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5 Energetics

5 Energetics. Year 11 DP Chemistry R Slider. 5.1 Exothermic and Endothermic. Chemical Energy-Enthalpy. Kinetic energy is the energy of motion Particles that make up matter are constantly in motion. Potential energy is stored energy (no motion)

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5 Energetics

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  1. 5 Energetics Year 11 DP Chemistry R Slider

  2. 5.1 Exothermic and Endothermic

  3. Chemical Energy-Enthalpy • Kinetic energy is the energy of motion • Particles that make up matter are constantly in motion • Potential energy is stored energy (no motion) • Every bond between atoms and ions has stored energy within the bonds • This stored energy is known as chemical potential energy Total energy = Kinetic energy +Potential Energy (known as Enthalpy or heat content–symbol H)

  4. Change in Enthalpy (ΔH) Because it is difficult to measure the amount of energy in individual reactants or products, Chemists measure the change in enthalpy (ΔH)that occurs when a reaction takes place. ΔH is the difference in the enthalpy of the products and the enthalpy of the reactants. Enthalpy changes result from bonds breaking and new bonds reforming. This breaking and reforming of bonds results in energy changes within the system which are measured as temperature changes Note: Standard enthalpy changes are measured under standard temperature (298K) and pressure (101.3kPa)

  5. Endothermic or Exothermic? Endothermic Some chemical reactions absorb heat from the surroundings so the products of the reaction contain more heat. The surroundings feel colder, ΔH = + Examples: photosynthesis, cold packs Exothermic Some chemical reactions release heat to the surroundings so the products have less heat than the reactants The surroundings feel hotter, ΔH = - Examples: combustion, neutralisation reactions

  6. Heat exchange Another way to visualise ΔH is to think about energy being exchanged between the chemical system and the surroundings Heat loss in a chemical system=Heat gain to the surroundings (Temperature increases) Heat gain in a chemical system=Heat loss to the surroundings (Temperature decreases) Source diagrams: http://www.chemhume.co.uk/ASCHEM/Unit%203/13%20Enthalpy/13%20Enthalpyc.htm

  7. Neutralisation is exothermic Acid Base Salt Water Acid + base reactions are called neutralisation reactions because the low pH (<7) of an acid and the high pH (>7) of a base, results in a pH closer to neutral (7) These reactions release heat and are exothermic. For example: HCl + NaOH  NaCl + H20 + 58 kJ/mol

  8. Combustion is exothermic Combustion is the process of burning Most often, the combustion of a material involves its combination with oxygen gas. Because combustion reactions release energy in the form of heat and light, they are exothermic. Example: The combustion of methane (natural gas) in oxygen: CH4 + 2O2 CO2 + 2H2O ∆ Hrxn = -832 kJ/mol (exothermic) Fire in the Penola Forest (SA), Ash Wednesday 1983 (Source: www.austehc.unimelb.edu.au/ fam/1611_image.html)

  9. Combustion: why exothermic? Energy required to break bonds • Energy is required to break bonds(endothermic) • Energy is released when bonds form (exothermic) In the previous example involving the combustion of methane: CH4 + 2O2 CO2 + 2H2O + 832kJ Energy released when bonds formed A thermochemical equation Breaking bonds Absorbs energy Forming bonds Releases energy H reactants > H products In summary: ∆Hrxn = H (products) – H (reactants) This reaction releases more energy than it absorbs resulting in a negative value for ∆Hrxn

  10. Exothermic reactions The enthalpy of the reactants is higher than the products. Reactants Products Endothermic reactions The enthalpy of the reactants is lower than the products. Products Reactants ∆ H is -ve ∆ H is +ve Heat absorbed Heat released Enthalpy (H) Enthalpy (H) Reaction progression Reaction progression Energy changes in chemical reactions – Enthalpy diagrams

  11. Examples The combustion of methane is exothermic and releases energy to the surroundings The decomposition of calcium carbonate is endothermic and absorbs energy from the surroundings Source diagrams: http://www.chemhume.co.uk/ASCHEM/Unit%203/13%20Enthalpy/13%20Enthalpyc.htm

  12. Activation energy enthalpy Reactants ∆ H Products Reaction progress Activation Energy The activation energy Ea is the minimum amount of energy that is required by the reactants for bonds to be broken and the reaction to proceed to the products. High Ea means strong bonds in the reactants

