ENERGY CHANGES IN CHEMICAL AND PHYSICAL PROCESSES (THERMOCHEMISTRY) - Chemistry Notes Form 4

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Introduction

  • Most chemical and physical process are accompanied by energy changes which occur in the form of heat measured in joules (J) and kilojoules (KJ)
  • The heat results from the motion of atoms and molecules.

Specific Heat Capacity

  • Specific heat capacity of a substance is the number of joules required to raise the temperature of one gram of the substance by one degree Kelvin e.g. SHC for water is 4.18Jg-1K-1


The Heat Content (Enthalpy) of a System

  • Heat content is denoted by H; while heat change is denoted as ΔH1;
  • And; ΔH = H Products - H Reactants = H− H1
  • Heat changes in chemical reactions can either be exothermic and endothermic .

Endothermic and Exothermic Reactions

Endothermic Reactions

  • Is a reaction accompanied by a fall in temperature and energy is absorbed from the surroundings.
  • The enthalpy change is normally positive since heat of products (H Products) is higher than heat of reactants (H reactants ).
  • Thus H− H1 is a positive value; since H1 is less than H2 .
  • Graphically this can be denoted as follows;-
    endothermic reaction graph
    Examples:
  1. When NH4NO3 dissolves in water, the temperature of the solution drops.
    Procedure
    • A spatula end-full of ammonium nitrate is dropped into a test tube of water.
    • The bottom of the test tube is felt with the hand.

    Observations
    • The hand feels cold.

    Reason
    • Energy is absorbed by the products, cooling the test tube.

    Thus H1 is less than H2 giving a positive ΔH1 showing an endothermic reaction
  2. N2(g) + O2(g) → 2NO(g) ; ΔH= + 91 kJMol-1

Exothermic Reactions.

  • Are reactions accompanied by a rise in temperature and energy is liberated to the surroundings.
  • The enthalpy change is normally negative , since heat of reactants (H1) is normally higher than heat of products (H2).
  • Thus, H2 − H1 gives a negative value since H1 is higher than H2 .
  • Exothermic reactions can be represented graphically as follows;-
    exothermic reaction graph
  • In exothermic reactions H2 is lower than H1 .

    Examples:
    1. Manufacture of ammonia in the Haber process;
      i.e. N2(g) + 3H2(g) → 2NH3(g) ; ΔH=-46Kjmol-1
    2. Dissolving sodium hydroxide pellets in water,
      • The temperature of the resulting NaOH(aq) is higher than the temperature of water at room temperature.
      • This implies the internal temperature of products is lower than the reactants’ original temperature.
      • Thus (H products H reactants) = -ve value, an exothermic reaction.

Activation Energy

  • Is the energy required to activate the reactants before a reaction can take place.
  • Thus activation energy is the energy required to initiate a reaction.
  • The size of activation energy will differ from one reaction to another and so will be the gap between the energy of reactants and the energy is products.
    Examples:
    • Exothermic reactions
      exothermic reaction activation energy

      At A; bonds are being broken and the energy is absorbed.
      At B; bonds are now being formed, and so energy is evolved.
    • Endothermic reactions
      endothermic reaction activation energy
      At A; bonds are broken and energy is absorbed.
      At B; bonds are formed and energy is evolved.

Determination/Measurement of Enthalpy (Heat) Changes

  • Calorimeters are used and have to be insulated to reduce heat loss to the surrounding.
    Source of errors
    • Heat loss to the surrounding.
    • Absorption of heat by the calorimeter (vessels).
  • Main heat changes under consideration:
    • Enthalpy of combustion (ΔHc)
    • Enthalpy of neutralization (ΔHneut)
    • Enthalpy of solution (ΔHsoln)
    • Enthalpy of precipitation (ΔHprecip)
    • Enthalpy of displacement (ΔHDisp)
    • Enthalpy of formation (ΔHf )

Heat of combustion ΔHc

  • Is the heat changes when the mole of a substance is completely burned in oxygen, at one atmospheric pressure
  • Since heat is usually evolved and hence ΔH is usually negative.
    Example;
    C(s) + O2(g) → CO2(g) ; ΔH= -394 kJmol-1

    Apparatus for Finding ΔHc of a Fuel

    heat of combustion
  • Heat evolved = Specific heat capacity of water x Mass of water x Temperature rise = CMΔT; joules,
    where; C = specific heat capacity, M = mass of water, ΔT= temperature change.

    Examples
    1. Assume
      Volume of water in calorimeter = 100cm3
      Initial temperature of water = 225oC
      Final temperature of water = 50.5oC
      Change in temperature of water = 28.0oC
      Mass of water = 100g
      Mass of lamp before burning = 30.46g
      Mass of lamp after burning = 30.06g
      Mass of ethanol burnt = 30.46 - 30.06 = 0.40g
      1. Determine the heat evolved;
        = ΔH= CMΔT
        =4.2 x 100 x 28
        =11760 Joules
        =11.760 kJ
      2. Hence calculate the molar heat of combustion of ethanol;
        Moles of ethanol burnt = Mass of ethanol burnt/RFM of ethanol; 0.4/46 = 0.008695 moles.
        Thus if 0.008695 moles = 11.76 Kilojoules,
        Then 1mole = ?
        1 mole = 11.76 x 1/0.008695 = -1352.50 KJMol-1
        = -1352.5 kJMol-1 (negative value as the reaction is exothermic since heat is evolved)
    2. When ethanol was burnt in the apparatus shown (in example 1), the results were: mass of fuel burnt, M1 =1.50g; mass of water, M2 =500g; ΔT= 13.0oC. (C=12; H=1; O=16; SHC of water=4.18KJKg-1 K-1). Find the molar enthalpy of combustion of ethanol. Compare the experimental value with the listed value of –1368 KJMol-1 and explain any difference.

