Chemsheets AS006 (Electron arrangement) 12/09/2018 CALORIMETRY

Introduction to Thermodynamics 10.2 Thermodynamics is the study of the interconversion of heat and other kinds of energy. In thermodynamics, there are three types of systems: An open system can exchange mass and energy with the surroundings. A closed system allows the transfer of energy but not mass. An isolated system does not exchange either mass or energy with its surroundings.

Calorimetry Fuse school ( calorimeters and heats of combustion ) https://www.youtube.com/watch?v=ifTtdvF98T8

PPT - AS Calorimetry 1 g of water 12/09/2018 1 ºC hotter 1 g of water Energy required = 4.18 J 2 ºC hotter 1 g of water Energy required = 2 x 4.18 J = 8.36 J

Energy required = 4.18 J Energy required = 5 x 4.18 J 1 ºC hotter 1 g of water Energy required = 4.18 J 1 ºC hotter 5 g of water Energy required = 5 x 4.18 J = 20.9 J

Energy required = 4.18 J Energy required = 3 x 10 x 4.18 1 ºC hotter 1 g of water Energy required = 4.18 J 10 ºC hotter 3 g of water Energy required = 3 x 10 x 4.18 = 125.4 J

Energy required = 0.39 J Energy required = 10 x 5 x 0.39 1 ºC hotter 1 g of copper Energy required = 0.39 J 5 ºC hotter 10 g of copper Energy required = 10 x 5 x 0.39 = 19.5 J

energy needed to make 1 g of substance 1ºC hotter q = m c ∆T mass heated (m) x energy needed to make 1 g of substance 1ºC hotter temperature rise (∆T)

Calculate the heat to take 255 g of water from 25.2 C to 90.5 C. Strategy Get the givens c = 4.184 J/g∙°C, m = 255 g, ΔT = 90.5°C – 25.2°C = 65.3°C. Formula q=mc ΔT Solution q = 255 g × 4.184 J/g∙°C x 65.3°C q= 69669.876 J q= 69669.876 J / 1000 J/kJ q= + 69.7 kJ (3 figs) the heat is going INTO the water, so it is +ve for the water This is heat that was GIVEN to the water, but FROM where. A chemical must have given it’s heat away so it is –VE for the chemical – 96.7kJ NOW, molar heat

2 moles of ethanol burned Energy given out = 2734 kJ Energy given out by 1 mole = 2734 kJ 2 = 1367 kJ ∆H = –1367 kJ mol-1

q moles in reaction ∆H =

Experimental Determination of Enthalpy Changes by Calorimetry Calorimeter = a container used for measuring the temperature change of solution Determination of Enthalpy Change of Neutralization

Determination of Enthalpy Change of Combustion Experimental Determination of Enthalpy Changes by Calorimetry Determination of Enthalpy Change of Combustion

Cal Problems worked 4 min https://www.youtube.com/watch?v=Hs5x0-IU2F4

Excess powdered zinc was added to 100ml of 0 Excess powdered zinc was added to 100ml of 0.2 mol/L copper (II) sulphate solution. A temperature rise of 10oC was recorded. Find the MOLAR enthalpy change for the reaction. H = m x c x T H = 100g x 4.18 KJ kg-1 K-1 x 10 oC 1000 = 4.180 KJ This is for the no of moles of CuSO4 used in the experiment No of moles of CuSO4 = 0.2 x 100 = 0.02 moles H = 4.180 = 209 KJ mol-1 0.02 The reaction is exothermic so we need to put in a negative sign Hr = - 209 KJ mol-1 of CuSO4

Combustion To find the heat of combustion of a substance a known mass of the substance is burned, the heat released transferred to water and the enthalpy change found as before In an experiment to find the heat of combustion of ethanol the following results were obtained Initial mass of lamp + ethanol = 65.20g Final mass of lamp = 64.28g Final temperature of water = 47.1oC Initial temperature of water = 28.5oC Mass of the water = 300g What are the products of complete combustion of ethanol? What mass of ethanol was burnt? How many moles is this? What quantity of heat was transferred to the water? Find Hc of ethanol Identify any sources of error Is ethanol a good fuel?

