Chapter 5 Notes: Energetics Thermochemistry Chapter 5.1: The enthalpy changes from chemical reactions can be calculated from their effect on the temperature of their surroundings.

Important terms for this section Energy Work System Surroundings Open and closed systems Enthalpy Endothermic Standard enthalpy changes Absolute zero Kelvin scale Heat capacity

Energy and heat transfer energy Energy: measured in Joules = kg*m2/s2 = N*m Ability to do work Ability to move an object against an opposing force Some forms Heat, Light, Sound, Electricity, Chemical energy Focus is heat in this unit = E transfer with Temp diff. Inc. heat = inc. ave. K.E. of molecules in a disordered fashion

System and surroundings System: area of interest Surroundings: everything else Open system Can exchange energy AND matter with surroundings Closed system Can exchange energy ONLY with surroundings Total Energy cannot change during process only exchange

Enthalpy Enthalpy “heat inside”: heat content of a system Changes in enthalpy is noted: ΔH If heat is added to a system, enthalpy increases (is positive) If heat is released from system, enthalpy decreases (negative)

Exothermic & endothermic Exo = release (external) exothermic releases heat Might feel hot when reacting Giving out heat means a negative enthalpy change Ereactants > Eproducts Endo = take in endothermic needs heat Might feel cold when reacting Heat added to the system means a positive enthalpy change Ereactants < Eproducts

Standard enthalpy changes Symbol for standard enthalpy change: ΔHѳ Conditions Pressure of 100 kPa Concentration = 1 mol dm-3 for all solutions All substances in their standard states Temp is not part of definition, but is typically 298K

Thermochemical equations Combustion of methane: Releases 890 kj mol-1 of heat energy Photosynthesis: Absorbs 2802.5 kj mol-1 of energy

Kelvin and kinetic energy Average kinetic energy of molecules related to Kelvin Absolute zero Absolute temp directly proportional to average KE particles If same amt of heat is added to 2 different objects, will temp change be the same? Heat changes depend on Mass of object Heat added Nature of substance

Heat changes continued Specific heat capacity Amount of heat needed to increase the temp of unit mass by 1K Depends on Number of particles present in the sample And therefore the mass of individual particles Heat capacity Amount of heat needed to increase temp of an object 1K

Heat capacities So, to recap:

Enthalpy and direction of change Energy moves from higher to lower stored energy And might look different than you assume 

Measuring/calculating Enthalpy changes Measuring enthalpy changes of Combustion, ΔHcѳ The enthalpy change for the complete combustion of 1 mole of a substance in its standard state in excess oxygen under STP. Calorimetry

Measuring/calculating Enthalpy changes Calculating ΔH from Temperature changes For Exothermic RXN Temp of H2O increases ΔH is negative For Endothermic RXN Temp of H2O decreases ΔH is postitive

Measuring/calculating Enthalpy changes ΔHsol of RXN in solution The enthalpy change when 1 mole of solute is dissolved in excess solvent to form a solution of ‘infinite dilution’ under std. conditions The results are not always what you expect Loss of heat to environment/ materials/etc. vs water

Definitions ΔHѳ standard enthalpy change Change in enthalpy under std. conditions 100kPa, 298K ΔHѳr standard enthalpy change of reaction Enthalpy change when molar amounts of reactants, as shown in a stoichiometric equation react together under std. conditions to give products ΔHѳc standard enthalpy change of combustion Enthalpy change when 1 mole of a substance is completely burnt in oxygen under std. conditions

definitions ΔHn enthalpy change of neutralization Enthalpy change when 1 mole of water molecules are formed when an acid (H+) reacts with an alkali (OH-) under std. conditions ΔHsol of RXN in solution The enthalpy change when 1 mole of solute is dissolved in excess solvent to form a solution of ‘infinite dilution’ under std. conditions c specific heat capacity The energy required to raise the temperature of 1kg (or 1g) of a substance by 1K (1ºC) q heat energy How much heat energy must be supplied to raise the temp of mass m by ΔTºC q=mc ΔT

