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Question-1

State and explain Hess’s law of constant heat summation by taking a suitable example.

Solution:
The total heat change (ΔH) accompanying a chemical reaction is the same whether the reaction takes place in one step or more steps"

In means that the heat of a reaction depends only on the initial and final states of the system and independent of the path follwed by the system.

Carbon can be converted into carbon dioxide either directly or via the formation of carbon monoxide. Formation of CO2 directly.

C(S) + O2(g) CO2(g); ΔH = -94 Kcal

Formation of CO2 via formation of carbon dioxide.

C(s) + ½ O2(g) CO(g) ; ΔH1 = -26. 4 Kcal

CO(g) + ½ O2(g) CO2(g) ; ΔH2 = 67.6 Kcal


According to Hess's Law : ΔH = ΔH1 + ΔH2
                                        -94 = -26.4 + (-67.6)
                                        -94 = -94
.

Question-2

What is heat of enthalpy for the transition of S rhombic to S monoclinic (heat of combustion of S (rhombic) and S monoclinic are –71.1 kcal and –71.7 k cal respectively)?

Solution:
S(rhombic) + O2 (g) SO2(g) ; Δ H = -71.1 kcal                           ---- 1

S(monoclinic) + O2(g) SO2(g) ; Δ H = -71.7 kcal                        ---- 2
                                             Reverse equation (2),

SO2 S(monoclinic) + O2(g); Δ H = +71.7 kcal                           ---- 3

On adding 1 and 3 we get

S(rhombic) + O2(g)+ SO2(g) SO2(g) + S(monoclinic) + O2(g)

S(rhombic) S(monoclinic) + 0.6 kcal (+71.7 + (-71.1) kcal; (Δ H = +0.6kcal).

Question-3

Derive the mathematical formulae of first law of thermodynamics.

Solution:
Let q be the heat supplied to the system

A part of q may be used up by the system itself in increasing its internal energy by Δ E and rest is used for performing external work w.

As per first law, q = ΔE + w
                        
ΔE = q w

If work done is the pressure - volume, then w = p.dv where dv is the change in volume P is the external pressure.

ΔE = q - P ΔV.

Question-4

Show that the heat absorbed at constant volume is equal to the increase in the internal energy of the system, whereas that at constant pressure is equal to the increase in the enthalpy of the system.

Solution:
As per mathematical form of first law of thermodynamics, ΔE = q - PΔV.

At constant volume,

ΔV = 0; (ie) ΔE = q - o = ΔE = q ; That is heat absorbed at constant volume is equal to the increase in the internal energy of the system.

From definition,

H = E + PV ; ΔH = ΔE + Δ(PV)
ΔH = ΔE + PΔV+ V. ΔP

At constant pressure,       Δp = o;
ΔH = ΔE + PΔV

substitution ΔH = qp - PΔ V
                 
ΔH = qp - PΔV + PΔV
                
ΔH = qp.

Question-5

What is the relation between enthalpy of reaction and enthalpies of formation?

Solution:
Enthalpy of reaction

ΔH = [Sum of the standard enthalpies of formation of products] - [sum of the standard enthalpies of formation of the reactants]

= Σ ΔH0f products - Σ ΔH0f reactants.

Question-6

Given the standard enthalpies of formation of CO(g) and H2O(g) as –110.5 kJ and –241.8 KJ mol-1 respectively. Calculate ΔH0 and ΔE0 for the reaction.

Solution:
C(s) + H2O(g) CO(g) + H2(g)
ΔH0 = Σ ΔH0f products - Σ ΔH0f reactants
        = - 110.5 - (-241.8)


ΔH0 = +131.3 ;
 
Δn = 2-1 = 1;
ΔH0 = ΔE0 + (Δn) RT
ΔE0 = ΔH0 - (Δn) RT
      = +131.3 - (1) (8.314
×3 k Jk-)(298 k)
      = +131.3 - 2.478 KJ
      = 128.82 kJ
.

Question-7

(a) What is the relationship between enthalpy of reaction and bond energies of reactants and products?

(b) Calculate the enthalpy change (
ΔH) of the following reaction:
2C2H2(g) + 5O2(g)
4CO2(g) + 2H2O(g) , given average bond energies of various bonds
(i.e.) C – H, C
C, O = O, O – H as 414, 810,499, 724 and 460 kJ mole-1 respectively.
One C2H2 contains two C – H bonds and one C
C bond
One O2 contains one O=O bonds
One CO2 contains, two C=C bonds
One H2O contains two O – H bonds.

Solution:
 

(a) Enthalpy of reaction = [Sum of bond energies of reactants] - [sum of bond energies products]

 

(b) Bond energy due to one C2H2 molecule

= (2 × C - H bond energy) + (1 × C C bond energy)

 

= (2 × 414) + (1 × 810) kJmol-1

B. E of one O2

= (1 × 0 = 0 bond energy) = 499 kJmol-1

B. E of one CO2

= (2 × 724 kJmol-1)

B. E of one H2O

= 1 × 460 kJ mole-1

 

ΔH = [Total energy required to break the bonds in reactants] - [Energy given out in forming the bonds in products]

ΔH =(4 ΔH C - H + 2ΔHC C + 5 ΔH O = 0) - [8 ΔC = 0 + 4 ΔH O-H]
     =(4
× 414 + 2 × 810 + 5 × 499) - (8 × 724 + 4 × 460)
     = 5771 - 7632 = -1861 KJ.

Question-8

Calculate the heat of reaction for the reduction of ferric oxide by aluminium (thermite reaction) at 250C. The heat of formation of Fe2O3 and Al2O3 are –197.3 and 400.5 kcal respectively.

Solution:
The thermite reaction is as below,
2Al + Fe2O3

2Fe + Al2O3
 

ΔH (in kcal)

 

2(0) - 197.3 2(O) - 400.5
ΔHf0 for any element in its standard state is zero.

ΔH0   = ΔHp - ΔHR
         = -400.5 - (-197.3)

ΔH0   = - 203.2kcal = -850kJ
(1kcal = 4.18 J).

Question-9

Calculate the difference between heats of reaction at constant pressure and constant volume for the reaction at 250C in kJ.

Solution:
2C6H6(l) + 15 O2(g) 12 CO2(g) + 6H2O(l)

Δn = 12 - 15 = -3;

ΔH = ΔE + ΔnRT; ΔH- Δ E = ΔnRT
                                   = (-)3
× 8.314 JK-1mole-1
                                   = (-3)
× 8.314 × 10-3 × 298 KJ mole-1
                                   = -7.443 kJ mole-1.

Question-10

The heat liberated on complete combustion of 7.8g benzene is 327kJ. This heat has been measured at constant volume and at 270C. Calculate the heat of combustion of benzene at constant pressure. (R = 8.3J mole-1 le-1).

Solution:
C6H6(R) + 71/2 O2 (g) 6 CO2(g) + 3H2O(l)

Heat liberated on complete combustion of 7.8 g or benzene = 327 kJ

Heat liberated on complete combustion of 78 g (1mol) of benzene = 327 kJ

Δn = 6-7 ½ = -1.5
   R = 8.3
× 10-3 kJ mole-1 K-1
   T = 300 K

ΔH = ΔE + ΔngRT;
      = -327kJ + (-1.5) (8.314
× 10-3) (300) kJ mole-1
                 -327 -3.7413 = 330.7413 KJ





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