Solved Problems 1 

Subjective: 

Prob 1. Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero?    

Sol. A substance has a perfectly ordered arrangement of its constituent particles only at absolute zero. Hence entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in the formation of one mole of the substance from its elements. An element formed from itself means no heat change.  

Prob 2. Out of carbon (diamond) and carbon (graphite), whose enthalpy of formation is taken as zero and why? 

Sol. The enthalpy of formation of graphite is taken as zero because it is a more commonly found stable form of carbon.      

Prob 3. Justify, an exothermic reaction is always thermodynamically spontaneous. 

Sol. Exothermic reactions are generally thermodynamically spontaneous because even if it is accompanied by decrease of randomness, the heat released is absorbed by the surroundings so that the entropy of the surroundings increases to such an extent that  is positive.      

Prob 4. Is q_p always greater than q_v? Explain why or why not? 

Sol. q_p is not greater than q_v always. It depends upon whether Δn_g is +ve or -ve.    

Prob 5. Justify, many thermodynamically feasible reactions do not occur under ordinary conditions.    

Sol. Under ordinary conditions, the average energy of the reactants may be less than threshold energy. They require some activation energy to initiate the reaction.    

IIT Level Questions 

Prob 6. The standard heats of formation at 298 K for CCl_­4 (g), H₂O (g), CO₂ (g) and HCl (g) are-25.5, -57.8, -94.1 and -22.1 Kcal/mole. Calculate the ΔH^o₂₀₉ for the reaction.    CCl₄ (g) + 2H₂O(g) ———> CO₂(g) + 4HCl (g)   ΔH = ?  

Sol.     ΔH = [ΔHCO₂ + 4 × ΔGHCl] – [ΔHCCl₁ + 2 × ΔH₂O]  

               = [–94.1 + 4 ×22.1] – [–25.5 + 2 × – 57.8] => +135.4 Kcal    

Prob 7. The molar heats of combustion of C₂H₂ (g), C(graphite) and H₂ (g) are 310.62 Kcal, 94.05 Kcal and 68.32 Kcal respectively. Calculate the standard heat of formation of C₂H₂ (g).  

Sol. The required equation is  

               2C + H₂ ——> C₂H₅; ΔH = ?

Writing the thermochemical equation of the given data  

               (i) C₂H₂ + 5/2 O₂ ——> 2CO₂ + H₂O      ΔH = –310.62 kcal 

               (ii) C + O₂ ——> CO₂     ΔH = –94.05 kcal

               (iii) H₂ + 1/2 O₂ ——> H₂O     ΔH = –68.32 kcal 

               (iii) + 2 × (ii) - (i)  

               2C + H₂ —— C₂H₂         ΔH = 54.20 kcal 

Prob 8. The molar heat of formation of NH₄NO₃ (s) is -367.54 KJ and those of N₂O (g) and H₂O are 81.46 and -285.8 KJ respectively at 25°C and 1 atmospheric pressure. Calculate ΔH andΔE of reaction.   NH₄NO₃(s) —— N₂O(g) + 2H₂O (l)  

Prob 9. Determine the value of ΔH and ΔE for the reversible isothermal evaporation of 90.0 gm of water at 100°C. Assume that water vapour behaves as an ideal gas and heat of evaporation of water is 540 cal/gm.  

Sol.        ΔH = 90 × 540 = 48.6 Kcal  

               ΔH = ΔE + PΔV  

               Volume of liquid is negligible as compared to volume of vapour  

               So ΔV = V_vapour  

               ΔH = ΔE + nRT  

               ΔE = 48600 – 90/18 × 2 × 373 = 44.87 kcal  

Prob 10. An athelete is given 100 gm glucose of energy equivalent to 15600 kJ. He utilizes 50% of this gained energy in an event. In order to avoid storage of energy in body, Calculate the wt of water he would need to prespire. Enthalpy of H₂O for evaporation is 44 kJ / mole.  