      Solution:

      Heat evolved = CMΔT
      = 4.18 x 500/1000 x 13 
      = 27.17 kJ
      Molar mass of ethanol, C2H5OH = [(12 x 2) + (1 x 6) + (1 x 16)] = 46
      Thus if 1.5g = 27.17kJ
      Then 46 g = ?
      = 46 x 27.17/1.5
      = 833.213 kJMol
    3. In an experiment to determine the heat of combustion of methanol, CH3OH a student used a set up like the one shown in the diagram below. Study the set up and the data and answer the questions that follow.
      Data:
      Volume of water = 500 cm3
      Final temperature of water = 27.0oC
      Initial temperature of water = 20.0oC
      Final mass of lamp + menthol = 22.11g
      Initial mass of lamp + methanol = 22.98g
      Density of water - 1gcm-3
      (Heat change = mass x temperature change x 4.2Jg-1oC-1

      Questions:
      1. Write an equation for the combustion of methanol. (1mark)
      2.  calculate
        1. The number of moles of methanol used in the experiment (2marks)
        2. The heat change for this experiment. (1mark)
        3. The heat of combustion per mole of methanol. (2marks)
      3. Explain why the molar heat of combustion for methanol obtained in this experiment is different from the theoretical value. (2marks)
    4. When 0.6g of element J were completely burned in oxygen, all the heat evolved was used to heat 500 cm3 of water, the temperature of the water rose from 23.0oC to 32.0oC. Calculate the relative atomic mass of element J given that the specific heat capacity of water = 4.2 Jg-1K-1 , density of water=1.0gcm-3 and molar heat of combustion of J is 380JMol-1 (3marks)
    5. Hydrogen peroxide contained in 100cm3 of solution with water was completely decomposed. The heat evolved was 1380 J. Determine the rise in temperature due to the reaction. (SHC of water = 4.2Jg-1K-1 ; density of water = 1gcm-3; O =16; H = 1) (3 marks)

Fuels

  • Are compounds which produce a high heat of combustion.
  • Fuels can be:-
    • Solids; such as charcoal, wood, coal
    • Liquids; such as ethanol, gasolin
    • Gaseous; such as methane, water, gas etc.

Basic Concepts

Calorific Value

  • Is the energy content of a fuel;
  • Is the heat evolved when a given mass of fuel is completely burnt in oxygen;
    Note:
    • Sometimes fuels may undergo incomplete combustion .
    • Incomplete combustion of fuel is disadvantageous in that:
      -It reduces the energy content.
      -It leads to pollution.

Heating Value

  • Is the amount of heat energy given out when a unit mass or unit volume of a fuel is completely burnt in oxygen.

Fuel pollution

  • Is commonly caused by internal combustion engine.
  • Fuel in engine (e.g. petrol) burns completely to water and carbon (IV) oxide, only under ideal conditions.

Causes of Fuel Pollution

  1. Incomplete combustion which causes production of (produces) CO and unburnt carbon (soot).
  2. Some fuels contain sulphur and nitrogen and on combustion release SO2 and NO2 . These gases are acidic, resulting to acidic rain which corrodes buildings and affects trees and animals in various ways.
  3. Fuel additives; e.g. tetraethyl lead, Pb (C2H5)4 added to petrol to enhance burning efficiency produces volatile lead compounds in the exhaust fumes. Lead is very poisonous and affects the nervous system and the brain in children.

Factors to Consider when Choosing a Fuel.

  1. Heating value
  2. Ease and rate of combustion.
  3. Availability
  4. Ease of transportation
  5. Ease of storag
  6. Environment effects.

Heat of Neutralization

  • Is the heat evolved when acid and a base react to form and mole of water.
    Alternatively;
  • It is the heat evolved when one mole of hydrogen ions from an acid reacts with one mole of hydroxide ions from an alkali to form\ give one mole of water.
    Equation:
    OH-(aq) + H+(aq) → H2O(l)
  • Neutralization reactions are exothermic .
  • Are determined by measuring the temperature rise produced when a known volume of acid is neutralized by a known volume of alkali.

    Examples
    1. Strong acids reacting with strong alkalis.
      • The ΔH is always about –57KJMol-1 and higher than that for weak acids – weak alkalis.
        Reason:-
      • Strong acids and alkalis are already fully ionized in aqueous solution and no heat is lost in ionizing the acid or the alkali.

        Consider:
        Reaction between sodium hydroxide and hydrochloric aci
        NaOH(aq) → Na+(aq) + OH-(aq) 
        HCl(aq) → H+(aq) + Cl-(aq)

        On reacting; OH-(aq) + H+(aq) → H2O(l)

        Diagrammatically (energy level diagram)
        reaction of sodium hydroxide and hydrochloric acid energy level diagram
    2. Neutralization reactions involving weak acids or weak alkalis.
      •  The ΔH is lower than expected, e.g. only –52 kjmol-1; and hence lower than that for strong acids-strong alkalis.
        Reason:
      • Weak acids and weak alkalis are NOT fully ionized in aqueous solutions and some heat is used in ionizing them.
        Consider; reaction between ethanoic acid and ammonia solution.
        energy level diagram for ethanoic acid and ammonium solution1
        Therefore;
        ΔH is given by;
        ΔH = ΔHI + ΔHII
        = 5 + (-57)
        = -52 kJMol-1
        Note
      • Dibasic acids e.g. H2SO4 contains two replaceable hydrogen atoms hence on incomplete neutralization, they form two moles of water.
      • Therefore, H neutralization = ½ x ΔH reaction.

