Thermochemistry

“Thermochemistry is a branch of physical chemistry which is concerned with energy changes accompanying chemical transformation. It is also termed as chemical energetics. It is based on the first law of thermodynamics.”

Thermochemical equations

A balanced chemical equation which tells about the amount of heat evolved or absorbed during the reaction is called a thermochemical equation. A complete thermochemical equation supplies the following information’s.

(1)     It tells about the physical state of the reactants and products. This is done by inserting symbol (s), (l) and (g) for solid, liquid and gaseous state respectively along side the chemical formulae.

(2)     It tells about the allotropic form (if any) of the reactant by inserting the respective allotropic name, for example, C (graphite) (s).

(3)     The aqueous solution of the substance is indicated by the word aq.

(4)     It tells whether a reaction proceeds with the evolution of heat or with the absorption of heat, i.e. heat change involved in the system.

Remember that like algebraic equations, thermochemical equations can be reversed, added, subtracted and multiplied.

 

Exothermic and Endothermic reactions

(1)     Exothermic reactions : The chemical reactions which proceed with the evolution of heat energy are called exothermic reactions. The heat energy produced during the reactions is indicated by writing +q or more precisely by giving the actual numerical value on the products side. In general exothermic reactions may be represented as, A + B → C + D + q (heat energy)

In the exothermic reactions the enthalpy of the products will be less than the enthalpy of the reactants, so that the enthalpy change is negative as shown below

          ΔH = H_p – H_r ; H_p < H_r ; ΔH = – ve

Examples :

(i)      C(s) + O₂(g) → CO₂(g) + 393.5 kJ

          (at constant temperature and pressure)

          or C(s) + O₂(g) → CO­2₂(g); ΔH = – 393.5 kJ

(ii)     H₂(g) +   O₂(g) → H₂O(l);   ΔH = – 285.8 kJ

(iii)    N₂(g) + 3H₂(g) → 2NH₃(g);  ΔH = – 92.3 kJ

(iv)    2SO₂(g) + O₂(g) → 2SO₃(g);  ΔH = – 694.6 kJ

(v)     CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O;  ΔH = – 890.3 kJ

(vi)    Fermentation is also an example of exothermic reaction.

 

(2)     Endothermic reactions : The chemical reactions which proceed with the absorption of heat energy are called endothermic reactions. Since the heat is added to the reactants in these reactions, the heat absorbed is indicated by either putting (–) or by writing the actual numerical value of heat on the reactant side

                    A + B → C + D – q (heat energy)

The heat absorbed at constant temperature and constant pressure measures enthalpy change. Because of the absorption of heat, the enthalpy of products will be more than the enthalpy of the reactants. Consequently,  will be positive (+ ve)  for the endothermic reactions.

                   ΔH = H_p – H_r ; H_p > H_r ;  ΔH = + ve

Example :

          (i)      N₂(g) + O₂(g) + 180.5 kJ → 2NO(g)

                    N₂(g) + O₂(g) → 2NO(g) ; ΔH = + 180.5 kJ

          (ii)     C(s) + 2S(s) → CS₂(l)  ΔH = + 92.0 kJ

          (iii)    H₂(g) + I₂(g) → 2HI(g) ;  ΔH = +53.9 kJ

          (iv)    2HgO(s) → 2Hg(l) + O₂(g) ; ΔH = + 180.4 kJ

          (v)     SnO₂(s) + 2C(s) → Sn(s) + 2CO₂(g) ; ΔH = +360.0 kJ

          (vi)    Fe + S → FeS

(vii)   Preparation of ozone by passing silent electric discharged through oxygen is the example of endothermic reaction. 

(viii) Evaporation of water is also the example of endothermic reaction.

The enthalpy changes for exothermic and endothermic reactions are shown in figure.

Similarly, if we consider heat change at constant volume and temperature, DE is – ve, now it may be concluded that

          For exothermic reaction : ΔH or ΔE = – ve

          For endothermic reaction : ΔH or ΔE = + ve

 

Heat of reaction or Enthalpy of reaction

Heat of reaction is defined as the amount of heat evolved or absorbed when quantities of the substances indicated by the chemical equation have completely reacted. The heat of reaction (or enthalpy of reaction) is actually the difference between the enthalpies of the products and the reactants when the quantities of the reactants indicated by the chemical equation have completely reacted. Mathematically,

Enthalpy of reaction (heat of reaction) = ΔH = ∑H_p – ∑H_R

(1)     Factors which influence the heat of reaction : There are a number of factors which affect the magnitude of heat of reaction.