  13. Chemical stability • Endothermic reactions • Products are less stable than reactants due to a higher enthalpy content • The higher the enthalpy the less stable • The higher the enthalpy, the weaker the bonds • Exothermic reactions • Products are more stable than reactants due to lower enthalpy content • The lower the enthalpy the more stable • The lower the enthalpy, the stronger the bonds Source: http://www.teachmetuition.co.uk/Energetics/chemical_energetics1.htm

  14. Summary Source diagrams: http://www.chemhume.co.uk/ASCHEM/Unit%203/13%20Enthalpy/13%20Enthalpyc.htm

  15. Exercises • Classify each of the reactions below as endothermic or exothermic • C6H12O6(aq) + 6O2(g) 6CO2(g) + 6H2O(l)ΔH = -2801 kJ mol-1 • 6CO2(g) + 6H2O(l) C6H12O6(aq) + 6O2(g)ΔH = +2801 kJ mol-1 • H2(g) + 1/2 O2(g) H2O(l) + 286 kJmol-1 • When a spark is introduced into a vessel containing a mixture of hydrogen and oxygen gases, they react explosively. • What is the role of the spark in this reaction? • What bonds are broken when this reaction is initiated? • What bonds are generated when the products are formed? • Is the reaction endothermic or exothermic? Justify your answer. • 10.0g of ammonium nitrate is dissolved in 100cm3 of water and the temperature of the solution decreases from 19.00C to 10.500C. • Which is greater, the enthalpy of the reactants or the products? • Is the reaction endothermic or exothermic? • What chemical bonds were broken when the ammonium nitrate dissolved in water? • Consider the reaction represented by the following equation: A + 2B  C • In this reaction, the total enthalpy of the reactants is 80kJmol-1, the total enthalpy for the products is -90kJ.mol-1and the activation energy for the forward reaction is 120kJ.mol-1. • Draw a diagram for the energy profile for this reaction. Label the diagram clearly to show ΔH and Ea. • State whether the forward reaction is endothermic or exothermic • Calculate the change in enthalpy • Calculate the amount of energy released as the products are formed • Calculate the activation energy of the reverse reaction: C  A 2B

  16. Exercises - Answers a) exothermic b) endothermic c) exothermic a)spark is to overcome the activation energy b) H-H and O=O covalent bonds broken c) O-H bonds are made during the reaction d) exothermic as energy is released a) products>reactants b) endothermic c) ionic bonds between NH4+ and NO3- a) diagram should show exothermic forward rxn b) exothermic c) ΔH = -90 – 80 kJ mol-1 = -170 kJ mol-1 d) 290 kJ mol -1 e) 290 kJ mol-1

  17. 5.2 Calculating Enthalpy

  18. Measuring Enthalpy Changes Recall that changes in temperature are related to changes in the amount of heat energy that is being absorbed or released in a system. This quantity of heat energy, q in joules (J), can be calculated using the following equation and often involves water being heated by reactions: q = m C ∆T Change in Temperature (final – initial) Heat energy released or absorbed in Joules (J) Mass (g) Specific heat Capacity (4.18 J g-1K-1 for water) Example: What quantity of energy is required to raise the temperature of 0.5L of water from 200C to 1000C? q = m C ∆T = 500g X 4.18 J g-1K-1 X 800C or K = 167,200 J = 167 kJ NB: 0C = K - 273

  19. Heat Capacity Values Heat capacity is a measure of how much energy a substance can absorb without changing temperature. These values vary greatly among substances. Specific heat capacity (C), is the amount of heat energy in Joules (J), required to raise the temperature of 1g of a substance by 1 Kelvin (K).

  20. Practice Questions • A Bunsen burner was used to heat 100 cm3 of water for 10 min. The temperature of the water increased from 15.00C to 80.00C. Determine the heat energy change of the water. (note: the density of water is 1.0 g cm-3) • The same Bunsen burner was used to heat a 500g block of copper for 10 minutes. Assuming the same amount of energy is transferred from the Bunsen burner to the block as in question 1, determine the highest temperature that the block could reach if the starting temperature was 15.00C. • Explain the difference in the temperatures considering the same amount of energy was put into the water and the copper Solutions: 1)27.2kJ 2) 1560C 3) The water has a much higher heat capacity than the copper which means the water can absorb much more heat energy before changing temperature.

