HEAT OF REACTION, THERMOCHEMISTRY AND HEAT SUMMATIONS
HEAT OF REACTION
HOW TO MEASURE THE HEAT/ ENTHALPY OF REACTION
Chemical reaction involves breaking of old bonds and formation of new bonds between atom in molecules. We need energy to break bonds and we release energy when we form bonds.
Example:
H2 + Cl2 ______ > 2HCl
Enthalpy of reaction = Enthalpy [bonds broken] – enthalpy [bonds formed]
Bonds broken H-H = 436.4 kJ/mol
Cl-Cl = 242.7 kJ/mol
Bonds formed H-Cl x 2 = 431.9 kJ/mol x 2
Enthalpy of reaction = [436.4 + 242.7] kJ/mol – [431.9 x 2] kJ/mol
= – 184.7 kJ/mol = enthalpy of formation of HCl gas
Calculate an estimate value for the enthalpy of reduction of ethyne to ethane using the information below.
2H2 + HC=CH _____> CH3-CH3
Given the following bond energies.
H-H H-H 436kJ/mol
HC=CH C=C 962kJ/mol
CH3-CH3 C-C 368kJ/mol [not needed to get the answer]
CH4 C-H 435 kJ/mol
Calculate the bond energy of the π-bonds in ethyne and ethene.
H2C=CH2 C=C 720 kJ/mol
EXOTHERMIC PROCESS OR REACTION = reaction involving overall release of energy to surrounding atmosphere. E.g. Acid reaction with base in a neutralization reaction to form salt and water.
Example ; H2SO4 (l) + 2NaOH (l) ______> Na2SO4 (aq) + 2H2O
ENDOTHERMIC PROCESS OR REACTION = Involves gain or drawing energy from the surroundings. E.g. dissolving Ammonium chloride in water.
Enthalpies of formation at 25oC (ΔHfo) has been calculated as standard. This can be used to calculate enthalpy of reactions.
Enthalpy values of elemental substance such as Cu solid, Carbon (graphite or diamond) H2, O2 or H+ is ZERO.
ΔH reaction = ΔH formation [products] – ΔH formation [reactants]
Examples;
NH4Cl + H2O _______> NH4+ + Cl–
-314.43 -285.83 -132.51 -167.16 kJ/mol
ΔH reaction = {-132.51 + -167.16} – {-314.43 + -285.83} kJ/mol
= 300.59 kJ/mol
Al (s) + 3HCl (aq) ___________> AlCl3 (s) + 3H2(g)
0 -167.16 -704.2 0
ΔH reaction = {3 x 0 + -704.2} – {3 x (-167.16) + 0} kJ/mol
=-202.72 kJ/mol
Calculate the enthalpy for the reactions below.
2H2 + O2 ________> 2H2O (g)
0 0 -241.82 kJ/mol
H2O(g) _________> H2O(l)
-241.82 -285.83 kJ/mol
2HCl (aq) + CaCO3 _______> CaCl2(s) + H2O(l) + CO2(g)
-425.61 -1206.9 -795.8 -285.83 -393.51 kJ/mol
ΔS = Entropy = amount of disturbance or chaos in a system at a particular temperature.
ΔS= Entropy of reaction = ΔS [formation products] – ΔS [formation reactants]
ΔG = free energy of reaction = ΔG [formation products] – ΔG [formation reactants]
= ΔH – TΔS
The higher negative value of ΔG implies reaction is favorable or spontaneous. This means we need a high negative ΔH and a high positive ΔS and high temperatures to favor reaction to occur spontaneously. However, overall reaction is favorable if ΔG is negative or lower, regardless of the value or sign of T, ΔS or ΔH.
For example, a chemical reaction can occur at very low temperatures because the enthalpy is highly negative, or entropy is highly positive.
Le Chatelier’s Principle: Moles of substance, pressure, volume and temperature can determine the direction of an equilibrium reaction.
Thermochemistry
A substance can hold heat and release heat energy with time to its surrounding till heat exchange rate with surroundings reaches equilibrium point. Rate of heat emitted or absorbed is equal on both sides.
Heat of fusion: (ΔH fusion): Heat energy absorbed to melt solid to liquid or heat given out to change liquid to solid. Melting point = freezing point.
Heat of vaporization: (ΔH vaporization): Heat energy absorbed to convert liquid to vapor or gas . This usually occur at boiling point of a liquid.
No temperature change occurs at melting or boiling point because the energy absorbed only separated the particles at the temperature of transformation. Therefore, the temperature will be constant till all transformation is done.