C2H5OH + 3O2  2CO2 + 3H2O q=mc∆t H = 300 x 4.18 KJ kg-1 K-1 x 18.6K = 23.3KJ 1000 H lost = h gained (so 23.3 KJ gained by calorimeter water was lost in burning) Mass of ethanol used = 0.92g 0.92g = 0.92 g = 0.020 mol 46g/mol H=mc∆t / n Hc = - 23.3 = -1160 KJ/mol 0.020

Errors Heat lost to surroundings (air, can thermometer) Errors in measuring temperature change (unavoidable error in reading thermometer) Errors in measuring masses (unavoidable error in reading balance) The enthalpy change of combustion is high ethanol is a good fuel

Chemistry in Action: Fuel Values of Foods and Other Substances C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) DH = -2820 kJ/mol Per gram Sugar = 2999 KJ / 180 g = 15.7 KJ/gram 1 cal = 4.184 J 1 Cal = 1000 cal = 4184 J =4.184KJ If 1.0 gram sugar is = 15.7 KJ Or 15.7 KJ / 4.184 KJ/Cal = 3.75 Cal So sugar burned is approx 4Cal/g Here 26g x 4 Cal/g = 104 Cal from sugar

Foods and Fuels Cheetos burned v Cereal https://www.youtube.com/watch?v=UDgeaAMdYIY Foods 1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal. Energy in us comes from carbohydrates and fats TED X (3 min, Calories) https://www.youtube.com/watch?v=VEQaH4LruUo Intestines: carbohydrates converted into glucose: C6H12O6 + 6O2  6CO2 + 6H2O, DH = -2816 kJ Fats break down as follows: 2C57H110O6 + 163O2  114CO2 + 110H2O, DH = -75,520 kJ Fats contain more energy; are not water soluble, so are good for energy storage.

moles of propane = mass / Mr = 1.00 / 58.0 = 0.01724 1) In an experiment, 1.00 g of propanone (CH3COCH3) was completely burned in air. The heat evolved raised the temperature of 150 g of water from 18.8C to 64.3C. Use this data to calculate the enthalpy of combustion of propanone (the specific heat capacity of water is 4.18 J g-1 K-1). q = mc∆T m = 150 c = 4.18 ∆T = 45.5 q = 150 x 4.18 x 45.5 = 28530 J ∆H = q / mol moles of propane = mass / Mr = 1.00 / 58.0 = 0.01724 ∆H = –28.53 / 0.01724 = -1650 kJ mol-1 (3 sig fig)

moles of propane = mass / Mr = 1.00 / 86.0 = 0.01163 2) In an experiment, 1.00 g of hexane (C6H14) was completely burned in air. The heat evolved raised the temperature of 200 g of water from 293.5 K to 345.1 K. Use this data to calculate the enthalpy of combustion of hexane (the specific heat capacity of water is 4.18 J g-1 K-1). q = mc∆T m = 200 c = 4.18 ∆T = 51.6 q = 200 x 4.18 x 51.6 = 43140 J ∆H = q / mol moles of propane = mass / Mr = 1.00 / 86.0 = 0.01163 ∆H = –43.14 / 0.01163 = -3710 kJ mol-1 (3 sig fig)

moles of propane = mass / Mr = 1.56 / 60.0 = 0.02600 3) In an experiment, 1.56 g of propan-1-ol (CH3CH2CH2OH) was completely burned in air. The heat evolved raised the temperature of 0.250 dm3 of water from 292.1 K to 339.4 K. Use this data to calculate the enthalpy of combustion of propan-1-ol (the specific heat capacity of water is 4.18 J g-1 K-1). q = mc∆T m = 250 c = 4.18 ∆T = 47.3 q = 250 x 4.18 x 47.3 = 49430 J ∆H = q / mol moles of propane = mass / Mr = 1.56 / 60.0 = 0.02600 ∆H = –49.43 / 0.02600 = -1900 kJ mol-1 (3 sig fig)

4). 25 cm3 of 2. 0 mol dm-3 nitric acid was added to 25 cm3 of 2 4) 25 cm3 of 2.0 mol dm-3 nitric acid was added to 25 cm3 of 2.0 mol dm-3 potassium hydroxide solution. The temperature rose by 13.7C. Calculate the enthalpy of neutralisation for this reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1. q = mc∆T m = 50 c = 4.18 ∆T = 13.7 q = 50 x 4.18 x 13.7 = 2863 J ∆H = q / mol Mol HNO3 = conc x vol = 2.0 x 25/1000 = 0.050 Mol KOH = conc x vol = 2.0 x 25/1000 = 0.050 HNO3 + KOH → NaNO3 + H2O ∆H = –2.863 / 0.050 = -57.3 kJ mol-1 (3 sig fig)

