Numerical related Hess’s Law
Numerical related Hess’s Law
Q 1. Calculate the heat of formation of napthalene from the following data i. C(S) + O2(g) ? CO2(g) H = – 94.405 Kcal ii. H2(g) + 1/2 O2(g) ? H2O (l) H = – 68.3 Kcal iii. GoH8(S) + 12O2(g) ? 10CO2(g) + GH2O (l) H = – 94.405 Kcal Napthalene Sol: The required r × n is, 10C(S) + 4H2 (g) GoH8(s) The required r × n is obtained by multiplying r × n (1) by 10m r × n (ii) by 4, reversing r × n (iii) and then adding, we get, 10 C(S) + 10O2 (g) 10CO2 (g) H = – 944.05 Kcal 4H2 (g) + 2 O2 (g) 4H2O (l) H = – 273.2 Kcal 10 CO2 (g) + 4H2O (l) GoH 8 (S) + 12O2 (g) H = + 1231.6 Kcal 10 C(S) + 4 H2 (g) GoH8(S) H = 14.35 Kcal Heat of formation of Naphalene is 14.35 Kcal mol-1 Alternate solution: Considering combustion of Napthalence, we have, H reaction = H formation products – H formation preactants. -1231.6 = (10 × H formation CO2 + 4 × H formation O2) -1231.6 = {10 × (-94.405) + 4 × (-68.3)} – H formation of GoH8-O -1231.6 = -944.05 + (-273.2) – H formation of GoH8 -1231.6 + 944.05 + 273.2 = – H formation of GoH8 -14.35 = – H formation of GoH8 H formation of GoH8 = 14.35 Kcal Jxole-1 Q 2. Calculate the standard heat of formation of CH4 (g) from the following informations. i. CH4 (g) + 2O2 (g) ? CO2 (g)+ 2H2O (l) H = – 890.3 KJ ii. C(s) + O2 (g) ? CO 2 (g) H = – 939.5 KJ iii. 2H2 (g) + O2 (g) ? 2H2O (l) H = – 571.7 KJ The required eqn is, C(s) + 2H2 (g) ? CH4 (g) The required eqn is obtained by reversing eqn (i) and adding with others. CO2 (g) + 2H2O (l) ? CH4 (g) + 2O2 (g) H = – 890.3 Kj C(s) + O2 (g) ? CO 2 (g) H = – 939.5 KJ 2H2 (g) + O2 (g) ? 2H2O (l) H = – 571.7 KJ C(s) + 2H2 (g) ? CH4 (g) H = -74.9 KJ The heat of formation of CH4 (g) is -74.9KJ mol-1 Q 3. Calculate the enthalpy of formation of ethane at 298k, if the enthalpies of combustion at CH2 and C2 H6 are -94.14, -68.47 and -373.3 Kcal respectively. Sol: Given thermochemical reactions are i. C(s) + O2 (g) ? CO 2 (g) H = – 94.14Kcal ii. H2 (g) + 1/2O2 (g) ? H2O (l) H = – 68.47 Kcal iii. C2H6 (g) + 7/2O2 (g) ? 2CO2 (g) + 3H2O (l) H = – 373.3 Kcal The required reaction is 2C(s) + 3H2 (g) ? C2H6 (g) The required r×n is obtained by r × n (i) by 2, (ii) by 3 and reversing (iii) and then adding, we get, 3C(s) + 2O2 (g) ? 2CO 2 (g) H = – 188.28Kcal 3H2 (g) + 3/2O2 (g) ? 3H2O (l) H = – 205.41 Kcal 2CO2 (g) + 3H2O (l) ? C2H6 (g) + 7/2O2 (g) H = – 373.3 Kcal 2 C(s) + 3H2 (g) ? C2H6 (g) H = -20.39 Kcal Enthalpy of formation of ethane at 298k is -20.39 kcal/mol Q 4. Standard enthalpy of formation of H2O (l), CO2 (g) and C6H6 (l) are -286, -393.5 and +49.02 KJ mol-1 respectively at 298K. Calculate the standard enthalpy of combustion of C6H6 (l) at the given temperature. Sol: The given thermochemical reactions are, i. H2 (g) + 1/2O2 (g) ? H2O (l) H = -286 KJ ii. C(s) + O2 (g) ? CO 2 (g) H = – 393.5 KJ iii. 6C(s) + 3H2 (g) ? C6H6 (l) H = + 49.02 KJ Required reaction is C6H6 (l) + 15/2 O2 (g) ? 6CO2 (g) + 3H2O (l) H =? The required reaction is obtained by multiplying (i) by 3, (ii) by 6 and reversing (iii) and adding, we get, 3H2 (g) + 3/2O2 (g) ? 3H2O (l) H = -858 KJ 6C(s) + 6O2 (g) ? 6CO 2 (g) H = – 2361 KJ C6H6 (l) ? 6C(s) + 3H2 (g) H = – 49.02 KJ C6H6 (l) + 15/2 O2 (g) ? 6CO2 (g) + 3H2O (l) H = -3265.02KJ Standard enthalpy of combustion of C6H6 (l) is -3268.02 KJ Q 5. Calculate the heat of combustion of glucose from the following data, C(s) + O2 (g) ? CO2 (g) H = – 395 KJ H2 (g) + 1/2O2 (g) ? H2O (l) H = -269 KJ 6C(s) + 6H2 (g) + 3O2 (g) ? C6H12O6(s) H = -1269 KJ Required reaction is 6C (s) + 6H2 (g) + 3O2 (g) ? C6 H 12 O6(s) H = 1169 KJ The required reaction is 6O2 (g) + C6 H 12 O6(s) ? 6H2 O (g) + 6C2 The required reaction is obtained by multiplying (i) by 6, (ii) by 6 reversing (iii) and adding, we get, 6C(s) + 6O2 (g) ? 6CO2 (g) H = – 2370 KJ 6H2 (g) + 3O2 (g) ? 6H2O (l) H = -1614 KJ C6H12O6(s) + O2 (g) ? 6C(s) + 6H2 (g) + 3O2 (g) H = 1169 KJ 6H12O6 + 6O2 ? 6CO2 + 6H2O H = 2815 KJ