Experiment: - To Determine the Heat of Neutralization of Hydrochloric Acid by Sodium Hydroxide.

Procedure:

  • A clean dry 250 cm3 glass or plastic beaker is wrapped with a newspaper leaf.
  • Exactly 50 cm3 of 2.0M hydrochloric acid solution is transferred into the beaker.
  • The temperature T1 of the acid solution is noted
  • Using another clean dry measuring cylinder, exactly 50 cm3 of 2.0M NaOH solution is measured and its steady temperature T2 is noted.
  • The contents of the beaker (acid), are carefully stirred with a thermometer while adding NaOH
  • The highest temperature T4 attained by the resulting mixture is noted.

Apparatus
apparatus to determine the heat of neutralisation1

Results

  • Temperature of the acid, T1 = oC
  • Temperature of the hydroxide, T2 = oC
  • Average temperature of the two solutions; T1 + T2/2 = T3 = oC
  • The highest temperature of the mixture; T4 = oC
  • The temperature change, ΔT= (T4 – T3 ) = oC;

Sample Calculations

  • Given: Temperature of hydrochloric acid solution T1 = 22.75oC;
  • Temperature of sodium hydroxide solution T2 = 22.80oC;
  • Average temperature of acid and alkali T1 + T2/2= T 3 = 22.78oC
  • Highest temperature of alkali and acid mixture, T4 = 36.40oC;
  • Temperature change, ΔT= T4 – T3 = (36.40 – 22.78)oC; =13.62oC;
  • Assumption: the specific heat capacity of the solution= 4.2 kJKg-1K-1
  • In the experiment, 50 cm3 of 2M HCl are neutralized by 50 cm3 of NaOH, thus; volume of the mixture= (50 + 50) cm3 = 100 cm3 .
  • Taking density of the resultant solution to be 1gcm-3, then;-
    Mass of solution M; =density x volume
    =1g/cm 3 x 100cm3 = 100g.
    =0.1 kg i.e. (100/1000)
    Thus heat evolved,
    Mass of solution x specific heat capacity x temperature change
    = MCΔT,
    = 0.1Kg x 4.2kJKg-1K-1 x 13.62oC
    =- 5.7 kJ.
    But 50 cm3 of 2M HCl contains 2 x 50/1000 moles, = 0.1 moles of H+ ions.
    Similarity 50 cm3 of 2M NaOH contains 2 x 50/1000 = 0.1 moles of OH - ions.
    This implies that the two solutions neutralize each other completely.
  • Therefore;
    Molar heat of neutralization (heat liberated when one mole of each reagent is used;-
    0.1 Mole = 5.72 kJ
    1mole = 1 x 5.72/0.1 = -57.2 kJMol-1
  • Since heat is evolved, the reaction is exothermic, the molar heat of neutralization= -57.2 kJMol-1 .
  • Thus, enthalpy change;
    H+(aq) + OH-(aq) → H2O(l);  ΔH(neut) = -57.2 KJMol-1 . (Thermochemical equation)
    1 mole      1 mole         1 mole
    Note:
  • Thermochemical equation : refers to a chemical equation which shows the enthalpy change.

Worked Examples

  1. Given, T1 = 21.0oC; T2=22.0oC; T4 = 34.5oC; volume of hydrochloric acid= 100 cm3 , volume of NaOH(aq) =100 cm3 , molarity of the solutions are each 2M. Calculate the heat of neutralization for the two reagents. (Assumptions, density of the mixture=1gcm-3 and SHC=4.2KJKg-1 K-1)
  2.  
    1. When 100g of water at 94.0oC were added to a calorimeter at 17.5oC, the temperature rose to 80.5oC. Determine the heat capacity of the calorimeter. What assumption did you make in your calculations?
      Solution:
      Heat given out by water = heat received by calorimeter of heat capacity C.
      Heat = MCΔT,
      0.100 x 4180 x (94.0 – 80.5) = C (80.5 – 17.5)
      C = 0.100 x 4180 x 13.5/62.5
      = 90.288Jg-1K-1
      Assumption; specific heat capacity of water is 4180 Jg-1K-1
    2. 250 cm3 of sodium hydroxide were added to 250 cm3 of hydrochloric acid in the calorimeter. The temperature of the two solutions was 17.5oC initially and rose to 20.1o C. Calculate the standard enthalpy of neutralization.
      Solution:
      - Assuming the specific heat capacities of the solutions is the same as that of water, 4180 Jg-1K-1 .
      Heat from neutralization = heat by calorimeter + solutions.
      = (CΔT) + MCΔT,
      Mass of solutions = (250+250) =500 cm3
      = density x volume; = 1gcm-3 x 500 cm3 =500g
      Hat from neutralization = 90(20.1 − 17.5) + (0.500 x 4180 x (20.1 − 17.5)
      = 5670J
  3. The following results were obtained in an experiment to determine the heat of neutralization of 50 cm3 2M HCl and 50 cm3 2M sodium hydroxide.
    Mass of plastic cup = 45.1g
    Initial temperature of acid = 27.0oC
    Initial temperature of alkali = 23.0oC
    Mass of plastic cup + HCl + NaOH = 145.1g
    Temperature of the mixture of acid and alkali = 38.5oC
    1. Define heat of neutralization; (1mark)
      Solution:-
      • Enthalpy change when one mole of water is formed from a reaction between an acid and a base
    2. Write an ionic equation for the neutralization of hydrochloric acid and sodium hydroxide. (1mark)
      Solution:
      H+(aq) + OH-(aq) → H2O(l)
    3. Calculate;
      1. The amount of heat produced during the experiment. (3marks) (Specific heat capacity of solution=4.2kjkg-1K-1 , density of solution= 1gcm-3).
        Solution:
        Amount of heat = MCΔT
        Mass of solution = (145.1 – 45.1) = 100g
        Temperature change; ΔT = (38.5) − (27.0 +23.0)/2 = 38.5 – 25 =13.5oC;
        Heat produced ΔH = 100g x 4.2kjg-1K-1 x 13.5oC
        =5670 joules
        =5.67K
      2. Molar heat of neutralization for the reaction. (3marks)
        Solution:
        Number of moles involved; taking only NaOH or HCl
        1000 cm3 = 2moles
        50 cm3 = 2 x 50/1000 = 0.1moles
        If 0.1 mole produced 5.67 kg; 1 mol = ?
        0.1=5.67kg
        1mol= (5.67 x 1)/0.1 = 56.7 KJMol-1
    4. Explain why the molar heat of neutralization of NaOH and ethanoic acid of equal volume and molarity would be less than the value obtained in c (ii) above. (1mark)
      Solution:
      • Some of the heat produced during neutralization is used up by the weak acid to dissociate fully hence the lower value.
    5. Write down the Thermochemical equation for the reaction between NaOH and dilute hydrochloric acid above.
      Solution
      NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) ; ΔH= -56.7 kJMol-1
    6. Draw an energy level diagram for the neutralization reaction in 4(c) above.
      reaction of sodium hydroxide and hydrochloric acid energy level diagram1