(i)      Physical state of reactants and products : Heat energy is involved for changing the physical state of a chemical substance. For example in the conversion of water into steam, heat is absorbed and heat is evolved when steam is condensed.  Considering the following two reactions

                   H₂(g) +   O₂(g) = H₂O(g);  ΔH = – 57.8 kcal

                   H₂(g) +   O₂(g) = H₂O(l);  ΔH = – 68.32 kcal

It is observed that there is difference in the value of ΔH if water is obtained in gaseous or liquid state. ΔH value in second case is higher because heat is evolved when steam condenses. Hence, physical sate always affects the heat of reaction.

(ii)     Allotropic forms of the element : Heat energy is also involved when one allotropic form of an element is converted into another. Thus, the value of ΔH depends on the allotropic form used in the reaction. For example, the value of ΔH is different when carbon in the form of diamond or in amorphous form is used.

                   C (diamond) + O₂(g) → CO₂ (g) ; ΔH = – 94.3 kcal

                   C (amorphous) + O₂(g) → CO₂(g) ; ΔH = – 97.6 kcal

The difference between the two values is equal to the heat absorbed when 12g of diamond is converted into 12g of amorphous carbon. This is termed as heat of transition.

(iii)    Temperature : Heat of reaction has been found to depend upon the temperature at which reaction is occurring. The variation of the heat of reaction with temperature can be ascertained by using Kirchhoff’s equation.

= ΔC_p

Kirchhoff’s equation at constant volume may be given as,

= ΔC_v

(iv)    Reaction carried out at constant pressure or constant volume : When a chemical reaction occurs at constant volume, the heat change is called the internal energy of reaction at constant volume. However, most of the reactions are carried out at constant pressure; the enthalpy change is then termed as the enthalpy of reaction at constant pressure. The difference in the values is negligible when solids and liquids are involved in a chemical change. But, in reactions which involve gases, the difference in two values is considerable.

                    ΔE + ΔnRT = ΔH or q_v + ΔnRT = q_p               

ΔE = q_v heat change at constant volume; = q_p heat change at constant pressure,

Δn = total number of moles of gaseous product – total number of moles of gaseous reactants.

 

(2)     Types of heat of reaction

(i)     Heat of formation : It is the quantity of heat evolved or absorbed (i.e. the change in enthalpy) when one mole of the substance is formed from its constituent elements under given conditions of temperature and pressure. It is represented by ΔH_f. When the temperature and pressure are as 25^oC and 1 atmospheric pressure. The heat of formation under these conditions is called standard heat of formation. It is usually represented by ΔH^o_f.

The standard heat of formation of 1 mole of NH₃ (g) and 1 mole of HCl(g).

                     N₂(g) +    H₂(g) → NH₃(g);  ΔH_(g) = – 11 kcal

                    H₂(g) +   Cl₂(g) → HCl;   ΔH_f = – 22 kcal

It may be calculated by

ΔH^o = [Sum of standard heats of formation of product] – [Sum of standard heats of formation of reactants]

ΔH^o = [∑ΔH^o_(Products) – ∑ΔH^o_(Reactants)]

                            

(ii)     Heat of combustion : It is the amount of heat evolved or absorbed (i.e. change in enthalpy) when one mole of the substance is completely burnt in air or oxygen. For example

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ; ΔH = – 192 kcal

C₂H₆(g) + 3.5O₂(g) → 2CO₂(g) + 3H₂O(l) ; ΔH = – 372.8 kcal

It may be calculated by             

ΔH^o = [Sum of standard heats of Combustion of products] – [Sum of standard heats Combustion of reactants]

ΔH^o = [∑ΔH^o_f(Products) – ∑ΔH^o_f(Reactants)]

Note :  

• Heat of combustion increases with increase in number of carbon and hydrogen.
• Heat of combustion of carbon is equal to the intrinsic energy of CO₂.

The enthalpy or heat of combustion have a number of applications. Some of these are described below,

(a)    Calorific value of foods and fuels : Energy is needed for the working of all machines. Even human body is no exception. Coal, petroleum, natural gas etc. serve as the principal sources of energy for man-made machines, the food which we eat serves as a source of energy to our body.

