5.1 ENERGETIC CALCULATIONS IB SL Chemistry Mrs. Page

Daily Objectives Understand the principle of experimental methods for determining enthalpy changes Work out enthalpy changes from experimental data

Enthalpy Change of Combustion Heat given off in combustion rxn is used to heat a substance of known specific heat capacity The mass and temp change of water must be measured c H 2 O = 4.18 J g -1 K Note: ∆1°C = ∆1K

Calculating Enthalpy from Combustion Rxn The mass of a spirit lamp containing ethanol (C 2 H 5 OH) was determined accurately. The lamp was then lit and placed under a beaker containing 150. g of pure water. The water was continually stirred. After the temperature of the water had increased by 12.0  C the flame was extinguished and the mass of the lamp containing the unburned ethanol was determined. Use the following data to determine the enthalpy change when 1 mol of ethanol is burnt completely. Mass of lamp + ethanol before burning = g Mass of lamp + ethanol after burning = g Mass of ethanol burned: = g Molar Mass of ethanol: g mol-1

Calculating Enthalpy from Combustion Rxn How much heat energy required to raise the temp of 150. g of water by 12.0  C q=mc  T = 150. (4.18) (12.0) = 7524 J It take 7524 J to burn moles of ethanol so how much to burn 1 mol? Combustion Rxns: Exothermic so  H= kJmol g C 2 H 5 OH g 1 mol = mol C 2 H 5 OH mol C 2 H 5 OH 7524 J = 895,714J/mol = 896,000J/mol or 896 kJ/mol

Enthalpy Change of Combustion Possible Sources of Error: Incomplete combustion (build up of soot, occurring in air, yellow flame) Loss of heat (using a beaker vs. insulated container) Heat loss to system (heating not only water but also beaker, thermometer, stirring rod, ring stand, etc) Loss of alcohol to evaporation Errors will always be lower than actual value

Enthalpy Change of Combustion Include the specific heat capacity of the container you are heating the water in as well Use Insulating can Use Draught Shield Use a Bomb Calorimeter Possible Improvements:

Enthalpy Change of Combustion Lit Value = -1371kJmol -1  T = 45.7 – 19.5 = 26.2°C q = mc  T = (4.18)(26.2) = J = 164 kJ Mol Ethanol = 1.05 g (1mol/46.08 g) = mol  H= -721 kJmol -1 Mass of Ethanol = – = 1.05 g  H = J/mol = kJ/ mol

Enthalpy Changes in Solution (  H sol ) In Lab: Measure known amounts of reagents Record initial temp Mix reagents in styrofoam cup Record max/min temp Specific heat of final solution assumed to = specific heat of water

Enthalpy Change of Solution(  H sol )  H sol when 1 mole of a solute is dissolved in excess solvent to form a solution of “infinite dilution” under STP May be endo or exothermic Example:

Example of Enthalpy Change of Solution When 50.0 cm 3 of 1.00 mol dm -3 sodium hydroxide solution (NaOH (aq)) at 25.0  C was added with stirring to 50.0 cm 3 of 1.00mol dm -3 hydrochloric acid solution HCl(aq), also at 25.0  C, in a polystyrene coffee cup, the temperature of the resulting solution rose very quickly to 31.8  C. What is the change in enthalpy of this reaction?  T = 31.8 – 25.0 = 6.8  C Total Volume = cm 3 Assume: resulting solution has same density & c of water q=mc  T = 100.0(4.18)(6.8) = J Note: system heated up so exothermic and  H is negative

Example of Enthalpy Change of Solution When 50.0 cm 3 of 1.00 mol dm -3 sodium hydroxide solution (NaOH(aq)) at 25.0  C was added with stirring to 50.0 cm 3 of 1.00 mol dm -3 hydrochloric acid solution (HCl(aq)), also at 25.0  C, in a polystyrene coffee cup, the temperature of the resulting solution rose very quickly to 31.8  C. What is the change in enthalpy of this reaction? Amt of HCl = amt NaOH (balance equation  H = / = Jmol -1 = kJmol cm cm 3 1 dm 3 = mol NaOH 1.00 mol 1 dm 3 q = mc∆T= 50.0(4.18)(6.8) = J Positive or negative? Why?