5) 50 cm3 of 0.10 mol dm-3 silver nitrate solution was put in a calorimeter and 0.2 g of zinc powder added. The temperature of the solution rose by 4.3C. Work out which reagent was in excess and then calculate the enthalpy change for the reaction (per mole Limiting reagent . Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1. Ignore the heat capacity of the metals. q = mc∆T m = 50 c = 4.18 ∆T = 4.3 q = 50 x 4.18 x 4.3 = 898.7 J ∆H = q / mol Mol AgNO3 = conc x vol = 0.1 x 50/1000 = 0.0050 Mol Zn = mass / Mr = 0.2 / 65.4 = 0.00306 XS 2 AgNO3 + Zn → ZnNO3 + 2 Ag ∆H = –0.8987 kJ / 0.0050AgNO3 = -180 kJ mol-1 (3 sig fig)

EXTRA PROBLEMS

moles of propane = mass / Mr = 0.600 / 44.0 = 0.01364 1) In an experiment, 0.600 g of propane (C3H8) was completely burned in air. The heat evolved raised the temperature of 100 g of water by 64.9C. Use this data to calculate the enthalpy of combustion of propane (the specific heat capacity of water is 4.18 J g-1 K-1). q = mc∆T m = 100 c = 4.18 ∆T = 64.9 q = 100 x 4.18 x 64.9 = 27130 J ∆H = q / mol moles of propane = mass / Mr = 0.600 / 44.0 = 0.01364 ∆H = –27.13 / 0.01364 = -1990 kJ mol-1 (3 sig fig)

2). 50 cm3 of 1. 0 mol dm-3 hydrochloric acid was added to 50 cm3 of 1 2) 50 cm3 of 1.0 mol dm-3 hydrochloric acid was added to 50 cm3 of 1.0 mol dm-3 sodium hydroxide solution. The temperature rose by 6.8C. Calculate the enthalpy of neutralisation for this reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1. q = mc∆T m = 100 c = 4.18 ∆T = 6.8 q = 100 x 4.18 x 6.8 = 2842 J ∆H = q / mol Mol HCl = conc x vol = 1.0 x 50/1000 = 0.050 Mol NaOH = conc x vol = 1.0 x 50/1000 = 0.050 HCl + NaOH → NaCl + H2O ∆H = –2.842 / 0.050 = -57 kJ mol-1 (2 sig fig)

3) 100 cm3 of 0.200 mol dm-3 copper sulphate solution was put in a calorimeter and 2.00 g of magnesium powder added. The temperature of the solution rose by 25.1C. Work out which reagent was in excess and then calculate the enthalpy change for the reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1. Ignore the heat capacity of the metals. q = mc∆T m = 100 c = 4.18 ∆T = 25.1 q = 100 x 4.18 x 25.1 = 10490 J ∆H = q / mol Mol CuSO4 = conc x vol = 0.200 x 100/1000 = 0.020 Mol Mg = mass / Mr = 2.00 / 24.3 = 0.0823 XS CuSO4 + Mg → MgSO4 + Cu ∆H = –10.49 / 0.020 = -525 kJ mol-1 (3 sig fig)

4). 50 cm3 of 2. 0 mol dm-3 hydrochloric acid was added to 50 cm3 of 2 4) 50 cm3 of 2.0 mol dm-3 hydrochloric acid was added to 50 cm3 of 2.0 mol dm-3 ammonia solution. The temperature rose by 12.4C. Calculate the enthalpy of neutralisation for this reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1. q = mc∆T m = 100 c = 4.18 ∆T = 12.4 q = 100 x 4.18 x 12.4 = 5183 J ∆H = q / mol Mol HCl = conc x vol = 2.0 x 50/1000 = 0.100 Mol NH3 = conc x vol = 2.0 x 50/1000 = 0.100 HCl + NH3 → NH4Cl ∆H = –5.183 / 0.100 = -51.8 kJ mol-1 (3 sig fig)

2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O 5) 50 cm3 of 1.0 mol dm-3 nitric acid was added to 20 cm3 of 1.0 mol dm-3 barium hydroxide solution. The temperature rose by 7.9C. Calculate the enthalpy of neutralisation for this reaction (per mole of nitric acid reacting). Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1. q = mc∆T m = 70 c = 4.18 ∆T = 7.9 q = 70 x 4.18 x 7.9 = 2312 J ∆H = q / mol Mol HNO3 = conc x vol = 1.0 x 50/1000 = 0.050 XS Mol Ba(OH)2 = conc x vol = 1.0 x 20/1000 = 0.020 2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O ∆H = –2.312 / 0.040 = -57.8 kJ mol-1 (3 sig fig)

