Topic 5.2 Calculations of enthalpy changes

 Specific heat (s) – the amount of heat necessary to raise the temperature of 1 g of a substance by 1 o C.  Heat capacity (C)- the amount of heat required to raise the temperature of a given quantity of a substance by 1 o C. So, C = m· s

 The heat(q) that is absorbed or released in a given process can be calculated by the following equation:  q= m · s · Δ T  Δ T= T final – T initial  Since C = m· s, the amount of heat absorbed or released can also be defined as:  q= C x Δ T

 According to the law of conservation of energy, if an object is dropped into a cold object, or vice versa, the heat given up by the hot object is absorbed by the cold object; this can be expressed as follows:  q hot = q cold  (q= m · s · Δ T) hot = (q= m · s · Δ T) cold  This equation can be used to find the specific heat of an object.

 The heat produced when one mole of a substance is burned in oxygen is called the enthalpy change of combustion.  When the heat released by exothermic reaction is absorbed by water, the temperature of water increases, so the heat produced by the reaction can be calculated if it is assumed that all the heat is absorbed by the water.  Heat change of reaction = heat change of water  Heat rxn = m H 2 O · s H 2 O · Δ T H 2 O

 In an exothermic reaction, the heat produced by the reaction is absorbed by the water so its temperature increases and the heat change is negative.  In an endothermic reaction, the heat absorbed by the reaction is taken from the water so its temperature decreases and the heat change is positive.  Since the heat changed observed depends on the amount of reaction, enthalpy changes are usually expressed in kJ mol -1

 Example  Calculate the enthalpy of combustion of ethanol from the following data. Assume all the heat from the reaction is absorbed by the water. Compare your answer with the IB Data booklet and suggest reasons for any differences. Mass of water in copper calorimeter (g) Temperature increase in water ( o C) Mass of ethanol burned (g)0.45

 Solution  1 st – calculate the heat released by the reaction.  Heat rxn = m H 2 O · s H 2 O · Δ T H 2 O  Heat rxn = · · = J  J = kJ  2 nd – divide the heat released by the number of moles to express the enthalpy change in kJ mol -1  Number of moles of C 2 H 5 OH = mass of C 2 H 5 OH molar mass of C 2 H 5 OH Moles of C 2 H 5 OH = 0.45 g C 2 H 5 OH/46.08 g mol -1 = moles kJ/ mol = kJ mol -1

 Solution (cont’d)  The IB Data booklet value is – 1371 kJ mol -1. Suggested reasons for the difference:  Not all the heat produced by the combustions is transferred to the water. Some heat is absorbed by the calorimeter and some has passed to the surroundings.  The combustion of the ethanol is not complete due to insufficient oxygen supply.

 The enthalpy changes of reaction in solution can be calculated by carrying out the reaction in an insulated system, for example, a polystyrene cup.  The heat released or absorbed by the reaction can be measured from the temperature change of the water.

 Example  50.0 cm 3 of 1.00 mol dm -3 hydrochloric acid solution was added to 50.0 cm 3 of 1.00 mol dm -3 sodium hydroxide solution in a polystyrene beaker. The initial temperature of both solutions was 16.7 o C. After stirring and accounting for heat loss, the highest temperature reached was 23.5 o C. Calculate the enthalpy change for this reaction.  q neutralization rxn = q solution

 Step 1- write equation for the reaction  HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l)  Step 2- Calculate molar quantities  Mol HCl = 50.0/1000 x 1.00 = mol  Mol NaOH = 50.0/1000 x 1.00 = mol  Therefore, the heat evolved will be for mol  Step 3- Calculate the heat evolved  Total volume of solution = sum of both solutions = mL solution  Mass of solution – assume density of solution is the same as water, so mL = g solution

 Specific heat of solution = assume specific heat of solution is the same as water = J/g o C  q neut = (m x s x Δ T) soln  q neu = 100 x x 6.8 = 2.84 kJ  Δ H = kJ / mol = kJ mol -1  The negative value indicates that this reaction is exothermic.

 One of the largest sources of error in experiments conducted in a polystyrene cup is heat losses to the environment.  For example, in the exothermic reaction between zinc and aqueous copper sulfate some of the heat produced in the reaction is lost to the environment, so the maximum temperature recorded is lower than the true valued obtained in a perfectly insulated system. We can make some allowance for the heat loss by extrapolating the cooling section of the graph to the time when the reaction started.

 For an exothermic reaction, Δ Hreaction is negative as heat has passed from the reaction into the water. Heat rxn = m H 2 O · s H 2 O · Δ T H 2 O  In order to determine the molar enthalpy change for the reaction, the limiting reactant must be identified.  The molar heat change of reaction = Heat rxn moles of limiting reactant

 Example  50.0 cm 3 of mol dm -3 copper(II) sulfate solution was placed in a polystyrene cup. After two minutes 1.20 g of powdered zinc was added. The temperature was taken every 30 seconds. Calculate the enthalpy change for the reaction taking place.  Step 1- write the net ionic equation for the reaction:  Cu 2+ (aq) + Zn(s)  Cu(s) + Zn 2+ (aq)  Step 2 – determine the limiting reactant  Amount of Cu 2+ (aq) = 50.0/1000 x = mol  Amount of Zn(s) = 1.20/65.37 = mol

 Step 3- extrapolate the graph to compensate for heat loss and determine Δ T.  In this case, Δ T= 10.4 o C  Step 4 – Calculate the heat evolved in the experiment for mol of reactants  Heat evolved = 50.0 x x 10.4/1000 = 2.18 kJ  Step 5 – Express this as the enthalpy change for the reaction, Δ H = kJ/ = -218 kJ mol -1