Form 5 Chemistry – PHYSICAL CHEMISTRY – Energetics

11. HEAT OF DILUTION
Is the heat change when one mole of a substance is dissolved in a given mole of water

METHODS OF FINDING HEAT OF FORMATION OF A GIVEN COMPOUNDS

The calculation in thermochemistry are categorized in the following;-

(i).    Calculation based on combustion Data.

(ii).   Calculation based on bond energies.

(iii).  Calculation based on atomization Data.

(iv).  Calculation based on calorimetry information.

(v).   Determination of molecular formular by combustion Data.

(vi).  Calculation based on Born–Haber cycle.

CALCULATIONS BASED ON COMBUSTION DATA

Calculations involving combustion data has got the following Steps:-

(i).   Identify the required equation.

(ii).  Data presentation.

(iii). Data manipulation.

(iv). Conclusion.

Example 1

a) With one example in each briefly define the following terms;-

(i).   Standard enthalpy of combustion.

(ii).  Sublimation energy.

(iii).  Standard enthalpy of formation.

(iv).  Atomization energy.

(v).   Lattice energy.

Solution

(i).   Standard enthalpy of combustion: is the heat energy given out when 1mole of a certain substance is completely burnt in a given moles of oxygen at standard state.

(ii).  Sublimation energy: Is the energy absorbed when one mole of solid atom is converted to gaseous atom.

(iii). Standard enthalpy of formation: is the heat change which occur when 1 mole of substance is formed from its element.
eg. C + O₂  →   CO_2
(g)    ΔH°f = – 393 KJmol^-1

(iv). Atomization energy: is the energy absorbed when a given molecule or element is converted into gaseous atom.

(v).  Lattice energy: Is the energy given out when 1 mole of ionic compound is formed from it ions.
eg.    Na⁺_(g) + Cl^–_(g) → NaCl

b) Calculate the enthalpy of formation of methane given that,

Enthalpy of combustion are;-

i.  Carbon 394 KJmol^-1.

ii.  Hydrogen 286 KJmol^-1.

iii. Methane 891 KJmol^-1.

Solution

Required equation

C + 2H₂
CH₄

Data presentation                             KJ mol ^-1

i.    C + O₂ CO₂                                – 394

ii.   H₂ + O₂ H₂O                – 286

iii.CH₄ + 2O₂ CO₂ + 2H₂O        891

Data manipulation                             KJmol^-1

i.    C + O₂
CO₂                         – 394

ii.   H₂ + O₂ H₂O                – 286) 2

       2H₂ + O₂ 2H₂O                   – 572

iii.2H₂ + O₂ CH₄ + 2O₂          891

       C + 2H₂
CH₄                      – 75

.^..       The enthalpy of formation of methane is -75 KJmol^-1.

Example 2

Calculate the standard enthalpy of formation of ethane given that enthalpy of combustion.

i)  C = -394 KJmol^-1

ii) hydrogen = -216 KJmol^-1

iii) Ethane = -1561 KJmol^-1

Solution

Required equation

   2C + 3H₂
C₂H₆

   Given                                                          KJmol^-1

   (C + O₂ CO₂                                 -394

   H₂ + O₂
2CO₂                        – 788

          C₆H₆ + 7/2 O₂
2CO₂ + 3H₂O      -1561

Data Manipulation

(C + O₂ CO₂                               -394 )×2
2C + 2O₂
2CO₂                          -788

                    (H₂ + O₂ H₂O                          -216) 3

                    3H₂ + O₂3H₂O                      -648

                    2CO₂ + 3H₂O CH₃ CH₃ + O_{2       1561}

2C +3H₂ C₂H₆                         + 125

.^..       The standard enthalpy of formation of C₂H₆ is 125 KJmol^-1.

Example 3

a) Define;-

i.Heat of combustion

Is the heat given out when one mole of a substance is completely burnt in given moles of O₂ at standard state.

ii. Heat of neutralization

                    Is the heat energy given out when one mole of water is formed the reaction between acid and base at standard state.

iii.Heat of dilution

                   Is the heat change when one mole of a compound is dissolved in a given moles of water.

b) Calculate the heat of formation of ethanoic acid , if the enthalpy of combustion are C = -394 H₂ = – 284 ethanoic acid -876 ( All in KJ mol^-1).

Solution

Required equation

2C + 2H₂ +O₂
CH₃COOH

Data presentation                                KJ mol^-1

C + O₂ CO₂                                  -394

H₂ + O₂ H₂O                       -284

CH₃ + 2O₂
2CO₂ + 2H₂O            -876

Data manipulation

2C + 2O₂
2CO₂                              – 788

2H₂ + O₂
2H₂O                         – 568

2CO₂ + 2H₂O CH₃COOH + 2O₂    876

The equation obtained

2C + 2H₂ + O₂
CH₃COOH           -480

.^.. The heat of formation of ethanoic acid is -480 KJmol^-1.

Example 4

From the following Thermodata at 298k.

(i) +

(ii)

(iii)

Calculate for the reaction

Solution

Required equation

Data presentation

+

Data manipulation

+                                           – 288

For the reaction is

Example 5

Calculate the enthalpy change for the reaction.

Given

Enthalpies of combustions are;-

Heat of formation of water is

Solution

Required equation

Data presentation                                  

Data manipulation

Example 6

Calculate the enthalpy change for the reaction.

Given

Heat of combustion of

Heat of combustion of

Heat of combustion of

Heat of combustion of

Solution

Required equation

Data presentation

Data manipulation

Example 7

(a) Define the following terms;-

(i) Enthalpy of sublimation.

Is the energy absorbed when one mole of solid atom is converted to gaseous atom.

(ii) Enthalpy of atomization.

Is the energy absorbed when a certain molecule or element is covered into gaseous atom.

(iii) Standard enthalpy of combustion.

Is the head given out when one mole of a substance is burnt in a given moles of Oxygen.

(b) The combustion of carbon disulphide is exothermic and the enthalpy of combustion of the compound is 1180^.

Given that carbon dioxide and sulphur dioxide are exothermic compounds with enthalpies of formation of 405 and 293 respectively.

(i) Calculate the heat of formation of carbon disulphide.

(ii) comment on the stability of this compound at various temperature considering the results obtained in the light of Le-chaterlier`s principle.

Solution

Required equation

2S + C

Data presentation                            

      – 1108

C +

S +

Data manipulation

                

The heat of formation of C is 117

ii) In high temperature the compound will be move stable.

Example 8

Given the following reaction;-

               Calculate the enthalpy change for the reaction

Solution

Required manipulation