6) 25 cm3 of 1.00 mol dm-3 copper sulphate solution was put in a calorimeter and 6.0 g of zinc powder added. The temperature of the solution rose by 50.6C. Work out which reagent was in excess and then calculate the enthalpy change for the reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1. Ignore the heat capacity of the metals. q = mc∆T m = 25 c = 4.18 ∆T = 50.6 q = 25 x 4.18 x 50.6 = 5288 J ∆H = q / mol Mol CuSO4 = conc x vol = 1.0 x 25/1000 = 0.025 Mol Zn = mass / Mr = 6.0 / 65.4 = 0.0917 XS CuSO4 + Zn → ZnSO4 + Cu ∆H = –5.288 / 0.025 = -212 kJ mol-1 (3 sig fig)

HCl + NaHCO3 → NaCl + H2O + CO2 7) 3.53 g of sodium hydrogencarbonate was added to 30.0 cm3 of 2.0 mol dm-3 hydrochloric acid. The temperature fell by 10.3 K. Work out which reagent was in excess and then calculate the enthalpy change for the reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1. q = mc∆T m = 30 c = 4.18 ∆T = 10.3 q = 30 x 4.18 x 10.3 = 1292 J ∆H = q / mol Mol HCl = conc x vol = 2.0 x 30/1000 = 0.060 XS Mol NaHCO3 = mass / Mr = 3.53 / 84.0 = 0.0420 HCl + NaHCO3 → NaCl + H2O + CO2 ∆H = 1.292 / 0.0420 = +30.8 kJ mol-1 (3 sig fig)

∆H = q / mol q = ∆H x mol Mol CH3OH = mass / Mr = 2.00 / 32.0 = 0.0625 8) A calorimeter was calibrated by burning 2.00 g of methanol (CH3OH) whose enthalpy of combustion is -715 kJ mol-1. The temperature of the calorimeter rose from 19.6C to 52.4C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol. 1.50 g of propan-2-ol CH3CH(OH)CH3 raised the temperature by from 19.8C to 56.2C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol. ∆H = q / mol q = ∆H x mol Mol CH3OH = mass / Mr = 2.00 / 32.0 = 0.0625 q = ∆H x mol = 715 x 0.0625 = 44.69 kJ q = mc∆T mc = q/∆T = 44.69 / 32.8 = 1.363 kJ K-1

q = mc∆T mc = 1.363 ∆T = 36.4 q = 1.363 x 36.4 = 49.61 kJ ∆H = q / mol 9) A calorimeter was calibrated by burning 2.00 g of methanol (CH3OH) whose enthalpy of combustion is -715 kJ mol-1. The temperature of the calorimeter rose from 19.6C to 52.4C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol. 1.50 g of propan-2-ol CH3CH(OH)CH3 raised the temperature by from 19.8C to 56.2C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol. q = mc∆T mc = 1.363 ∆T = 36.4 q = 1.363 x 36.4 = 49.61 kJ ∆H = q / mol Mol C3H7OH = mass / Mr = 1.50 / 60.0 = 0.0250 ∆H = –49.61 / 0.0250 = -1984 kJ mol-1 (3 sig fig)

PRACTICE enthalpy change of solution ΔHS Enthalpy change of solution ΔHS – is when 1 mole of substance is dissolved Place a clean, dry polystyrene cup into an outer plastic cup to increase insulation and stability. Use the measuring cylinder to transfer 25 cm3 of water into the polystyrene cup. Place a thermometer into the cup through a cardboard lid and record this temperature as an initial temperature Weight 1 g NH4Cl of and put it into polystyrene cup. Measure the temperature of the mixture in the cup and record the greatest temperature change that you observed as a final temperature; Rinse the plastic cup with water and shake to dry. Repeat same steps with NaOH (keep this solution for next experiment) Calculate the enthalpy change for both using the equation q = mcΔT NH4Cl NaOH Mass (g) Volume of water is measured (cm3) Initial temperature (K) Final temperature (K) ΔH ( kJmol-1 )

PRACTICE enthalpy change of neutralisation ΔHN The standard enthalpy change of neutralisation is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water. Place a thermometer into the cup with NaOH through a cardboard lid and record this temperature as an initial temperature; Use the measuring cylinder to transfer 25cm3 of HCl into the polystyrene cup. Measure the temperature of the mixture in the cup immediately record the greatest temperature change that you observed as a final temperature; Calculate the enthalpy change for reaction using the equation q = mcΔT