The energy released by the combustion of foods or fuels is usually compared in terms of their combustion energies per gram. It is known as calorific value. The amount of heat produced in calories or Joules when one gram of a substance (food or fuel) is completely burnt or oxidised.

When methane burns, 890.3 kJ mol^–1 of energy is released.

  + 2O₂(g) → CO₂(g) + 2H₂O(l) ;  ΔH_CH4 = – 890.3 kJ

So, the calorific value of methane = –   = – 55.6 kJ/g

 

Calorific values of some important food stuffs and fuels

Fuel	Calorific value (kJ/g)	Food	Calorific value (kJ/g)
Wood	17	Milk	3.1
Charcoal	33	Egg	6.7
Kerosene	48	Rice	16.7
Methane	55	Sugar	17.3
L.P.G.	55	Butter	30.4
Hydrogen	150	Ghee	37.6

 

Out of the fuels isted, hydrogen has the highest calorific value.  The calorific value of proteins is quite low.

(b)    Enthalpies of formation : Enthalpies of formation of various compounds, which are not directly obtained, can be calculated from the data of enthalpies of combustions easily by the application of Hess’s law.

Heat of reaction = ∑ Heat of combustion of reactants – ∑ Heat of combustion of products.                  

(iii)    Heat of neutralisation : It is the amount of heat evolved (i.e., change in enthalpy) when one equivalent of an acid is neutralised by one equivalent of a base in fairly dilute solution, e.g., Neutralisation reactions are always exothermic reaction and the value of  ΔH is .

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O ; ΔH = – 13.7 kcal

The heat of neutralisation of a strong acid against a strong base is always constant (13.7 kcal or 57 kJ mol^–1). It is because in dilute solutions all strong acids and bases ionise completely and thus the heat of neutralisation in such cases is actually the heat of formation of water from H⁺ and OH^– ions, i.e.,

H⁺ + OH^– → H₂O ; ΔH = – 13.7 kcal

In case of neutralisation of a weak acid or a weak base against a strong base or acid respectively, since a part of the evolved heat is used up in ionising the weak acid or base, it is always less than 13.7 kcal mole^–1 (57 kJ mole^–1). For example, heat of neutralisation of HCN (a weak acid) and NaOH (a strong alkali) is –2.9 because 10.8 kcal of heat is absorbed for the ionisation of HCN (i.e., the heat of dissociation or ionisation of HCN is 10.8 kcal) Similarly Heat of neutralization of NH₄OH and HCl is less then 13.7 kcal. 

                   HCN(aq) + NaOH(aq) → NaCN(aq) + H₂O ; ΔH = – 2.9 kcal

                   HCN(aq) ⇌  H⁺ + CN^–; DH = 10.8 Kcal

(iv)    Heat of solution : It is the amount of heat evolved or absorbed (i.e., change in enthalpy) when one mole of the solute is dissolved completely in excess of the solvent (usually water). For example,

NH₄Cl(s) + H₂O(l) → NH₄Cl (aq) ; ΔH = + 3.90 kcal                         

BaCl₂(s) + H₂O(l) → BaCl₂ (aq) ; ΔH = – 2.70 kcal

(v)     Heat of hydration : It is the amount of heat evolved or absorbed (i.e change in enthalpy) when mole of an anhydrous or a partially hydrated salt combines with the required number of moles of water to form a specific hydrate. For example,

CuSO₄(s) + 5H_2­O(l) → CuSO₄. 5H₂O(s) ; ΔH = – 18.69

(vi)    Heat of vapourisation : When a liquid is allowed to evaporate, it absorbs heat from the surroundings and evaporation is accompanied by increase in enthalpy. For example: 10.5 kcals is the increase in enthalpy when one mole of water is allowed to evaporate at 25^oC. When the vapours are allowed to condense to liquid state, the heat is evolved and condensation of vapour is accompanied by decrease in enthalpy. The value for the condensation of one mole of water vapour at 25^oC is also 10.5 kcals.

The evaporation and condensation can be represented as,

                   H₂O(l) → H₂O(g) ; ΔH = + 10.5 kcals (+43.93 kJ)

                   H₂O(g) → H₂O(l) ; ΔH = – 10.5 kcals (–43.93 kJ)

Thus the change in enthalpy when a liquid changes into vapour state or when vapour changes into liquid state is called heat of vapourisation.