Enthalpy Change of Neutralization (  H n )  H n when 1 mole of water molecules are formed when an acid (H + ) reacts with an alkali (OH - ) under STP Neutralizations are ALWAYS exothermic (-  H)

You Try A 25.0 g sample of an alloy was heated to  C and placed in a beaker containing 90.0g of water at  C. The temperature of the water rose to a final value of  C. Neglecting heat losses to the room and the heat capacity of the beaker itself, Calculate the specific heat of the alloy.

You Try Assuming no heat loss to the surroundings or to the container, calculate the final temperature when 100g of silver at 40.0  C is immersed in 60.0 g of water at 10.0  C. The specific heat capacity of silver is Jg  C

You Try When sulfuric acid dissolves in water, a great deal of heat is given off. The enthalpy change for this process is called the enthalpy of solution. To measure it, 175 g of water was placed in a coffee-cup calorimeter and chilled to 10 o C. Then 49.0 g of pure sulfuric acid, also at 10.0 o C was added, and the mixture was quickly stirred with a thermometer. The temperature rose rapidly to 14.9 o C. Assume that the value of the specific heat of solution is J/g o C. You may assume that the specific heat of the resulting sulfuric acid solution will also be J/g o C. Calculate q for the formation of this solution, and calculate the enthalpy of solution in kilojoules per mole of H 2 SO 4.

Molar Enthalpy Change for a Reaction 1.Write equation for reaction 2.Determine the limiting reagent 3.Measure mass of aqueous solution 4.Use calorimeter and measure initial temperature (probe should measure temp. every 30 seconds until a constant temperature is reached) 5.Add other reactant and continue measuring temperature for about 5 minutes after maximum temperature is reached 6.Produce a temperature vs time graph (Excel) 7.Look at cooling section of curve to extrapolate back to the point of introduction of reactant. 8.Calculate the heat evolved in the experiment. 9.Find the enthalpy change for the reaction

Molar Enthalpy Change for a Reaction Equation: Determine Limiting Reagent: 50.0 cm 3 of mol dm -3 copper (II) sulfate solution was placed in a polystyrene cup. After 2 minutes 1.20 g of powdered zinc was added. The temperature was taken every 30 seconds and the following graph was obtained. Calculate the enthalpy change for the reaction taking place cm cm 3 1 dm 3 = mol CuSO mol 1 dm g Zn g 1 mol = mol Zn LIMITING CuSO 4(aq) + Zn (s)  ZnSO 4(aq) + Cu (s)

Molar Enthalpy Change for a Reaction Extrapolate from Graph: Calculate Heat: Calculate Change in Enthalpy: 50.0 cm 3 of mol dm -3 copper (II) sulfate solution was placed in a polystyrene cup. After 2 minutes 1.30 g of powdered zinc was added. The temperature was taken every 30 seconds and the following graph was obtained. Calculate the enthalpy change for the reaction taking place. q = mc ∆T = 50.0(4.18)(10.5) = J = kJ ∆T = 27.5 – 17 = 10.5 °C ∆H = kJ/mol = kJ/ mol = 2.19 kJ/mol Endo or Exo? ANSWER kJ/mol

STANDARD ENTHALPY CHANGE OF REACTION ∆H ө Ө = at standard conditions Temperature 25°C/ 298K Pressure 100 kPa All species in standard states Standard enthalpy change of formation = ∆H ө f is the amount of energy change to form 1 mole of a compound from elements in standard states at 298 K and 100 kPa Standard enthalpy change of combustion = ∆H ө c is amount of enthalpy change when 1 mole of a substance is completely combusted in oxygen under standard conditions. Data booklet Section 12 – ∆H ө f for common compounds. Section 13 - ∆H ө c

HOMEWORK Quick Questions pp 145 & 146