Miscellaneous Exercises - I

Exercise 1: State, whether the following statements are True or False?

(i) ΔHfusion = ΔHsub – ΔHvap 

(ii) Bond formation is always exothermic

(iii) Enthalpy of neutralisation of NH4OH with HCl is higher than enthalpy of NaOH with HCl.

(iv) Heat of reaction is independent of temperature.

(v) Heat of combustion of a fuel = caloric value of fuel

Exercise 2: A solution of 5 gm of Haemoglobin (molecular weight = 64000) in 100 cc of solution shows a temperature raise of for complete oxygenation. Each mole of haemoglobin binds 4 mole of oxygen. If the heat capacity of the solution is calculate per gm mole of oxygen bond.

Exercise 3: At 25oC, the enthalpy change for the reaction

H2SO4 + 5H2O = H2SO4.5H2O (all liquids) is -580.32 kJ / mole. Calculate the temperature change if 1 mole of is dropped into 5 mole of  Assume no heat loss to the surroundings and that the specific heat capacity of solution is 4.184 Jk–1 g–1 

Exercise 4: The 1073_equation.JPG for the neutralisation reaction


                        Compute C2H6(q) for the reaction CO2

Exercise 5: The.2091_equation.JPG The first and second ionization energies of Ca are 590 and 1145 kJ / mole. The enthalpy of formation of Ca is 178 kJ / mol. The bond enthalpy of is 193 kJ / mole and enthalpy of vapourisation of is 31 kJ / mole. The electron affinity of Br(g) is 325 kJ / mole. Calculate the lattice energy of CaBr2(s).

Exercise 6: The heats of combustion of yellow phosphorus and red phosphorous are –9.19 kJ and – 8.78 kJ respectively, and then calculate the heat of transition of yellow phosphorus to red phosphorous. 

Exercise 7: Calculate the enthalpy change when infinitely dilute solutions of CaCl2 and Na2CO3 mixed  for Ca2+ (aq), (aq) and CaCO3(s) are – 129.80, – 161.65, – 288.5 kcal mole–1respectively.

Exercise 8: Calculate the heat of formation of ethyl acetate from ethyl alcohol and acetic acid. Given that heat of combustion of ethyl alcohol is 34 kcal and of acetic acid, it is 21 kcal and of ethyl acetate, it is 55.4 kcal.

Exercise 9: For the given heat of reaction,

                        (i)  C(s) + O2(g) = CO2(g) + 97 kcal 

                        (ii) CO2(g) + C(s) = 2CO(g) – 39 kcal

Calculate the heat of combustion of CO(g). 

Exercise 10: 1 mole of an ideal gas undergoes reversible isothermal expansion from an initial volume V1 to a final volume 10V1 and does 10 KJ of work. The initial pressure was 1 × 107 Pa.

                        (i) Calculate V1

                        (i) If there were 2 mole of gas, what must its temperature have been?

Answer to Miscellaneous Exercises

Exercise 1:       (i) True   

                      (ii) True 

                      (iii) False 

                      (iv) False 

                      (v) False 

Exercise 2:         -41.47 kJ

Exercise 3:         73.7oC

Exercise 4:        14.0 kJ / mol

Exercise 5:         -2162 kJ / mole                                       

Exercise 6:   (i)    P4 (yellow) +5O2(g) ——> P4O10 + 9.19 kJ

                        (ii)  P4(red) + 5O2(g) ——> P4O10 + 8.78 kJ


                                subtracting, P4(yellow) – P4 (red) = 1.13 kJ

                              => P4(yellow) = P4(red) + 1.13 kJ

So, heat of transition of yellow to red phosphorus is – 1.13 kJ 

Exercise 7:         On mixing CaCl2 (aq) and Na2CO3 

                        CaCl2 + Na2CO3 ——> CaCO3¯ + 2NaCl

Solutions are very dilute and thus 100% dissociation occurs 

Ca2+(aq)+2Cl-(aq)+2Na+ (aq) + (aq) —> CaCO3¯ + 2Na+(aq) +2Cl-(aq) or Ca2+(aq)+ + CO32– (aq) —> CaCO3(s) 

                        ΔH = ΣH°products  – ΣH°reactants

                        or ΔH = 1456_dissociation.JPG

                        ΔH° of a compound = ΔH° formation  = –288.5 – (–129.8 – 161.65)

                        = 2.95 kcal/mole

Exercise 8:         -400 cals

Exercise 9:         Subtracting equation (ii) from equation (i), we get

                        C(s) + O2(g) = CO2(g) + 97 kcal

                        CO2(g) + C(s) = 2CO(g) – 39 kcal

                         or, –CO2(g) + O2(g) = CO2(g) – 2CO(g) + 136 kcal

                        or, 2CO(g) + O2 = 2CO2(g) + 136 kcal

                        or, CO(g) + 1/2 O2(g) = CO2(g) + 68 kcal

                        Required value = 68 kcal

Exercise 10: (i) 4.34 × 10–4 m3 

                   (ii) 261.13 K

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