(vii)   Heat of fusion : When a solid is allowed to melt, it changes into liquid state with the absorption of heat (increase in enthalpy) and when a liquid is allowed to freeze, it changes into solid with the evolution of heat (decrease in enthalpy). The change in enthalpy of such type of transformations is called enthalpy of fusion. For example,

H₂O (ice) → H₂O (liquid) ; ΔH = + 1.44 kcals (+6.02 kJ)

H₂O (liquid) → H₂O (ice) ; ΔH = – 1.44 kcals (– 6.00 kJ)

Note :  The enthalpy of fusion of ice per mole is 6 kJ.

(viii) Heat of precipitation : It is defined as the amount of heat liberated in the precipitation of one mole of a sparingly soluble substance when solutions of suitable electrolytes are mixed, for example

                    Ba²⁺ + SO₄^2–(aq) → BaSO₄(s) : ΔH = – 4.66 kcal

(ix) Heat of sublimation : Sublimation is a process in which a solid on heating changes directly into gaseous state below its melting point.

Heat of sublimation of a substance is the amount of heat absorbed in the conversion of 1 mole of a solid directly into vapour phase at a given temperature below its melting point.

                   I₂(s) → I₂(g);   Δ H = + 62.39 kJ

Most solids that sublime are molecular in nature e.g. iodine and naphthalene etc.

 ΔH_sub = ΔH_fusion + ΔH_vaporisation

 

(3)     Experimental determination of the heat of reaction : The heat evolved or absorbed in a chemical reaction is measured by carrying out the reaction in an apparatus called calorimeter. The principle of measurement is that heat given out is equal to heat taken, i.e., Q = (W +m) × s × (T₂ – T₁).

Where Q is the heat of the reaction (given out), W is the water equivalent of the calorimeter and m is the mass of liquid in the calorimeter and s its specific heat, T₂ is the final temperature and  the initial temperature of the system. Different types of calorimeters are used but two of the common types are,

 

(i)      Water calorimeter and (ii) Bomb calorimeter       

Bomb calorimeter : This is commonly used to find the heat of combustion of organic substances. It consists of a sealed combustion chamber called a bomb. A weigh quantity of the substance in a dish along with oxygen under about 20 atmospheric pressure is placed in the bomb which is lowered in water contained in an insulated copper vessel. The vessel is fitted with a stirrer and a sensitive thermometer. The arrangement is shown in fig.

The temperature of the water is noted and the substance is ignited by an electric current. After combustion the rise in temperature of the system is noted. The heat of combustion can be calculated from the heat gained by water and calorimeter.

Since the reaction in a bomb calorimeter proceeds at constant volume, the heat of combustion measured is ΔE

ΔE =  × M  kcal

Where M is the molecular mass of the substance, w₁ is the weight of substance taken, W is the water equivalent of calorimeter, m is the mass of liquid in the calorimeter and s is the specific heat of liquid.

ΔH can be calculated from the relation, ΔH = ΔE + ΔnRT

 

Examples based on Heat of reaction

Example ­1 :      The heat of formations of CO(g) and CO₂(g) are –26.4 kcal and –94.0 kcal respectively. The heat of combustion of carbon monoxide will be

(a) + 26.4 kcal    (b) – 67.6 kcal              (c) –120.6 kcal    (d) + 52.8 kcal

Solution(b) :     CO +   O₂ → CO₂ ; ΔH = ?

                             ΔH_CO2 = – 94.0 kcal, ΔH_CO = – 26.4 kcal

                             ΔH = ΔH_(CO2) – ΔH_(CO) = – 94.0 – (–26.4) = – 67.6 kcal

 

Example ­2 : Reaction H₂(g) + I₂(g) → 2HI(g) ; ΔH = – 12.40 kcal. According to this, the heat of formation of HI will be

(a) 12.4 kcal       (b) –12.4 kcal               (c) –6.20 kcal      (d) 6.20 kcal

Solution(c) :      H₂(g) + I₂(g) → 2HI(g)

                                For 2M, ΔH = – 12.40 kcal

                            1M, = – 6.20 kcal

 

