Transcript Slide 1
Enthalpy = heat produced/absorbed during reactions (kJmol-1) Exothermic Endothermic • Energy out • ∆rH = -ve • Reactants have more energy than products • Bonds forming • e.g. condensing • Energy in • ∆rH = +ve • Reactants have less energy than products • Bonds breaking • e.g. evaporating Specific heat capacity (water is 4.18Jg-1C-1) Change in energy (J) ∆E = m s ∆T Change in enthalpy (kJmol-1) Temperature Mass ofchange (ignore -) liquid (g) ∆H = ∆E n No. moles Molar heat of fusion (∆fusH) Energy required to change one mole of a substance from a solid to a liquid at the melting point Indicates the strength of the force holding the particles together in a solid H2O(solid at 0C)→H2O(liquid at 0C) ∆fusH = 6.0kJmol-1 Molar heat of vaporisation (∆vapH) Energy required to change one mole of a substance from a liquid to a gas at the boiling point Indicates the strength of the force holding the particles together in a liquid H2O(liquid at ∆vapH is always than 100C) higher → H2O(gas at ∆ 100C) fusH ∆vapH = 41.0kJmol-1 Use the following information to answer the question below: Ethanal Propanal Butanal Ethanoic acid CH3C CH3CH2C CH3CH2CH2C CH3C 30 34 52 ∆vapH/kJm 26 ol-1 Discuss the trend in ∆vapH of the compounds in the table above in terms of the attractive forces between the particles and the factors affecting those forces. ∆vapH is the amount of heat energy required to change a mole of liquid into a gas. In ethanal, propanal & butanal there is a polar bond between the C and the O so there will be a permanent dipole-dipole interaction between the molecules. Also, they all have a nonpolar chain so they will have temporary dipole-dipole, however, as this attraction increases with molar mass, so the amount of energy required to overcome the forces will be small and in this order ethanal<propanal<butanal. Ethanoic acid has a much smaller molar mass so the temporary dipole-dipole interactions resulting from the non-polar tail will be less than the others. Despite this, ∆vapH of ethanoic acid is much higher than the others as it has a polar bonds resulting in permanent dipole-dipole interactions. Also, due to the O-H bond, ethanoic acid forms hydrogen bonds so it will take much more energy to overcome these stronger forces. Enthalpy change when one mole of the substance is completely burnt with all reactants and products in their standard states. Room temp (25C) and ∆cH Always exothermic pressure Be careful when balancing... must start with 1mole of reactant C2H6(g) +3½ O2(g) →2CO2(g) + 3 H2O(l) ∆cH= -1557kJmol-1 When 1 mole of 1-propanol is ignited under standard conditions according to the equation CH3CH2CH2OH(l) + 9/2O2(g) → 3CO2(g) + 4H2O(l) 2016kJ of energy is released. Only mole in a) Give the value (with units) of ∆cH for 1this reaction equation. All under these conditions. exo... Sketch a diagram showing how the total energy content of this reaction system changes during the course of this reaction. Yr12 diagram to show products and reactants Energy CH3CH2CH2OH + O2 ∆Hc = -2016kJmol-1 H2O + CO2 Time Enthalpy change when one mole of the substance is formed from its elements, with all reactants and products in their standard states. ∆fH Be careful when balancing... must end with 1mole of product 2 C(s) + 3 H2(g) + ½O2 → C2H5OH(l) ∆fH= -278kJmol-1 ∆rH = ∑∆fH(products) - ∑∆fH(reactants) Be careful with your minus signs. Minus a minus = +! ∆fH for elements are always zero Calculate ∆H for the highly exothermic You must set ‘thermite’ reaction this out very Cr2O3(s) + 2Al(s) → 2Cr(s) + Al2Ocarefully! 3(s) Given ∆fH (Cr2O3) = -1128.4 kJmol-1 ∆fH (Al2O3) = -1669.3 kJmol-1 When 1 mole of 1-propanol is ignited under standard This reactant is conditions according to the equation unknown....we CH3CH2CH2OH(l) + 9/2O2(g) → 3CO2(g) + 4H2O(l) will have to 2016kJ of energy is released. rearrange the Given the additional data at 25C calculate the formula standard heat of formation for 1-propanol. This gives us ∆fH (kJmol-1) -285 ∆HrH it is 2O(l) CO2(g) -393 negative To get∆fH on its own we will have to minus the 2016. This makes it positive on the other side • Write data in the form of equations • Rewrite equations so desired species on the correct side, reversing signs if necessary • Add equations (cancelling terms) and ∆H together The smelting of zinc ores involves the reaction of Zn and ZnS with oxygen gas according to This the following equations: is what you Zn(s) + ½O2(g) → ZnO(s) ∆rH = -348kJmol-1 are aiming for. -1 ZnS(s) + 1½O2(g)→ZnO(s) + SO2(g) ∆rH = -441kJmol Look for the bits Using the following information and the data above, you want and flip S(s) + O2(g) → SO2(g) ∆rH = -297kJmol-1 the equation if Calculate the enthalpy change, ∆rH for the reaction: necessary Zn(s) + S(s) → ZnS(s) You may need to multiply individual equations to match the desired equation Using the result of the calculation, describe, with a reason, whether the heat of formation of ZnS is endothermic or exothermic Bond enthalpy = change in energy upon breaking one mole of the respective bonds. Bond breaking = endothermic Bond making = exothermic We can estimate the enthalpy of a reaction by adding the energy of the bonds broken to the energy of the bonds formed Can only do this if the reactants and products are in the gaseous state Carbon monoxide is reacted with steam to produce hydrogen gas. CO(g) + H2O(g) → H2(g) + CO2(g) ∆rH = -41.2kJmol-1 a) The bond enthalpies for the carbon bonds in CO2 an is the Use the bond enthalpies in the table CO areThis different. unknown so the and the enthalpy of the reaction to calculate the bond equation enthalpy of will the need carbon to oxygen bond in carbon This is the total to be rearranged monoxide. for the reaction...total Bond Bond enthalpy for Firstand draw out bonds breaking (kJmol-1) forming the bonds O-H 463 H-H 436 C=O 743