation ΔH D θ (aka bond enthalpy) – the enthalpy change when 1mole 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) THE GREATER THE LATTICE ENTHALPY, THE MORE STABLE THE COMPOUND

14 If we want to find the ΔH f θ 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 Enthalpy change of formation for NaCl using a Born-Haber cycle 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 NaCl lattice (lattice energy – exothermic) Step 6 – solve for ΔH f θ = ΔH at θ + ½ΔH D θ + ΔH I θ + ΔH EA θ + ΔH latt θ = (-349) + (-787) = -412 kJ mol -1

15 Another Example: Construct a Born-Haber cycle for the formation of MgO Note: Mg forms Mg 2+, so ΔH I θ will be the sum of the first two ionisation enthalpy changes Mg (s) + ½O 2(g) Mg (g) + O (g) Mg 2+ (s) + O 2- (g) MgO (s) ΔH EA θ ½ΔHDθ½ΔHDθ ΔH I θ = ΔH I θ (1)+ ΔH I θ (2) ΔH at θ ΔH latt θ ΔHfθΔHfθ

16 EXERCISE Construct a Born-Haber cycle and find the enthalpy change for the following reaction: Mg (s) + Cl 2(g)  MgCl 2(s) Use the following data: ΔH at θ = +148 kJ/mol ΔH IE θ (Mg1)= +736 kJ/mol ΔH IE θ (Mg2)= kJ/mol ΔH D θ (Cl 2 )= +244 kJ/mol ΔH EA θ (Cl)= -349 kJ/mol ΔH latt θ = kJ/mol Mg (s) + Cl 2(g) Mg (g) + 2Cl (g) Mg 2+ (s) + 2Cl - (g) MgCl 2(s) 2ΔH EA θ = -698kJ/mol ΔH D θ = +244kJ/mol ΔH IE θ = ΔH IE θ (1)+ ΔH IE θ (2) = +2187kJ/mol ΔH at θ = +148kJ/mol ΔH latt θ = kJ/mol ΔHfθΔHfθ ΔH f θ = = -661kJ mol -1

17 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. LiCl (845 kJ/mol) versus NaCl (787 kJ/mol) The charges are the same (+1 & -1) Since Li +1 is smaller than Na +1, LiCl 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 kJ/mol) versus CaCl 2 (2258 kJ/mol) K +1 and Ca +2 are similiar in size (same period next to each other). Since Ca +2 has higher charge, CaCl 2 has greater L.E.

18 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 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

19 Comparing ΔH latt θ for some ionic compounds Compound Lattice Enthalpy (kJ mol -1 ) ExperimentalTheoreticalDifference (%) NaF NaCl NaBr NaI AgF AgCl AgBr AgI 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 the theoretical values. Since theoretical values are based on charge interactions (ionic bonding), this implies that there is some other bonding influencing the lattice enthalpy values. Secondly, notice that the silver halides have a much greater % difference than sodium halides. The experimental values that are closest to the theoretical value fit the ionic model more closely. For example, the ionic model for NaCl fits quite well, so we say it has a large ionic character. AgCl 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.

20 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?

21 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.

22 ENTROPY (S)

23 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: ProcessΔS (+/-) Ethanol evaporating Water freezing A salt dissolving in water Decomposition: CuCO 3(s)  CuO (s) + CO 2(g) Heating water from 15 0 C to 45 0 C 2Al (s) + 3S (s)  Al 2 S 3(s) CH 4(g) + H 2 O (g)  3H 2(g) + CO 2(g)

24 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: ProcessΔS (+/-) Ethanol evaporating+ Water freezing- A salt dissolving in water+ Decomposition: CuCO 3(s)  CuO (s) + CO 2(g) + Heating water from 15 0 C to 45 0 C+ 2Al (s) + 3S (s)  Al 2 S 3(s) - CH 4(g) + H 2 O (g)  3H 2(g) + CO 2(g) +

25 STANDARD ENTROPY CHANGE The standard entropy change, ΔS rxn θ is calculated just as the standard enthalpy change using standard entropy values (all positive) at 101.3kPa and 298K. Many of these values can be found in the IB Chemistry Data Booklet. ΔS rxn θ = Σ n i ΔS θ (products) – Σ n j ΔS θ (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, ΔS rxn θ, 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 ΔS rxn θ = (( ) - ( *131)) J (K mol) -1 = -216 J (K mol) -1

26 ENTROPY AND PROBABILITY Entropy is often equated with probability: 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. 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? So, we can conclude that a positive entropy change is more probable ΔS = + (more favourable)

27 SPONTANEITY We have now looked at two values that describe whether a reaction is spontaneous or not. The following values predict spontaneity: ΔH rxn θ = - (negative enthalpy) ΔS rxn θ = + (positive entropy) Consider the reaction: N 2(g) + 3H 2(g)  2NH 3(g) ΔH rxn θ ΔS rxn θ kJ mol J K -1 Predicted spontaneousPredicted non-spontaneous Which is correct? Is it spontaneous or not??

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

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

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

31 GIBBS FREE ENERGY EXAMPLE ΔS = J·K -1 mol -1 = kJ·K -1 mol -1 K = °C = = 298 K ΔG = (298)( ) ΔG = (-84.9) ΔG = kJ mol -1 Solution We will calculate ΔG using the formula ΔG = ΔH - TΔS but first we need to convert units for ΔS and temperature to Kelvin: Now we can solve our equation: Since ΔG < 0 the reaction will be spontaneous.


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