Example ­3 :      C_(diamond) + O₂(g) → CO­₂(g) ;  ΔH = –395 kJ

                              C_(graphite) + O₂ (g) → CO₂(g) ; ΔH = – 393.5 kJ

From the data, the ΔH when diamond is formed from graphite is

(a) –1.5kJ           (b) + 1.5kJ                     (c) +3.0kJ            (d) –3.0kJ

Solution (b) :    C_(graphite) → C_(diamond)

                             ΔH­_(diamond) = – 395 kJ ; ΔH_(graphite) = – 393.5 kJ

                             ΔH = ΔH_graphite – ΔH_diamond = – 393.5 (–395.0)        

                             ΔH = + 1.5 kJ

 

Example ­4 :      The enthalpy of combustion of benzene from the following data will be

(a) + 3172.8 kJ  (b) –1549.2 kJ             (c) –3172.8 kJ    (d) –3264.6 kJ

                              (i) 6C(s) + 3H₂(g) → C₆H₆(l) ; ΔH = + 45.9 kJ

                              (ii) H₂(g) +   O₂(g) → H₂O(l) ; ΔH = – 285.9 kJ

                              (iii) C(s) + O₂(g) → CO₂(g) ; ΔH = – 393.5 kJ

Solution (d) :    C₆H₆ +   O₂ → 6CO₂ + 3H₂O ; ΔH = ?

ΔH = ΔH_(product) – ΔH_(reactant) = [(6 × –393.5) + (3 × –285.9)] – (45.9) = 3264.6 kJ

 

Example ­5 :      The enthalpy of formation of H₂O(l) is – 285.77 kJ mol^–1 and enthalpy of neutralisation of strong acid and strong base is –56.07 kJ mol^–1, what is the enthalpy of formation of OH^–  ion

(a) + 229.70kJ   (b) –229.70kJ              (c) +226.70kJ     (d) –22.670kJ

Solution (b) :   H⁺(aq) + OH^–(aq) → H₂O(l) ; ΔH = – 56.07 kJ

                             ΔH = ΔH_f(H₂O) – [ΔH_f(H⁺) + ΔH_f(OH)^–][ ΔH_f(H⁺) = 0]

                             – 56.07 = – 285.77 – (0 + x)

                             x = – 285.77 + 56.07 = – 229.70 kJ

                            

Example ­6 :      The heat of combustion of glucose is given by C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O ; ΔH = – 2840 kJ. Which of the following energy is required for the production of .18 gm of glucose by the reverse reaction.

(a) 28.40 kJ        (b) 2.84 kJ                     (c) 5.68 kJ           (d) 56.8 kJ

Solution (b) :    ΔH_comb = – 2840 Kj for 180 g of C₆H₁₂O₆

                             ΔH_comb for 0.18 gm glucose =   × 0.18 = – 2.84 kJ

                             For the reverse reaction ΔH_comb of glucose = 2.84 kJ

 

Example ­7 :      One gram sample of NH₄NO₃ is decomposed in a bomb calorimeter. The temperature of the calorimeter increases by 6.12K. The heat capacity of the system is 1.23kJ/g/deg. What is the molar heat of decomposition for NH₄NO₃.        

(a)– 7.53 kJ/mole                           (b) – 398.1 kJ/mole    

(c) – 16.1 kJ/mole                           (d) – 602 kJ/mole

Solution  (a) :   Heat evolved = 1.23 × 6.12

                              Molar heat capacity = 1.23 × 6.12 = 7.5276

                             Molar heat of decomposition = – 7.53 kJ/mole

 

Example 8 :      If ΔH_f⁰ for H₂O₂ and H₂O are – 188kJ/mole and –286 kJ/mole What will be the enthalpy change of the reaction 2H₂O₂(l) → 2H₂O(l) + O₂(g)

(a) – 196 kJ/mole                            (b) 146 kJ/mole 

(c)– 494 kJ/mole                                       (d) – 98 kJ/mole

Solution  (a) :   H₂ + O₂ → H₂O₂ ; ΔH_f⁰ = – 188 kJ/mole

                             H₂ +  O₂ → H₂O ΔH_f⁰ = – 286 kJ/mole

                             ΔH = Δ H^o_(product) – Δ H^o_(Reactants)

                             = (2 × – 286) – (2 × –188) = – 572 + 376 = – 196

 

Example 9 :      The heat of combustion of carbon is –94 kcal at 1 atm pressure. The intrinsic energy of CO₂ is

(a) + 94 kcal       (b) –94 kcal                  (c) + 47 kcal        (d) –47 kcal

Solution  (b) :   C_(s) + O_2(g) → CO₂(g) ; ΔH = – 94 kcal

                             ΔH = ΔE + Δn_gRT ; ΔE = ?