  21. Calorimetry What is calorimetry Water can be used to measure the change in heat energy in a chemical reaction due to its ability to absorb heat. This is known as calorimetry. A calorimeter is a device that is used to measure the enthalpy change that occurs during a chemical reaction. Measuring heat in the laboratory If a known quantity of water is placed in the calorimeter and a reaction is carried out, the change in temperature due to the reaction is transferred to the water and is measured using a thermometer placed in a hole in the lid. The heat energy change is: q = m C ∆T The amount of heat lost or gained in the reaction is equal in size but opposite in sign to the amount of heat lost or gained by the water. Coffee cup calorimeter The simplest calorimeter makes use of two polystyrene (Styrofoam) cups, one inside the other with a lid on top. This minimises the amount of heat lost to the surroundings. This is the most significant source of error.

  22. Calorimetry example When a student added excess Mg is to 100cm3 of 2.00 mol dm-3 CuSO4, the temperature rose from 20.0oC to 65oC. i)  The energy change: Q=mxcxΔT Q=100x4.18x(65 - 20)Q=100x4.18x45Q=18810 J Q=18.81 kJConvert the energy calculated into KJ (divide by 1000) ii)  Calculate the number of moles used:                                                   No moles = c  x  V                                                                          1000                                                   No moles = 2  x  100                                                                          1000                                                   No moles =  0.2 iii)  Calculate the amount of energy exchanged per mole , this is the enthalpy:                                                               Enthalpy = Energy                                                                                     moles                                                               Enthalpy = - 18.81                                                                                        0.2                                                               Enthalpy = - 94.05 kJ mol-1 iv) Finally write the equation with the enthalpy change: Mg(s)+CuSO4(aq) MgSO4(aq)+Cu(s)ΔH=- 94.05 kjmol-1 It is exothermic, therefore negative

  23. Heat of solution In order for a solute to dissolve in a solvent, the solute particles and the solvent particles must separate, which involves energy changes. In addition, solute particles interact with solvent particles, which also can result in energy changes. Overall: ΔHsoln = ΔH1 + ΔH2 + ΔH3 The energy profile diagrams on the right show net exothermic and endothermic dissolutions Question: How are these processes similar to a chemical reaction? How are they different? Diagrams Source: http://wps.prenhall.com/wps/media/objects/3312/3391718/blb1301.html

  24. Heat of Combustion (ΔHc) • Heat of Combustion of a substance is the heat liberated when 1 mole of the substance undergoes complete combustion with oxygen at constant pressure. • Combustion is always exothermic, ΔH is negative. • By definition, the heat of combustion is minus the enthalpy change for the combustion reaction. • By definition, the heat of combustion is a positive value. • Heat of Combustion can be measured experimentally.

  25. Heat of Combustion (ΔHc) The diagram to the right shows a typical school laboratory setup to test the heat of combustion of a fuel such as ethanol. A known quantity of water is placed in a flask or beaker (calorimeter) A thermometer is positioned with bulb near the middle of the volume of water A known quantity of fuel, such as an alcohol (alkanol), is placed in the spirit burner The initial temperature of the water is measured and recorded (Ti) The wick on the spirit burner is lit, burning the fuel, and heating the water When the temperature has risen an appreciable amount, the spirit burner is extinguished and the final temperature recorded (Tf) The final quantity of fuel is measured and recorded Source: http://www.ausetute.com.au/heatcomb.html

  26. Heat of Combustion Example A student used the apparatus on the previous slide to determine the heat of combustion of ethanol. Their results for one trial are below: Use the results above to determine the heat of combustion for ethanol in kJ/mol. (Solution on the next slide)

  27. Heat of Combustion Example • Calculate moles (n) of fuel used • molecular mass (MM) of ethanol = 46.1g/mol mass ethanol used = 1.75g • n = mass ÷ MM = 1.75 ÷ 46.1 = 0.0380 mol • 2. Calculate energy required to change temperature of water • energy = mass of water x specific heat capacity of water x change in water temp (mCΔT) • energy = 200g x 4.184 JK-1g-1 x 55oC = 46024 J = 46.024 kJ • 3. Calculate the heat of combustion of ethanol • Assume all the heat produced from burning ethanol has gone into heating the water, ie, no heat has been wasted. • 0.0380 mole ethanol produced 46.024 kJ of heat. • Therefore 1 mole of ethanol would produce 46.024 kJ ÷ 0.0380 mol = 1211 kJ/mol • The heat of combustion of ethanol is 1211 kJ/mol • The experimentally determined value for the heat of combustion of ethanol is usually less than the accepted value of 1368 kJ/mol because some heat is always lost to the atmosphere and in heating the vessel.

