Unit 6 – Energy and Thermochemistry Energy – the capacity to do work Two major types - PowerPoint PPT Presentation

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Unit 6 – Energy and Thermochemistry Energy – the capacity to do work Two major types
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Unit 6 – Energy and Thermochemistry Energy – the capacity to do work Two major types

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  1. Unit 6 – Energy and Thermochemistry • Energy – the capacity to do work • Two major types • Kinetic – E of motion • Potential – E stored-up (can do work, but not doing it yet)  W=F x d • First Law of Thermodynamics • Three Parts: • E of universe is constant • E cannot be created or destroyed • E can be transformed • Average KE of particles  temperature • So how are temperature and heat related?

  2. 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

  3. 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) 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-PΔV • This 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 ΔH = ΔE – (-PΔV)

  4. Exothermic - Energy is given off (lost) by the system Endothermic - Energy is absorbed (gained) by the system • Here’s a table to help with endothermic/exothermic ideas: • AP ♥’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.

  5. Examples of endothermic and exothermic reactions Ba(OH)2. 8H2O(s) + NH4NO3(s) → Ba(NO3)(aq) + NH3(g) + H2O(l) Cl- NH4NO3(s) → NH4NO3(l) → N2O(g) + H2O(g) Zn(s) + NH4NO3(s) → N2(g) + ZnO(s) + 2H2O(g)

  6. 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…in normal lab conditions (can Δ volume )  (not in a • bomb calorimeter), q≈ΔH  ΔH=CΔT • C=heat capacity (not Cp – specific heat capacity) C = m x Cp • If Δ temp easily (ex – metals ) lower Cp • If d/n/Δ temp easily (ex – plastic)  higher Cp Q=mCpΔT • Some values to know… • Cp H2O = 4.184 J/g0C • Cm H2O = 75.3 J/mol0C (molar heat capacity  J/molK)

  7. 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?

  8. Constant Volume (Bomb) Calorimetry A combustible substance is burned in pure O2 in a sealed metal container or ‘bomb’ calorimeter. Because the volume cannot change, no expansion work can be done and ∆E = q that is, all energy is released as heat. The heat released warms the bomb and the water that is surrounding it. The amount of heat released by the reaction can be determined using the equation: q (released by rxn) = q (absorbed by H2O) + q (absorbed by bomb) qrxn = mCp∆T(H2O) + Cbomb∆T A 1.00 g sample of sucrose is burned in a bomb calorimeter. The temperature of 1.50 x 103 g of water in the calorimeter rises from 25.000C to 27.320C. The heat capacity of the bomb is 837 J/K. Calculate the ∆Hcomb for sucrose in kJ/mol. Answer: 5650 kJ/mol

  9. Average Bond DissociationEnthalpies • Break substance up into individual gaseous diatomic molecules • Then, measure energy required to break them apart • Luckily for us, values already established listed on Handout • These values can be used to determine ΔH for a reaction • Let’s try…(drumroll)…combustion of methane! • CH4+ 2O2 CO2 + 2H2O Answer: -802 kJ (for the reaction as written)

  10. 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 (∆Hrxn0): ∆Hrxn = Σnp∆Hfө(products) - Σnr∆Hfө(reactants)

  11. 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.

  12. Jobs for thermite lab Mass 17.5 g Al powder Mass 55.0 g iron(III) oxide Combine and thoroughly mix the 2 solids in a container Add the mixture to the pots and form a cavern on top Mass 25.0 g potassium permanganate Thoroughly grind the permanganate into a very fine powder Carefully add the permanganate to the cavern in the pot and form a small cavern on top of it Using something longish and thin (like a thin glass rod) very very carefully poke about 4 holes in the permanganate cavern to create an opening to the mixture underneath Measure 10ish mL of glycerin in a 10 mL grad cyl (Outside) Pour as much as possible of the glycerin into the very center of the permanganate (girly, cry-baby, ‘fraidy cats need not apply) Video record n.b. All jobs require use of goggles; pouring glycerin also requires apron

  13. Hess’s Law Since energy and therefore enthalpy change is a state function, it is possible to determine ΔHfor reactions in which energy changes are difficult to measure (anything come to mind?). 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.

  14. 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

  15. 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: ΔH1 3 C(s) + 4 H2(g) 5 O2(g) C3H8(g) + 5 O2(g) ΔH3 ΔH2 3 CO2(g) + 4 H2O(g) 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).

  16. 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)

  17. Born-haber Cycle Na+ and Cl- form an electrosatic attraction (ionic bond) following transfer of electron from Na to Cl. Solid consists of a 3-D network in which each Na+ is surrounded by and attracted to a number of Cl- ions and the same for Cl- surrounded by Na+. When 1st IE for Na (+495.8 kJ/mol) is compared to E for electron affinity for Cl (-348.6 kJ/mol) the reaction should be unfavorable (+147.2 kJ/mol). Another factor must be involved i.e. large gain in stability due to the formation of ionic bonds. This is known as the Lattice Energy (enthalpy) = quantity of energy released in the formation of 1 mole of an ionic solid from its separated gaseous ions.

  18. Born-haber Cycle Although it occurs at once, the energy change, often given the symbol of ΔU, can be calculated by dividing the reaction into a series of 5 steps (Born-Haber Cycle) and using Hess’s law. • In general the steps of the cycle are: • Sublimation of the solid metal to gaseous atoms – endothermic • Dissociation of non-metal into individual atoms – endothermic • Ionization of metal atoms to cations (1st, 2nd, etc. IE) – endothermic • Anion formation from non-metal atoms (e- affinity) – exothermic • Formation of ionic crystal from gaseous ions (ΔU) – exothermic Adding the values of all 5 steps gives the value for ΔHf for the ionic crystal. Of course using the heat of formation, the other 3 values and Hess’s law allows us to calculate the lattice energy (enthalpy – or whatever they are calling it this week).

  19. For sodium chloride, the Born - Haber cycle is: For sodium chloride, the Born - Haber cycle is:

  20. Lattice energy comparison 2 main factors determine the strength of interaction among ions in a crystal; charge and ion size. This relationship is expressed by a form of Coulomb’s Law: The energy of attraction b/w 2 particles is directly proportional to the product of the charges and inversely proportional to the distance b/w them. E = k q1 x q2 d k – proportionality constant; q1 & q2 – ion charges; d - internuclear distance Determine which of each of the following pairs would have the greatest lattice energy: MgO or CaS NaCl or MgS NaCl or CaS When in doubt, go with charge.