                             Δn_g = 1 – 1 = 0 ; ΔH = ΔE ; ΔE = – 94 kcal

 

Example 10 :    When 1 mole of ice melts at 0°C and a constant pressure of 1 atm, 1440 cal of heat are absorbed by the system. The molar volume of ice and water are 0.0196 and 0.0180 L, respectively calculate

(a)1430.03 cal    (b) 1500.0 cal      (c) 1450.0 cal      (d) 1440.03 cal

Solution: (d)     q = 1440 kcal

                   Δ H = 1440 kcal (absorbed) given H₂O(s) ⇌ H₂O(l)

                    P Δ V = 1 × (0.0180 – 0.0196) = – 0.0016 L atm

                    (1L atm = 24.206 cal)

                    – 0016 L atm = – 0.039 kcal

                     ΔH = ΔE + PΔV ⇒ E = ΔH – PΔV  = 1440 –(–0.0039) = 1440.039 cal

 

Example 11 :    The difference between heats of reaction at constant pressure and constant volume for the reaction 2C₆H₆(l) + 15O₂(g) → 12CO₂(g) + 6H₂O(l) at 25^oC in kJ is                       

(a) –7.43              (b) + 3.72                       (c) – 3.72             (d) + 7.43

Solution (a) :    Δ n_g  = 12 – 15 = – 3

                              Q_p – q_v = Δn_gRT = – 3 ×    × 298 = – 7.43 kJ

 

Laws of thermochemistry

(1)     Levoisier and Laplace law : According to this law enthalpy of decomposition of a compound is numerically equal to the enthalpy of formation of that compound with opposite sign, For example,

          C(s) + O₂ → CO₂(g) ; ΔH = – 94.3 kcal ;         

          CO₂(g) → C(s) + O₂(g) ; ΔH = +94.3 kcal

(2)     Hess’s law (the law of constant heat summation) : This law was presented by Hess in 1840. According to this law “If a chemical reaction can be made to take place in a number of ways in one or in several steps, the total enthalpy change (total heat change) is always the same, i.e. the total enthalpy change is independent of intermediate steps involved in the change.” The enthalpy change of a chemical reaction depends on the initial and final stages only. Let a substance A be changed in three steps to D with enthalpy change from A to B, ΔH₁ calorie, from B to C, ΔH₂ calorie and from C to D, ΔH₃ calorie. Total enthalpy change from A to D will be equal to the sum of enthalpies involved in various steps,

                   Total enthalpy change ΔH_steps = ΔH₁ + ΔH₂ + ΔH₃

Now if D is directly converted into A, let the enthalpy change be ΔH_direct According to Hess’s law ΔH_steps + ΔH_direct = 0 i.e. ΔH_steps must be equal to ΔH_direct numerically but with opposite sign. In case it is not so, say ΔH_steps (which is negative) is more that ΔH_direct (which is positive), then in one cycle, some energy will be created which is not possible on the basis of first law of thermodynamics. Thus, ΔH_steps must be equal to ΔH_direct numerically.

(i)      Experimental verification of Hess’s law

(a)     Formation of carbon dioxide from carbon

First method : carbon is directly converted into

                    C(s) + O₂(g) = CO₂(g) ; ΔH = – 94.0 kcal        

Second method : Carbon is first converted into CO(g) and then CO(g) into CO₂(g), i.e. conversion has been carried in two steps,

                   C(s) +   O₂ = CO(g) ;   ΔH = – 26.0 kcal       

                   CO(g) +   O₂ = CO₂ (g) ;   ΔH = – 68.0 kcal

          Total enthalpy change C(s) to CO₂ (g) ΔH = – 94.0 kcal

 

(b)     Formation of ammonium chloride from ammonia and hydrochloric acid:

First method : NH₃(g) + HCl = NH₄Cl(g) ; ΔH = – 42.2 kcal

                             NH₄Cl(g) + aq= NH₄Cl (aq) ; ΔH = + 4.0 kcal

                             NH₃(g) + HCl(g) + aq = NH₄Cl (aq) ; ΔH = – 38.2 kcal 

Second method : NH₃(g) + aq = NH₃(aq) ; ΔH = – 8.4 kcal

                             HCl(g) + aq = HCl (aq) ; ΔH = – 17.3 kcal

                             NH₃(aq) + HCl (aq) = NH₄Cl (aq) ; ΔH = –12.3 kcal

                             NH₃(g) + HCl(g) + aq = NH₄Cl (aq) ; ΔH = – 38.0 kcal

­­­­­­

Conclusions

• The heat of formation of compounds is independent of the manner of its formation.
• The heat of reaction is independent of the time consumed in the process.
• The heat of reaction depends on the sum of enthalpies of products minus sum of the enthalpies of reactants.
• Thermochemical equations can be added, subtracted or multiplied like algebraic equations.

(ii)    Applications of Hess’s law

(a)     For the determination of enthalpies of formation of those compounds which cannot be prepared directly from the elements easily using enthalpies of combustion of compounds.

(b)     For the determination of enthalpies of extremely slow reactions.

(c)      For the determination of enthalpies of transformation of one allotropic form into another.

(d)     For the determination of bond energies.

ΔH_reaction = ∑Bond energies of reactants – ∑Bond energies of products.

(e)     For the determination of resonance energy.

(f)      For the determination of lattice energy.

         

Bond energy or Bond enthalpies

When a bond is formed between atoms, energy is released. Obviously same amount of energy will be required to break the bond. The energy required to break the bond is termed bond dissociation energy. The more precise definition is,

“The amount of energy required to break one mole of bond of a particular type between the atoms in the gaseous state, i.e., to separate the atoms in the gaseous state under 1 atmospheric pressure and the specified temperature is called bond dissociation energy.”

For example, H – H(g) → 2H(g) ;                ΔH = + 433 kJ mol^–1

                        Cl – Cl(g) → 2Cl(g) ;               ΔH = + 242.5 kJ mol^–1 

                        H – Cl(g) → H(g) + Cl(g) ;      ΔH = + 431 kJ mol^–1

                         I – I(g) → 2I(g) ;                     ΔH = + 15.1 kJ mol^–1

                        H – I(g) → H (g) + I(g) ;          ΔH = + 299 kJ mol^–1

The bond dissociation energy of a diatomic molecule is also called bond energy. However, the bond dissociation energy depends upon the nature of bond and also the molecule in which the bond is present. When a molecule of a compound contains more than one bond of the same kind, the average value of the dissociation energies of a given bond is taken. This average bond dissociation energy required to break each bond in a compound is called bond energy.

Bond energy is also called, the heat of formation of the bond from gaseous atoms constituting the bond with reverse sign.

                   H(g) + Cl(g) → H – Cl(g) ; ΔH = – 431 kJ mol^–1

Bond energy of H – Cl = –(enthalpy of formation)

                   = – (–431) = + 431 kJ mol^–1

Consider the dissociation of water molecule which consists of two  bonds. The dissociation occurs in two stages.

                   H₂O(g) → H(g) + OH(g) ;      ΔH = 497.86 kJ mol^–1

                   OH(g) → H(g) + O(g) ;          ΔH = 428.5 kJ mol^–1

The average of these two bond dissociation energies gives the value of bond energy of O – H

                  Bond energy of O – H bond  =  = 463.15 kJ mol^–1

Similarly, the bond energy of N – H bond in NH₃ is equal to one – third of the energy of dissociation of NH₃ and those of C–H bond in CH₄ is equal to one – fourth of the energy of dissociation of CH₄.

          Bond energy of C – H =   =  416 kJ mol^–1

                                [CH₄(g) → C(g) + 4H(g) ;      ΔH = 1664 kJ mol^–1]

 

Applications of bond energy  

(1)     Heat of a reaction = ∑Bond energy of reactants – ∑ Bond energy of products.

Note : 

• In case of atomic species, bond energy is replaced by heat of atomization.
• Order of bond energy in halogen Cl > Br > F₂ > I₂

(2)     Determination of resonance energy : When a compound shows resonance, there is considerable difference between the heat of formation as calculated from bond energies and that determined experimentally.