 

 

Heat of Solution/Enthalpy of Solution.

  • Is the heat change when a given mass (moles) of a substance is dissolved in a stated amount of solvent (water).

Molar Enthalpy of Solution

  • Is the heat change when one mole of a substance is dissolved in a stated amount of solvent (water).
    Alternatively;
  • It is the heat change when one mole of a substance dissolves in water to give an infinitely dilute solution i.e. (a solution which shows no change in its properties when more water is added).
  • Is determined as;
    ΔHsolution = - ΔHlattice + ΔHhydration; when ΔH lattice is given as a negative value;
    Alternatively;
    ΔHsolution = ΔHlattice + ΔHhydration ; where ΔH lattice is given as a positive value;
  • Where,
    Lattice energy
    - Is the leaf evolved when one mole of a compound is formed from its separate gaseous ions.
    Alternatively;
    - It is the energy required to break the ionic bonds within a crystal (solid) lattice.
    Hydration (solvation) energy
    - Is the heat evolved when one mole of ions are hydrated by water molecules.
  • Note: Water is a good solvent because it has a high negative(-ve) enthalpy of solvation resulting from powerful interaction between polar molecules and solute ions due to large dipoles on water molecules.
  • Therefore, heat of solution can either be exothermic or endothermic depending on the magnitude of hydration and lattice energies
  • Solution process occurs into two stages.
    Example:
  • Consider the dissolution of sodium chloride solid.
    1. Energy is taken in to break the crystalline lattice.
      NaCl(s) → Na+(g) + Cl-(g)
      ΔH = +ΔHlattice
      =+ 776 kJ
    2. Heat is evolved when one mole of icons are hydrated by water molecules.
      Na+(g) + Cl-(g) + H2O(l) → Na+(aq) + Cl-(aq)
      ΔHhydration= -771 kJ
      Illustrations:
      1. Energy level diagram for an endothermic dissolving process for a solid MX (s)
        energy level diagram for endothermic dissolving of solid mx
        The lattice energy is larger than the hydration energy hence;-
        ΔHsolution = +ve
      2. Energy level for the diagram for an exothermic dissolving process for solid MX (s)
        energy level diagram for exothermic dissolving of solid mx
        The lattice energy is smaller than hydration energy hence;
        ΔHsolution = -ve.
        Note: All gases dissolve with evolution of heat;
        Reason:
        - There are no intermolecular forces or bonds to break before hydration occurs.

Worked examples

  1. The equation below represents changes in physical states for iron metal.
    Fe(s) → Fe(l); ΔH = 15.4 kJMol-1
    Fe(l) → Fe (g); ΔH = 354 kJMol-1
    Calculate the amount of heat required to change 11.2g of solid into iron to gaseous iron. (Fe=56.0)
    Solution:
    Total heat needed to convert 1 mole of iron from solid to gas = (15.4Kj + 354Kj); = 369.4Kj;
    1 mole = 56g;
    Thus if 56g requires 369.4 kJ;
    Then 11.2g =?
    = 11.2 x 369.4/56 ; = 73.88 kJ;
  2. The lattice and hydration enthalpies for lithium chloride and potassium chloride are given below.
     Salt  ΔHlattice  ΔHhydration  
     LiCl  − 861  − 884
     KCl  − 719  − 695
    1. Calculate the enthalpy of solution, ΔH solution ; Kj mol - for;
      1. Lithium chloride. (2 marks)
        Solution:
        Hsolution = -ΔHlattice + ΔHhydration
        = - (-861) + (-884)
        = 861 - 884;
        = -23 kJmol-1
      2. Potassium chloride: (2 marks)
        Solution:
        Hsolution = -ΔHlattice + ΔHhydration
        = - (-719) + (-695)
        = 719 - 695;
        = -24 kJmol
    2. Which of these two salts above has a higher solubility in water? Give reasons for your answer.
      Solution:
      • Potassium chloride; the difference in ΔH solution is greater
  3.  Given that the lattice energy of NaCl(s) is +77 kJmol-1 and hydration energies of Na+(g) and Cl-(g) are -406kjmol-1 and -364kjmol-1 respectively. Calculate the heat of solution, ΔHsoln for one mole of NaCl(s)

Experiment: To Measure the Molar Enthalpy Change for the Dissolution (Enthalpy of Solution) for Various Compounds.