A system of the same substance at equilibrium e.g. ice and water or solid CO2 and gaseous CO2
Free energy or Gibbs free energy at temperature T is given by G = Go + RTInK. , Where K is equilibrium constant at temperature T.
G-Go = ΔG = RTInK = ΔH –TΔS
G=free energy at temperature T
Go = standard free energy at 298K
At equilibrium, there is no change in the system, therefore ΔG = 0 , then K=1
Therefore 0 = ΔH –TΔS
ΔH=TΔS
ΔS=ΔH/T
Heat capacity (Cp) : Ability of a substance to hold heat energy.
Specific heat capacity: Amount of heat energy that can be stored in specific amount of substance. (1kg, g, or mole).
Cp =ΔH /ΔT ,
ΔH = Cp x ΔT
In this case DH is heat transfer that occur at constant pressure and Cp is specific heat capacity at constant pressure.
On the other hand,
Cv = ΔH /ΔT
ΔH = Cv x ΔT
In this case ΔH is heat transfer that occur at constant volume and Cv is specific heat capacity at constant volume.
EXAMPLE 1.
1 mole of water at 14oC was heated to 75oC. Calculate the amount of heat energy absorbed by the water.
Molar het capacity of liquid water = 72.291JK-1mol-1 .
Change temperature from Celsius to Kelvin, but ΔT is just a difference and already takes care of that.
ΔH = Cp x ΔT , ΔH = 72.291JK-1mol-1 x [T 2 – T1 ] , ΔH = 72.291JK-1mol-1 x [75 – 14 ]K
ΔH = 72.291JK-1mol-1 x 61K = 4409.751Jmol-1
1mole will absorb 4409.751 J of heat energy.
EXAMPLE 2.
1 mol of ice at -5 oC and 1atm was heated to melt and then boiled to vapor. Calculate the energy involved in transformation. [ (Cpice ) Molar heat capacity of solid ice = 36.4JK-1mol-1].
Take freezing point of water to be 0oC and boiling point to be 100oC.
ΔH fusion = 6.0kJ at 0oC
ΔH vaporization = 40.8kJ at 100oC
Enthalpy change for the process is as follows;
ΔH = Cpice x ΔT [0-(-5)] K + ΔH fusion + Cp x ΔT [100-0] K + ΔH vaporization
= [36. 4JK-1mol-1 x 5K ] + 6.0kJ + [ 1.0 mol x 72.291JK-1mol-1 x 100 K ]+ 40.8kJ
= 182J + 6kJ + 7229.1 J + 40.8kJ
= [0.182 + 6 + 7.2291 + 40.8] kJ
=54.21kJ
Diagram sketch of the enthalpy changes from ice to water vapor.
HOW TO CALCULATE HEAT/ENTHALPY OF REACTIONS IN THE LAB: CALORIMETRY
Calorimeter: The container in which enthalpy change is measured.
This should not allow heat to transfer across it. Good insulation so that heat that is measure is accurate.
However, it absorbs some heat from and has heat capacity value Cc. Heat exchange from the walls of the calorimeter will be measured by the change in its temperature ΔT= [T2-T1] of the water or solution inside the calorimeter, during the experiment.
Heat absorbed by calorimeter ΔHc = CcΔT = Cc[T2-T1]
For a closed system, no heat goes out or comes in.
Total heat transfer = 0
Total heat transfer = Heat solution or water + heat of calorimeter + heat reaction = 0
Heat solution or water + heat calorimeter = -[heat of reaction]
heat of reaction=- [ Heat of solution or water + heat calorimeter]
Heat of solution or water = [C solution or water] x (mass or moles) x ΔT
(Where Csolution or water is Heat Capacity of the solution or water in the calorimeter).
For a constant pressure process (e.g. open cup calorimetric experiment) , heat of reaction = DH enthalpy
Therefore, [ Heat of solution or water + heat calorimeter] = – ΔH enthalpy
ΔH (J/mol) = heat of reaction / moles of solute
Example 1.
10 mol of Methane was burnt in excess oxygen. Calculate the heat energy [ in Joules and then in Calories] produced in this reaction if the temperature was measured by 10.0 g of water from 25 to 105oC. [1 cal = 4.184J ]
Assume: Heat capacity of water is 4.184J/oCg and the heat capacity of the Calorimeter = 1005J/oC
CH4 + 2O2 ___________> CO2 + 2H2O
Heat of reaction = – [ heat of water + heat of calorimeter]
heat of water = mass x Cwater x ΔT , where Cwater = heat capacity of water , ΔT = [105-25]oC
= 10g x 4.184J/oCg x 80oC
=3,347.2 J
heat calorimeter = CcΔT
=1005J/oC x 80oC
= 80400J
Heat of reaction = -[3,347.2 J + 80400J ] = -83747J
10 mol of methane was used for the reaction,
Therefore, heat generated per mole methane = 83747J/10mol = 8374.7J/mol
This value is the molar heat of combustion of methane. This is not the actual value but just to show how to work or calculate when given such a question.