Resonance energy = Experimental or actual heat of formation ~ Calculated heat of formation.

       

Examples based on Bond energy

Example ­12 :    If enthalpies of methane and ethane are respectively 320 and 360 calories then the bond energy of C – C bond is        

(a) 80 calories    (b) 40 calories              (c) 60 calories     (d)– 120 calories

Solution(d) :     CH₄ → 4C – H ; ΔH = 320

                            1E_C-H =   = 80 calories then 6E_C–H = 480 cal.

                             C₂H₆ → E_C–C + 6E_C–H ; ΔH = 360 cal.

                             360 = E_C–C + 480

                             E_C–C = 360 – 480 = – 120 cal

 

Example ­13 :    Given that C(g) + 4H(g) → CH₄(g) ;  Δ H = – 166 kJ. The bond energy C – H will be

(a) 208 kJ/mol                      (b) – 41.5 kJ/mol        

(c) 832 kJ/mol                       (d) None of these

Solution (b) :    C_(g) + 4H(g) → CH₄(g) ; ΔH = – 166 kJ

                             Bond energy for C – H = –  = – 41.5 kJ mole^–1

 

Example 14 :    Calculate resonance energy of N₂O from the following data. Observed ΔH^o_f (N₂O) = 82 kJ mol^–1

          B.E of N ≡ N ⇒ 946 kJ mol^–1 ; B.E of N = N 418 kJ mol^–1

          B.E of O = O ⇒ 498 kJ mol^–1; B.E of N = 0 607 kJ mol^–1

(a) – 88 kJ mol^–1                                         (b) – 44 kJ mol^–1

(c) – 22 kJ mol^–1                                  (d) None of these

Solution (a) :    N₂(g) + O₂(g) → N₂O

                             N ≡ N + O = O → N = N = O

Calculated ΔH^o_f (N₂O) = [B.E._(N≡N) + B.E (O = O)] – [B.E_(N=N) + B.E_N=O]

                             = – [418 + 607] = + 170 kJ/mole

                              Resonance energy = observed ΔH^o_f – calculated ΔH^o_f

                             = 82 – 170 = – 88kJ mol^–1

 

Example 15 :    If  at 298K the bond energies of C – H, C –C, C = C and H – H bonds are respectively 414, 347, 615 and 435 kJ mol^–1, the value of enthalpy change for the reaction H₂C = CH₂(g) + H₂(g) → H₃C – CH₃(g) at 298 K will be

(a) + 250 kJ        (b) –250 kJ                   (c) + 125 kJ        (d) – 125 kJ

Solution (d) :    CH₂ = CH₂(g) + H₂(g) → H₃C – CH₃(g)

                             4E_C–H ⇒ 414 × 4 =1656 6E_C–H ⇒ 414 × 6 = 2484

                             1E_C=C ⇒ 615 × 1 = 615        1E_C–C ⇒ 347 × 1 = 347

                            1E_H-H ⇒ 435 × 1 = 435

                            4ΔH_C–H + ΔH_C–C + ΔH_H–H = 2706 → 6ΔH_C–H + 1ΔH_C–C = 2831

                             ΔH = 2706 – 2831 = – 125 kJ

                   

Example ­16 :    The bond dissociation energies of gaseous H₂, Cl₂ and HCl are 104, 58 and 103 kcal respectively. The enthalpy of formation of HCl gas would be

(a) – 44 kcal       (b) 44 kcal                     (c) –22 kcal         (d) 22 kcal

Solution (c) :      H₂ + Cl₂ → HCl

                             ΔH = ∑B.E._(Reactants) – ∑B.E._(Products)

                             ΔH = [B.E.(H₂) + BE(Cl₂)] – B.E.(HCl)

                             =   (104) +   (58) – 103 = 81 – 103 = – 22 kcal

                             ΔH = – 22 kcal.

 

Example ­17 :    Given the bond energies N ≡ N, H – H and N – H  bonds are 945, 436 and 391 kJ mole^–1 respectively the enthalpy of the following reaction N₂(g) + 3H₂(g) → 2NH₃(g) is  

(a) – 93 kJ           (b) 102 kJ                     (c) 90 kJ              (d) 105 kJ

Solution (a) :   

Net energy (enthalpy) = BE_reactants – BE_products = 2253 – 2346 = – 93

                   ΔH = – 93 kJ