 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

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