Procedure

  • A clean 250 cm3 glass or plastic beaker is wrapped with a newspaper leaf
  • About 50 cm3 of tap water is measured into the beaker and the steady temperature noted.
  • The beaker is held in a tilted position and 2 cm3 of sulphuric acid added into the water.

Apparatus.
enthalp of solution apparatus

Caution

  • Concentrated sulphuric acid should always be added to water and never vise versa.
    Reason:
    Sulphuric acid (H2SO4) reacts very vigorously with water, in a highly exothermic reaction. 
    If you add water to concentrated sulphuric acid, it can boil and spit and you may get a nasty acid burn.
  • The mixture is then carefully but vigorously stirred using a thermometer and the highest temperature of the solution recorded.
  • The experiment is repeated with other compounds:
    • Ammonium nitrate solid
    • Potassium nitrate solid
    • Sodium hydroxide pellets
    • Sodium nitrate solid

 

 

 

Results and Observations

   Concentrated sulphuric acid  Ammonium nitrate solid  Potassium nitrate solid  Sodium hydroxide pellets
 Temperature of water(oC)        
 Highest temperature of solution (oC)        
 Change in temperature (oC)        

Calculate the enthalpy change (ΔHsoln) in each experiment done.

Additional information;

  • Density of concentrated Sulphuric acid =1.84gcm-3
  • Density of water =1gcm-3
  • No change in volume when any of the solids dissolved in water
  • The specific heat capacities of the solutions=4.2kJkg-1K-1
  1. Sulphuric acid:
    - Total volume of solution = (50 +2) =52 cm3
    - Temperature change =A rise of oC
    - Enthalpy change; ΔHsoln ; =MCΔT
    M = 52 cm3 x 1gcm-3 = 52g = 0.52kg.
    C = 4.2kJkg-1K-1
    ΔT = 5K.
    Thus enthalpy change = 0.52 x 4.2 x 1
    = - 2.184kJ
    Calculating molar enthalpy of solution of concentrated sulphuric acid:
    Mass of acid used = density x volume
    = 1.84 x 2 = 3.68g;
    Formula mass of Sulphuric acid = 98g
    3.6g of acid liberated 2.184 kJ
    Thus 98g of acid liberated; 2.198 x 98/3.68 =56Kj;
    Because heat was liberated (+ve ΔT), the ΔHsoln = -56 kJmol-1
  2. Sodium nitrate:
    Data:
    - Volume of water used = 100 cm3
    - Mass of sodium nitrate = 2.5g
    - Initial temperature of water = 21.0ºC
    - Final temperature of water =19.5oC
    - Temperature change; = A fall of 1.5ºC i.e. (19.5 - 21.0)oC;
    - Mass of 100 cm3 of water = 100g = 0.1kg;
    Enthalpy change i.e. heat absorbed = MCΔT
    = 0.1 x 4.2 x 1.5
    = 0.63 kJ.
    Molar heat of dissolution;
    - Molar mass of NaNO3 =23 + 14 + 48 = 85g;
    - 2.5g of NaNO3 absorbed 0.63 kJ;
    - 85g of NaNO3 will absorb =?= 0.63 x 85/2.5
    = +21.42 kJmol-1
    Since heat is absorbed (-ve heat change), the reaction is endothermic thus ΔHsoln = +21.42 kJmol-1
    Note: energy level diagram for the dissolution of sodium nitrate
    energy level diagram for endothermic dissolving of sodium nitrate
  3. Potassium hydroxide;
    • Calculate the molar enthalpy of dissolution of potassium hydroxide (based on values obtained).

Heat of Displacement

  • This refers to the enthalpy change that occurs when one mole of a substance is displaced from a solution of its ions.

Experiment: To Determine the Molar Enthalpy Change in the Reaction between Cu2+ ions and Zinc or Iron

Procedure

  • A plastic cup or glass beaker is wrapped with a newspaper leaf.
  • 25 cm3 of 0.2M copper (II) sulphate solution is transferred into the beaker.
  • The steady temperature of the solution is noted.
  • 0.5g of zinc powder are carefully transferred into the plastic cup and stirred with a thermometer
  • The highest temperature attained by the solution is recorded.

Apparatus
heat of displacemnent

Observations

  • The blue colors of copper (II) sulphate fades.
  • Brown deposits of copper metal are formed in the plastic cup.

 

Explanation:

  • Zinc is higher in the electrochemical series than copper;
  • Zinc therefore displaces copper ions from its solution.
    Equations:
    Zn(s) + CuSO4(aq) →ZnSO4(aq) + Cu(s)
    Ionically:
    Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
  • During the reaction the blue Cu2+ in the solution are replaced by the colorless Zn2+.
  • Consequently the blue colour of the solution fades, as brown deposits of copper metal are formed in the plastic cup.
  • Excess solid (zinc powder) was used to ensure complete displacement of Cu2+.

Results:

  • Initial temperature of copper sulphate solution, T1 = 23oC
  • Highest temperature of the mixture T2 = 33oC
  • Temperature change, ΔT; = T2 - T1 =10ºC
  • Volume of copper sulphate solution used = 25.0 cm3 .
  • Mass of zinc powder taken = 0.6g
  • Density of the solution = 1gcm-3.
  • Specific heat capacity of the solution = 4.2 kJkg-1K-1 .