Example 2.
Calculate the heat of reaction and the molar enthalpy reaction of 0.1g sulfuric acid and 0.2g NaOH in a Styrofoam cup containing 10ml of distilled water if the temperature rose from 25oC to 90oC. Assuming no change in volume occurred and no heat absorbed by Styrofoam cup.
Example 3.
Calculate the heat of reaction of 25 ml 0.5M HCl with 25ml 0.5M NaOH in a constant volume calorimeter at 25oC if the temperature rose to 85oC. This calorimeter was first calibrated with Methane (used in example 1) with temperature change from 25oC to 50oC.
INDIRECT CALCULATION OF Enthalpy (ΔH) = [Hess’s law of Heat Summation]
Summation of enthalpy of reactions can be used to find unknown enthalpies of reactions. We can use other reaction containing the formation or consumption of reactants A and B and product C to write a net reaction and net sum of enthalpies.
Example; A + HC ______> C + HA …..….ΔH unknown
HA + B ____> HB + A ΔH1 (known) ……….1
C + HB ____> HC + B ΔH2 (known)…………….2
Reversing equation 1 and 2 gives ;
HB + A _____> HA + B ……..equation 3 = -ΔH1
HC + B _____> C + HB …….equation 4 =-ΔH2
Adding equation 3 and 4 gives ;
NOTE:
- Same Items appear on OPPOSITE sides [e.g. HB and B] CANCEL out.
- Same items on SAME side add up.
- Reversing an equation reverse the sign on enthalpy of the reaction.
- Multiplying through an equation by a number also multiplies the enthalpy by the same number.
A + HC ______> C + HA ……..…ΔH reaction = -ΔH1 + -ΔH2
Find the enthalpy of reaction and hence enthalpy of formation of C in reaction equation A + B _______> C , given the information about the reactions below;
A + Y _____> Z …………………. ΔH = -219 kJ/mol
X + Z _____> B …………………. ΔH = -524 kJ/mol
Y + C _____> X + 2Z …………. ΔH = 200 kJ/mol
Find the enthalpy of reaction A + 2B ______> C
B + Z + A ____> Y ΔH = -219 kJ/mol
X + Z _____> B ΔH = -424 kJ/mol
C _____> X + Y ΔH = 200 kJ/mol
Find the enthalpy of reaction A + B ______> 2C
Y ____> Z + ½ C ΔH = 570 kJ/mol
A + 2Z _____> X ΔH = 305 kJ/mol
C _____> X + 2Y + B ΔH = 223 kJ/mol
EXAMPLE :
The enthalpy of reaction below is also the enthalpy of formation of the product compound.
CH3(CH2)2CBr2-CHBr2
2Br2 + CH3(CH2)2C= CH ______________> CH3(CH2)2CBr2-CHBr2
Possible intermediate reactions
1…… Br2 + CH3(CH2)2C = CH ______________> CH3(CH2)2CBr = CHBr + CH3(CH2)2C=CH
limiting amount excess amount bp = 201oC bp = 71oC
Brown colorless colorless mixture: Can be separated by distillation.
2…….. Br2 + CH3(CH2)2CBr = CHBr ______________> CH3(CH2)2CBr2-CHBr2
[excess amount] limiting amount product: [allow excess bromine to evaporate to obtain product]
We can measure the enthalpy of reaction 1 and 2 and add up the values to get the enthalpy of the reaction.
2Br2 + CH3(CH2)2C=CH______________> CH3(CH2)2CBr2-CHBr2
.
EXERCISE:
Bond energy calculation for estimating enthalpy of reactions.
| Bond | C = C | C= C | C- C | C-Br | Br – Br |
| Enthalpy (kJmol-1) | 835.1 | 610 | 345.6 | 293 | 192.80 |
- Try using bond enthalpies given above to estimate the enthalpies of intermediate step reactions below and add them to get the overall enthalpy.
Br2 + CH3(CH2)2C= CH ______________> CH3(CH2)2CBr = CHBr
Br2 + CH3(CH2)2CBr = CHBr ______________> CH3(CH2)2CBr2-CHBr2
- Try doing the calculation directly for the overall reaction equation and compare your answer to that in question 1.
2Br2 + CH3(CH2)2C=CH______________> CH3(CH2)2CBr2-CHBr2