  28. Heat loss compensation During exothermic reactions using a calorimeter in the laboratory, there are many errors associated with the loss of heat. Because reactions are not instantaneous, heat loss is often gradual. We can compensate for this heat loss by graphing temperature against time. Then, we can extrapolate back to what would have been the final temperature if no heat had been lost to the surroundings.

  29. 5.3 Hess’s Law

  30. Hess’s Law According to the Law of Conservation of Energy, the energy exchanges that take place in reactions must be conserved. Hess’s Law states that the enthalpy of a chemical process is the same, whether the process takes place in one step or several steps. H is the same regardless of how the reactants change into products The diagram above shows an enthalpy cycle and reflects Hess’s Law: ΔHx = ΔH1 + ΔH2 So, if we cannot measure the enthalpy for the overall reaction (x), but can measure the reactions (1) and (2), we can determine the enthalpy of reaction mathematically.

  31. Hess’s Law Analogy • Little Red Riding Hood wants to visit her grandmother, who lives at the top of the mountain. • Her grandmother’s house is 2000 meters above Red’s house. • Regardless of how Red climbs the mountain, her change in altitude will be the same: 2000 meters.

  32. The pathway doesn’t matter; the altitude change is the same. Grandma’s House Red’s House

  33. Hess’s Law ProblemsHelpful Hints • 1. If you reverse a reaction, the sign of H changes. • 2H2O2(l) 2H2O(l) + O2(g) • H = -196.4 kJ • 2H2O(l) + O2(g) 2H2O2(l) • H = +196.4 kJ

  34. Hess’s Law ProblemsHelpful Hints • 2. If you multiply the coefficients of a thermochemical equation by some number, you must multiply H by the same number. • 2H2O2(l) 2H2O(l) + O2(g) • H = -196.4 kJ • 4H2O2(l)  4H2O(l) + 2O2(g) • H = -392.8 kJ • H2O2(l) H2O(l) + ½O2(g) • H = -98.2 kJ

  35. Hess’s Law ProblemsHelpful Hints • 3 Thermochemical equations can be combined to give new thermochemical equations. • N2 + 2O22NO2 • H = 67.8 kJ • N2O4N2 + 2O2 • H = -9.6 kJ • N2 + 2O2 + N2O4 2NO2 + N2 + 2O2 • N2O4  2NO2 • H = (67.8 kJ) + (-9.6 kJ) = 58.2 kJ

  36. Hess’s Law Problems - summary Below are the steps to follow when performing Hess’s Law problems: Rearrange the balanced chemical equations so that the reactants and products are on the correct sides. If you reverse a chemical equation, then the sign of the enthalpy also changes. Check that you have the correct states, enthalpy changes will be different for species in the solid, liquid and gas states. Assign each rearranged equation the correct ΔH value.Remember, if you need to multiply each species in the chemical equation by 2, then the enthalpy change must also be multiplied by 2. Add the rearranged equations together to give the overall equation for the reaction. Add the ΔH values for these equations to calculate ΔH for the overall reaction.

  37. Example 1 When carbon combusts in an excess of oxygen, carbon dioxide is formed and 393.5 kJ of heat is released per mole of carbon. C(s) + O2(g) -----> CO2(g)ΔH = -393.5 kJ This overall reaction can also be produced as a two stage process: Carbon combusts in limited oxygen producing carbon monoxide: C(s) + ½O2(g) CO(g)ΔH = -110.5 kJ Carbon monoxide then combusts in additional oxygen: CO(g) + ½O2(g) CO2(g)ΔH = -283.0 kJ These two equations can be added together to calculate ΔH for the overall reaction: C(s) + ½O2(g) ---> CO(g)ΔH = -110.5 kJ CO(g) + ½O2(g) ---> CO2(g)ΔH = -283.0 kJ C(s) + O2(g) ---> CO2(g)ΔH = -393.5 kJ

  38. Example 1 - Graph This shows a graphical representation of the previous example. Using one step or two, the overall enthalpy change is the same. This follows Hess’s Law Source: http://www.ausetute.com.au/hesslaw.html

  39. Example 2 – you try Calculate ΔH for the reaction: NH3(g) + HCl(g)NH4Cl(s) Given that: ½N2(g) + 1½H2(g)NH3(g)ΔH = -46.1 kJ ½H2(g) + ½Cl2(g)HCl(g)ΔH = -92.3 kJ ½N2(g) + 2H2(g) + ½Cl2(g)NH4Cl(s)ΔH = -314.4 kJ