Assumptions

  • The volume of the solution remains unchanged after the reaction.
    (Cu = 63.5, S = 32, O =16, Zn = 65, Fe =56)

Question:

Using the above data, calculate, calculate the;

  1. Heat change for the above reaction.
  2. The molar heat of displacement of copper (II) ions.
    Solutions:
    (i) Heat change = MCΔT
    Mass = density x volume
    1gcm -3 x 25cm3 =25g.
    ΔH = 25kg/1000 x 4.2 kJkg-1K-1 x 10k
    = 1.050 kJ;
    (ii). Molar heat of displacement of Cu2+ displacement.
    1000 cm3 = 0.2 moles.
    25 cm3 = 0.2 x 25/1000 = 0.005moles
    Thus,
    0.005moles =1.050 kJ
    1mole = 1 x 1.050/0.005 = 1050/5 = 210 kJ;
  • Since the final temperature of the mixture is higher than initial temperature; it means the H products is lower than H reactants .
  • The reaction is thus exothermic.
  • Thus molar heat of displacement of Cu2+ = -210 kJmol -1
    Thermochemical equation:
    Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq); ΔH = -210kJmol-1
    Note:
  • The experimental value for the heat liberated in this reaction is lower than the theoretical value of 216Kjmol-1.
    Reasons:
  • The heat lost to the surroundings and the heat absorbed by the apparatus is not accounted for in the calculations.
  • This reaction is typically a Redox reaction.
    i.e. 
    redox reaction of zinc and copper

Heat of Precipitation

  • Is the heat change which occurs when one mole of a substance is precipitated from the solution.

Experiment: To determine the heat of precipitation of silver chloride.

Procedure:

  • 50 cm3 each of 2M AgNO3 and 2M NaCl are left in separate beakers and their constant temperatures noted and recorded.
  • The NaCl solution is added to the silver nitrate solutions.
  • The beaker is covered with a card board and shaken gently to allow mixing of the solution.
  • The highest temperature of the mixture is noted and recorded.

Apparatus
heat of precipitation

Data (results)

  • Temperature of AgNO3 solution = 21.0ºC
  •  Temperature of NaCl solution =19.0ºC
  • Mean temperature of the solutions = (19+21)/2 = 20.0oC
  • Final temperature after mixing = 34.0oC;
  • Temperature change = a rise of 14oC;
  • Volume of solution mixed =100 cm3
  • Mass of solution mixed =100g (assuming density=1gcm-3 )

Questions:

From the above data, calculate;-

  1. Energy change/ heat change for the reaction
  2. The molar ΔH precipitation for AgCl.
    Solution:
    (i) Heat change, = MCΔT
    = 100/1000 x 4.2 x 14
    = - 5.88 kJ
    (ii) Equation,
    AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
    Mole ratio 1 :       1 :               1         :        1
    Moles of AgNO3 ;
    1000 cm3 = 2 moles
    50 cm3 = 2 x 50 = 0.1mole;
    Since reaction ratio AgNO3 : AgCl = 1:1, then moles of AgCl(s) precipitated = 0.1mole
    If 0.1mole = 5.88 kJ
    1 mole = (1 x 5.88)/0.1 = 58.8kJmol-1
    Reaction being exothermic, ΔHprecip = -58.8kJmol-1
    Thermochemical equation:
    Ag+(aq) + Cl-(aq) → AgCl(s) ; ΔH= -58.8kjmol-1
    Suitable energy level diagram,
    heat of precipitation of silver chloride energy level diagram

Enthalpy of formation

  • Is the heat change when one mole of a substance is formed from its constituent elements under standard conditions.
  • Note: this will be dealt with under thermochemical cycles and Hess s law.

Standard Conditions for Measuring Enthalpy Changes.

  • Experimental evidence shows that enthalpy changes are affected by physical stats of the reactants, temperature, concentration and pressure.
  • Consequently, for comparison purposes certain conditions have been chosen as standard for measurement and determination of enthalpy changes.
  • These conditions are:
    • A temperature of 25oC (298 K)
    • Pressure of 1 atmosphere (101.325 kPa)

Standard Enthalpy Changes

  • Are values of enthalpy changes that are measured under standard conditions.
  • They are denoted by he symbol ΔH0
  • A subscript is also added to the symbol to indicate the type of enthalpy change involved.

Common Standard Enthalpy Changes.

  1. ΔH0f ; refers to the standard molar enthalpy change of formation.
    Example:
    H2(g) + ½O2(g) → H2O(l) ; ΔH0f(H2O) = -286 kJmol-1
  2. ΔH0c ; refers to the standard molar enthalpy change of combustion;
    Example:
    CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ; ΔH0c(CH4) = -890 kJmol-1
  3. ΔH0neut; refers to the standard molar enthalpy change of neutralization.
    Example:
    HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ; ΔH0neut = -58 kJmol-1
  4.  ΔH0soln; refers to the standard molar enthalpy change of solution (dissolution).
    Example:
    NaNO3(s) + H2O(l) → NaNO3(aq) ; ΔH0soln(NaNO3) = +21 kJmol-1
  5. ΔH0hyd; refers to the standard molar enthalpy change of hydration.
    Example:
    Na+(s) + Cl-(s) + H2O(l) → Na+(aq) + Cl-(aq); ΔH0hyd (NaCl) = -774 kJmol-1
  6. ΔH0latt ; refers to the molar enthalpy change of lattice formation.
    Example:
    Na+(g) + Cl-(g) → NaCl(aq); ΔH0latt(NaCl) = -774KJmol -
  7. ΔH0at ; refers to the standard molar enthalpy change of atomization.
    Example:
    Na(s) →Na(g); ΔH0at(Na) = 108.4 kJmol-1


Energy Level Diagrams,Thermochemical Cycles and Hess’s law

Hess’s Law (The Law of Constant Heat Summation)

