Learning objective: To measure enthalpy of combustion of alcohols To explain the trend in terms of molecules To identify and evaluate sources of error 22/11/2016 Enthalpy = heat energy (H) The enthalpy of a substance can’t be measured… …but the change can be. system reactants and products surroundings the rest of the universe!

Enthalpy change (ΔH) Enthalpy (H) reactants products Enthalpy of reactants > enthalpy of products Enthalpy lost to surroundings Surroundings heat up EXOTHERMIC ΔH -ve

Enthalpy change (ΔH) Enthalpy (H) reactants products Enthalpy of reactants < enthalpy of products Enthalpy taken from surroundings Surroundings cool down ENDOTHERMIC ΔH +ve

Standard enthalpy of combustion ΔH c Ө The standard enthalpy of combustion is the enthalpy change when one mole of a substance completely reacts with oxygen under standard thermodynamic conditions 1 atm and 298K

SAFETY GOGGLES must be worn; hair and loose clothing must be secured; remove excess papers; alcohols are flammable and TOXIC Always have calorimeter over the base of the stand

Calculation Enthalpy change (ΔH) of system (J) = - (mass water (g) x specific heat capacity x ΔT) Enthalpy per gram Enthalpy per mole: calculate amount of alcohol burned Comment on the trends in your results

Things to think about What is the relationship between number of chemical bonds and energy released per mole? Why do you need to light the fuel if the reactions are so exothermic? How are chemical bonds breaking and forming related to the energy change? methanolethanolpropanol butanol + O=O + O=O + O=O  O=C=O + O=C=O + H-O-H + H-O-H + H-O-H

Why won’t the burner light until the lit splint arrives? Is the splint supplying or taking energy from the alcohol and oxygen? + O=O + O=O + O=O  O=C=O + O=C=O + H-O-H + H-O-H + H-O-H What must happen to the bonds between the atoms for the reaction to start? Does breaking bonds require or release energy?

Bonds and energy changes Watch the MSS slideshow The changes can be represented in an energy level diagram: Heat energy H-H + H-H + O=O H-O-H + H-O-H EXOTHERMIC – energy released to surroundings BONDS FORMING ENDOTHERMIC – energy taken from surroundings BONDS BREAKING This is the ACTIVATION ENERGY H + H + H + H + O + O Heat energy to the surroundings – overall the reaction is exothermic Energy required to break bonds: 2 x H-H + 1 x O=O = 2 x x 496 = 1367 kJ/mol Energy released by forming bonds = 4 x O-H = 4 x 463 = 1852 kJ/mol Overall energy released = 1852 – 1367 = 485 kJ/mol

Practice calculation Try the bond energy method to calculate the energy released on burning 1 mole of ethanol Include an energy level diagram in your calculation + O=O + O=O + O=O  O=C=O + O=C=O + H-O-H + H-O-H + H-O-H Bond Energies (kJ/mol) C-CC-HC-OO-HO=OC=O Write down the number of each bond broken 2.Write down the number of each bond formed 3.Calculate the total energies of the broken and formed bonds 4.Subtract the energy of the broken bonds (ENDOTHERMIC) from the energy of the formed bonds (EXOTHERMIC)

+ O=O + O=O + O=O  O=C=O + O=C=O + H-O-H + H-O-H + H-O-H Bond Energies (kJ/mol) C-CC-HC-OO-HO=OC=O Write down the number of each bond broken 2.Write down the number of each bond formed 3.Calculate the total energies of the broken and formed bonds 4.Subtract the energy of the broken bonds (ENDOTHERMIC) from the energy of the formed bonds (EXOTHERMIC) 1.1 x C-C + 5 x C-H + 1 x C-O + 1 x O-H + 3 x O=O 2.4 x C=O + 6 x O-H 3.1 x x x x x 496 = 4835 kJ/mol 4.4 x x 463 = Overall reaction: = 915 kJ/mol

O=C=O + H-O-H + H-O-H + H-O-H H H H H H H O O O O O O O C C Heat energy 5750 kJ/mol BONDS FORMING 4835 kJ/mol BONDS BREAKING This is the ACTIVATION ENERGY 915 kJ/mol to surroundings + O=O + O=O + O=O

Sources of error Why is your experimental result so much lower than your calculated value? Estimate the percentage error Is this error likely to be systematic (the same each time)? Are you still confident in your conclusions about the relationship between bonds and enthalpy? How could we reduce the experimental error?

Bomb calorimeter

Actual values of ΔH c Ө Methanol = kJmol -1 Ethanol = kJmol -1 Propanol = kJmol -1 Butanol = kJmol -1 Why is this different from the bond energy calculation? Which one is right?