  40. Example 2 - Solution 1. Rearrange each balanced equation with reactants and products on the correct side, ie, NH3(g) and HCl(g) on the left, NH4Cl(s) on the right hand side. NH3(g) ½N2(g) + 1½H2(g) HCl(g) ½H2(g) + ½Cl2(g) ½N2(g) + 2H2(g) + ½Cl2(g)NH4Cl(s) 2. Assign correct ΔH values to each equation. NH3(g) ½N2(g) + 1½H2(g)ΔH = +46.1 kJ HCl(g) ½H2(g) + ½Cl2(g)Δ H = +92.3 kJ ½N2(g) + 2H2(g) + ½Cl2(g)NH4Cl(s)Δ H = -314.4 kJ 3. Add the rearranged equations together. NH3(g)½N2(g)+ 1½H2(g) ΔH = +46.1 kJ HCl(g)½H2(g)+ ½Cl2(g)Δ H = +92.3 kJ ½N2(g) + 2H2(g) + ½Cl2(g)NH4Cl(s)Δ H = -314.4 kJ NH3(g)+ HCl(g)NH4Cl(s) 4. Add the ΔH values. NH3(g)½N2(g) + 1½H2(g)Δ H = +46.1 kJ HCl(g)½H2(g) + ½Cl2(g)Δ H = +92.3 kJ ½N2(g)+ 2H2(g) + ½Cl2(g) NH4Cl(s)Δ H = -314.4 kJ NH3(g) + HCl(g)NH4Cl(s)Δ H = -176.0 kJ

  41. Exercise Calculate the enthalpy change (Δ H°), in kJ, of the reaction: N2H4 + 2 H2O2 N2 + 4 H2O using the following enthalpy of combustion data: a. N2H4 + O2 N2 + 2 H2O ΔH° = -622 kJ b. H2 + O2 H2O2Δ H° = -188 kJ c. H2 + ½ O2 H2O Δ H° = -286 kJ To solve this, leave equation (a) alone. Reverse equation (b) and double the coefficients. Then, double the coefficients for equation (c). When that is complete, equations a, b, and c add up to the desired equation and the overall DH° value equals -818 kJ.

  42. Finding Enthalpy changes using Hess’s Law cycles Standard enthalpy change of formation ΔH0f is the energy change when 1 mole of a compound forms from its elements at under standard conditions and constant pressure. If we want to determine ΔH0f the for a substance such as benzene C6H6, we can use the standard enthalpy changes of combustion and Hess’s Law cycles to help us. You can find the data below in your data booklet.

  43. Drawing the cycle 3. There are two possible routes to the same products. These routes are shown in blue. 1. The formation of benzene is drawn horizontally 2. The Hess’s Law cycle is completed showing the combustion products. (O2 has been left off for convenience only - it would be added in excess anyway) 4. Construct and solve the equation: ΔH - 3267 = 6(-394) + 3(-286) ΔH = 3267 + 6(-394) + 3(-286) ΔH = +45 kJ mol-1

  44. Hess’s Law cycle – example 2 Calculate the enthalpy change of the reaction between ethene and hydrogen chloride gases to form chloroethane gas using a Hess’s Law cycle. These standard enthalpy changes of formation can be found in your data booklet (except HCl) ΔH for this reaction can easily be found by the same method as in the previous question once the cycle has been drawn. Find the two pathways that lead to the same product and equate them. Then solve for the unknown. +52 - 92 + ΔH = -137 ΔH = -52 + 92 - 137 ΔH = -97 kJ mol-1 -137

  45. 5.4 Bond Enthalpies

  46. Bond enthalpy Also known as bond energies, this is the enthalpy change required to break a covalent bond when all species are in the gaseous state. These are positive values as breaking bonds is an exothermic process that requires energy. We can use bond enthalpy values to estimate the overall enthalpy changes in reactions. We simply take the sum of the individual bond energies for reactants and subtract the sum of the bond enthalpies for the products (negative because forming bonds releases energy – exothermic) Consider this example: CO(g) + H2O(g)  CO2(g) + H2(g)

  47. AVERAGE BOND ENERGIES OF COMMON BONDS The amount of energy required to break these bonds Use these bond energy values to determine the ∆ Hrxn of methane combustion ∆Hrxn = ∑ E (bonds broken) - ∑ E (bonds formed)

  48. Example - Solution CO(g) + H2O(g)  CO2(g) + H2(g) ΔH = 1077 + 2(464) – 2(746) – 436 ΔH = + 77 kJ mol -1

  49. 5 Energetics Compiled by: Robert Slider (2011) Please share this resource with others

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