  • States that:
    • If a reaction can occur by more than one route the overall change in enthalpy is the same, whichever route is followed.
  • Illustration:
    hess law illustrated1

Worked examples

  1. Use the information below to determine the enthalpy of combustion of carbon (formation of carbon (IV) oxide).
    C(s) + ½O2(g) → CO(g) ; ΔH = -110.45 kJ
    CO(g) + ½O(g) → CO2(g); ΔH = -282.0 kJ
    Solution:
    hess law for formation of carbon dioxide
    Therefore; ΔH1 = ΔH2 + ΔH3
    = -110.45 + (-282.0) = - 392.45 kJ
    ΔH1 = ΔHcombustion
    = -392.45 kJ;

Note: - Enthalpy of formation, ΔHf;

  • Is the heat absorbed or evolved when one mole of a substance is formed from its elements under standard conditions.
  • This can not be determined experimentally because reactions cannot take place under standard conditions.
  • In such cases the enthalpy change is determined theoretically using other measurable enthalpies.
  • In the case of enthalpy of formation, the enthalpies of combustion of the compound and that of its constituent elements are linked using an energy level or energy cycle diagram.
  • The enthalpy of formation is then calculated using the law of constant heat summation (Hess s law).

Worked examples:

  1. One mole of butane(C4H10) burns completely in oxygen and liberates 2877 kJ.
    1. Write an equation for the combustion of butane.
      2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)
    2. Draw an energy cycle, hence an energy level diagram for the reactions concerned.
      Solution:
      Equations for combustion:
      C(s) + O2(g)→ CO2(g); then 8C(s) + 8O2(g) →8CO2(g) (based on balancing for butane)
      H2(g) + ½O2(g) →H2O(g) ; then 10H2(g) + 5O2(g) → 10H2O(g); (based on balancing for butane)
      2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)
      Then, energy cycle for the changes,
      energy cycle for combustion of butane
      Thus energy level diagram for the reactions.
      combustion of butane energy level diagram
    3.  If the following heats of combustion are given
      ΔHºc (graphite) = -393kJmol-1
      ΔHºc (H2(g)) = -286KJmol-1
      ΔHºc (C4H10 ) = -2877 kJmol-1
      Calculate the heat of formation of butane, C4H10 from its elements in their normal states at S.T.P.
      Solution:
      ΔH1 =ΔH2 + ΔH3 where:
      ΔH2 = ΔH formation of C4H10;
      ΔH1 =ΔHºc(graphite) + ΔHºc(hydrogen) ;
      While ΔH3 = ΔHºc(butane)
      From the equation,
      ΔH1 = ΔH2 + ΔH3 ,
      ΔH2 = ΔH1 − ΔH3
      = [8(−393) +10(−286)]−[2(−2877)]
      Dividing all through by 2;
      = [4(−393) +5(−286)]−2877; i.e. in order to work with 1 mole of butane;
      = (−1572) + (−1430) − (−2877)
      = −3002 + 2877
      = −125KJmol−1
  2.  
    1. Define standard heat of combustion of a substance.
      Solution:
      • The heat change when one mole of a substance is completely burnt in oxygen under standard conditions;
    2. Ethanol, CH3CH2OH; cannot be prepared directly from its elements and so its standard heat of formation must be obtained indirectly.
      1. Write an equation for the formation of ethanol from its elements in their normal physical states at standard conditions of temperature and pressure.
        Solution:
        2C(s) + 3H2(g) + ½O 2(g) → C2H5OH(l)
      2. Draw an energy level diagram linking the heat of formation with its heat of combustion and the heats of combustion of its constituent elements.
        Solution:
        Equations
        (i). C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
        (ii). 2C(s) + 2O2(g) → 2CO2(g)
        (iii). 3H2O(g) + 1½O2(g) → 3H2O(l)
        Then, energy cycle for the changes,
        energy cycle diagram for ethanol formation
        Thus energy level diagram for the reactions.
        energy level diagram for formation of ethanol
      3.  If the following heats of combustion are given:
        ΔHc(graphite) = -393kJmol-1
        ΔHc(hydrogen) = -286kJmol-1
        ΔHc(ethanol) = -1368kJmol-1
        Calculate the heat of formation of ethanol, ΔH0f(ethanol)
        Solution:
        ΔH1 = ΔH2 + ΔH3
        ΔH2 = ΔH1 - ΔH3
        = [2(-393 + 3(−286)]−[−1368]
        = [(−786) + (-856)] − [−1368]
        = −164 + 1368
        = - 276kJmol−1
  3. Given the standard enthalpies of combustion as
    C(s) + O2(g) → CO2(g) ; ΔH1 = -394kJmol-1 ;
    H2(g) + ½O2(g) → H2O(l) ; ΔH2 = -286kJmol-1
    C2H2(g) + 1½O2(g) → 2CO2(g) + H2O(l) ; ΔH = -1300 kJmol-1.
    Find the standard enthalpy of formation of ethyne, C2H2(g).
    Solution:
    - Considering the Hc(ethyne) , 2 moles of CO2 and 1 mole of H2O(l) are produced;
    - Rebalancing the equations for Hc(carbon) and Hc(hydrogen) likewise:
    2C(s) + O2(g) → 2CO2(g);
    H2(g) + ½O2(g) → H2O(l);
    Then, energy cycle for the changes,
    ENERGY CYCLE DIAGRAM FOR THE FORMATION OF ETHYNE
    Thus energy level diagram for the reactions.
    energy level diagram for the formation of ethyne
    Thus ΔH1 = ΔH2 + ΔH3
    ΔH2 = ΔH1 - ΔH3
    = [2(−394) + (−286)] − [−1300]
    = (−788-286) + 1300
    = -1074 + 1300
    = +226kJmol−1
  • Since ΔHf is +ve, ethyne is described as an endothermic compound.
  • Note:
    - In the determination of ΔHf of ethyne, Hess law is used because attempts to make ethyne from carbon and hydrogen {2C(s) + H2(g) → C2H2(g)} will result to the formation of a mixture of hydrocarbons.