Sol.        Energy gained by athelete = 1560 kJ  

               Energy utilized in event = 50 / 100 × 1560 = 780 kJ  

               Energy left = 1560 – 780 = 780 kJ  

               Since 44 kJ energy used to evaporate = 18 gm H₂O  

               780 kJ energy used to evaporate = 18 × 780 / 44 = 319.09 of H₂O  

Prob 11. The specific heat at constant volume for a gas 0.075 cal / g and at constant pressure is 0.125 cal / gm. Calculate:  

               (i)   The molecular weight of the gas  

               (ii) Atomicity of gas  

               (iii) Number of atoms in its 1 mole  

Sol.          (i)   Specific heat at constant pressure C_p = 0.125 cal / gm  

                     Specific heat at constant volume, C_v = 0.125 cal / gm  

                     As we know, C_p – C_v = R/M  

                     or, M = R/C_p – C_v = 2/(0.125 – 0.075) = 40  

               (ii) For atomicity,  

                    Hence gas is mono atomic.Y = C_p / C_v = 0.125 / 0.075 = 1.66 

                 (iii) Since gas is mono atomic. Hence 1 mole of gas contains = 6.023 × 10²³ atoms  

Prob 12. When 1 mole of ice at and 4.6 mm of Hg is converted to water vapours at a constant temperature and pressure. Find ΔH and ΔE, if the latent heat of fusion of ice is 80 cal / gm and latent heat of vaporisation of liquid water at is 596 cal / gm. The volume of ice in comparison to volume of vapours may be neglected.  

Sol.        Ice ——> vapour  

               ΔH = ΔH_f + ΔH_v 

                   = 80 × 18 + 596 × 18  

                   = 12168 cal / mole  

               ΔH = ΔE + PΔV  

               ΔV = Volume of vapours at 4.6 mm and 0^oC (as V_ice = neglible) 

               Now applying PV = nRT  

               PΔV = nRT = 1 × 8.314 × 273 / .18 cal = 543 cal  

               ΔE = ΔH – PΔV  

                     = 12168 – 543  

                     = 11625 cal   

Prob 13. Calculate the standard enthalpy of reaction  

               ZnO(s) + CO(g) ———> Zn(s) + CO₂(g)  

               Given   

               ΔH^o_f (CO, g) = –110.53 kJmol^–1 

Sol.        We have   

               =  

Prob 14. Calculate the enthalpy of vaporization for water from the following  

               H₂(g) + 1/2 O₂(g) ——> H₂O (g)           ΔH = – 57.0 kcal  

               H₂(g) + 1/2 O₂(g) ——> H₂O(l)             ΔH = – 68.3 kcal  

               Also calculate the heat required to change 1 gm H₂O (l) to H₂O (g).  

Sol.        H₂(g) + 1/2 O₂(g) ——> H₂O (g);    ΔH = –57.0 kcal ----------------- (1)  

               H₂(g) + 1/2 O₂(g) ——> H₂O (l);     ΔH = – 68.3 kcal------------------(2)  

               Subtracting (2) from (1)  

               H₂O (l) ¾® H₂O (g) ;                    ΔH = 11.3 kcal                

               Enthalpy of vaporization for H₂O = 11.3 kcal  

               Also 18 g H₂O requires enthalpy of vaporization = 11. 3 kcal  

               1 g H₂O requires 11.3/18 kcal = 0.628 kcal  

Prob 15. The standard enthalpy of combustion of H₂, C₆H₁₀ and Cyclohexane (C₆H₁₂) are – 241,   – 3800, – 3920 kJ mole^–1 at 25°C respectively. Calculate the heat of hydrogenation of cyclohexene.  

Sol.       We have to find Δ H for  

               C₆H₁₀ + H₂ ——> C₆H₁₂  

               Given H₂ + 1/2 O₂ ——> H₂O                        ΔH = – 241 kJ --------------- (1)  

               C₆H₁₀ + 17/2 O₂ ——> 6CO₂ + 5H₂O             ΔH = – 3800 kJ ---------------(2)  

               C₆H₁₂ + 9O₂ ——> 6CO₂ + 6H₂O                   ΔH = – 3920 kJ ---------------(3)  

               Adding equation (1) and (2) and then subtracting equation 3  

               C₆H₁₀ + H₂ ——> C₆H₁₂                              ΔH = – 121 kJ  

               Heat of hydrogenation of cylohexene = – 121 kJ