Problem 1 Using the data, determine the ΔHѳc of ethanol (C2H5OH) given that the cH2O = 4.18 J g-1 ºC-1 (remember c is specific heat capacity) Mass of water = 150.00 g Initial temp of water = 19.5 ºC Maximum temp of water = 45.7 ºC Initial mass of spirit burner = 121.67 g Final mass of spirit burner = 120.62 g REMEMBER: ΔHѳc is the enthalpy change when 1 mole of a substance is completely burnt in oxygen under std. conditions

Problem 1 solution So if ΔHѳc is the enthalpy change when 1 mole of a substance is completely burnt in oxygen under std. conditions We need to find: The heat energy change (how many Joules are being produced) How many moles are being burned of ethanol

Problem 1 solution Plug in the values: Amount of energy supplied to the water: Use q =mcΔT c: given as 4.18 J g-1 ºC-1 ΔT: 45.7-19.5 = 26.2 ºC m: 150.00 g water Plug in the values: q = 150.00 x 4.18 x 26.2 = 16400 J And is the heat energy supplied by the burning of ethanol

Problem 1 solution How many moles of ethanol are burned Initial mass of spirit burner = 121.67 g Final mass of spirit burner = 120.62 g So, Mass burned = 121.67 – 120.62 = 1.05 g Convert to moles ethanol (46.08 g mol-1) 1.05 g (1 mol/46.08 g) = 0.0228 mol ethanol Therefore, when 0.0228 mol ethanol are burned, it produces 16400 J heat Convert to 1 mol: 16400 J/0.0228 mol = 721000 J mol-1 Thus, ΔH = -721 kJ mol-1

Problem 1 solution Now, let’s compare the calculated value with the literature value ΔH = -721 kJ mol-1 ΔHѳc = -1371 kJ mol-1 Why are they so different? Heat loss to surroundings Some heat goes to heating copper can and air Incomplete combustion of ethanol Evaporation of ethanol and water Improvements: Taken into account the c of the can in calculations Insulating the can Use other kind of calorimeter

Problem 2 Consider the experiment:100.0 cm3 of water are measured out and poured into a polystyrene cup and the temp of the water measured. Then 5.20 g ammonium chloride are measured out. The ammonium chloride was added to the water and the solution stirred vigorously until all the solute dissolved. The minimum temp was recorded. Results: Initial temp of water = 18.3 ºC Minimum temp = 15.1 ºC Find the enthalpy change of solution for ammonium chloride, ΔHsol Remember: The enthalpy change when 1 mole of solute is dissolved in excess solvent to form a solution of ‘infinite dilution’ under std. conditions

The specific heat capacity of the solution is the same as water Problem 2 hints What do you need to do? Find the enthalpy change of solution for ammonium chloride, ΔHsol The enthalpy change when 1 mole of solute is dissolved in excess solvent to form a solution of ‘infinite dilution’ under std. conditions Actual data you have: 100.0 cm3 water (so actually 100.0 g water) Initial temp of water = 18.3 ºC Minimum temp = 15.1 ºC 5.20 g ammonium chloride You will make the following assumption: The specific heat capacity of the solution is the same as water

Problem 3 The following experiment may be used to determine the enthalpy of reactions for: Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s) 50.0 cm3 of 0.200 mol dm-3 copper(II) sulfate solution are placed in a polystyrene cup. The temperature was recorded every 30s for 2 min. At 2 min every 30s for 2 min. At 2 min 30s, 1.20 g of powdered zinc are added. The mixture was stirred vigorously and the temp recorded every 30s for several minutes. The results obtained were then plotted to give the graph shown Use these data to determine the enthalpy change for this reaction.

Problem 3 hints Determine the enthalpy change for this reaction. Remember: ΔHѳr standard enthalpy change of reaction Enthalpy change when molar amounts of reactants, as shown in a stoichiometric equation react together under std. conditions to give products Data you actually have: Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s) 50.0 cm3 of 0.200 mol dm-3 CuSO4(aq) The graph to the right  By extrapolating back to the area above the start of the reaction, Temp change, ΔT = 10.3 ºC 1.20 g Zn(s) Assumption: density of CuSO4 solution = water therefore the m and c are the same as water 50.0 g and 4.18