Heat of Formation and Bond Energies

  • Chemical reactions involve:
    • Bond breaking ; this requires energy.
    • Bond formation ; releases energy.
  • Example:
    1. Formation of methane from carbon and hydrogen.
      C(s) + 2H2(g) → CH4(g); ΔH = -74.9kJmol-1
      Energy level diagonal for the formation of CH4 .
      energy level diagram for the formation of methane1
  • Energy changes involved are:
    • ΔHI; Enthalpy of atomization – this is the energy absorbed when a substance decomposes to form one mole of gaseous atoms.
    • ΔHII; enthalpy of atomization of H2; i.e. dissociation of hydrogen molecules to free hydrogen atoms.
    • ΔHIII; enthalpy of formation of four C – H bonds , during which heat is liberated;
      Therefore, enthalpy of reaction ΔHR = ΔHIV
      ΔHR = ΔH1 + ΔHII + ΔHIII
      = +715 + 4(218) + −4(C−H)
      −74.9 =+715 + 872 +−4 (bond energy)
      4 (C−H) =1587 + 74.9
      =1661.9
      C−H = 1661.9/4 = 415.4 kJmol
    •  This is the amount of energy released when one C−H bond is formed;

Worked Examples.

  1. Use the bond energies given below to calculate the heat of reaction for;-
    H2(g) + Cl2(g) → 2HCl(g)
     Bond  Bond Energy
     H - H  435 kJmol-1
     Cl - Cl  243 kJmol-1
     H-Cl  431 kJmol-1
    Solution:

    Bond breaking,
    H2(g) = H-H               2H(g); ΔH = +435 kJ;
    Cl2(g) = Cl-Cl             2Cl(g); ΔH= +243 kJ
    Bond formation,
    2H(g) + 2Cl(g) → 2(H-Cl)(g); ΔH = 2(−431) = −862 kJ;
    Heat of reaction:
    Energy absorbed in bond breakage + energy released in bond formation.
    = [+435 + 243] + [−862]
    = [+678 − 862] kJ
    = − 184 kJ
    Thus, H - H + Cl – Cl                2[H – Cl]; ΔH = −184 kJ
    Note: energy level diagram.
    energy level diagram for the formation of HCl
  2. Study the information given in the table below and answer the questions that follow;-
    1. Calculate the enthalpy change for the reaction
      CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
      Solution:
      Bond breaking
      (i). 4(C-H)                4CH(g); ΔH =4(+414) = +1656 kJ;
      (ii). Cl – Cl                2Cl(g); ΔH1 = +244 kJ
      Bond formation
      (i). 3(CH)                 3(C-H); ΔH= 3(-414) =-1242 kJ
      (ii). CCl                      (C-Cl); ΔH= -326 kJ
      (iii). HCl                     (H-Cl); ΔH= -431kJ
      Heat change = ΔHbond breakage + ΔHbond formation
      = [(+1656) + (+244)] + [(-1242) + (-326) + (-431)]
      = -1900 + (-1999)
      ΔH = -99 kJ
    2. Draw an energy level diagram for the reaction
      energy level diagram for the formation of chloromethane
  3. Study the information given in the table below and answer the questions that follow
     Bond  Bond Energy
     C- C  346 kJmol-1
     C = C  610 kJmol-1
     C - Br  280 kJmol-1
     Br - Br  193 kJmol-1
     C - H  413 kJmol-1
    1. Calculate the enthalpy change for the reaction.
      dibromoethane reaction
      Solution:
      Bond breaking,
      4(C-H)           4CH(g); ΔH = 4 (+413) = +1652 kJ
      Br2                2Br ΔH = +193 kJ
      C = C             C = C; ΔH= + 610 kJ;
      Bond formation:
      4(C-H),               4CH; = 4(-413) = -1662 kJ
      2(C-Br),              2CBr; = 2(-280) = -560 kJ
      (C-C )                 2C; = -346 kJ
      Heat change:
      ΔHbond breakage + ΔHbond formation
      = [+1652 + 193 + 610] + [-1652 + -560 + -346kJ]
      = [+2455] + (-2558)
      ΔH change = -103 kJ;

 

Molar heat of fusion, ΔHf

  • Is the amount of heat required to change one mole of a solid into a liquid at its melting point.
  • It is a measure of intermolecular forces between solid particles since it is the heat energy used to separate the solid particles.

Molar heat of vaporization, ΔHvap;

  • Is the amount of heat required to change one mole of liquid into vapor at its boiling point.
  • It is a measure of the intermolecular forces between the liquid particles since it is the heat energy used to separate liquid particles.

Relationship between heats of fusion and vaporization to structure.

  • They are;-
    1. Higher for substances with giant structures e.g.
      • Ionic solids
      • Metals
      • Giant covalent structures, e.g. diamond, graphite and silicon (IV) oxide
    2. Lower in simple molecular structures whose particles are held together by weak intermolecular forces.
      Examples:
      • Iodine crystals
      •  Sulphur
      • Hydrogen.

      Examples:
         NaCl  C(graphite)  Cu  H2O  I(s)  S
      Struture type   Giant atomic  Giant atomic  Giant metallic  Molecular  Molecular  Molecular
       ΔHfusion kJmol-1  28  -  13  6  8  -
       Δvap kJmol-1  171  717  305  41  21  10

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