# Hess’s law of Constant Heat Summation

 Science > > You are Here

#### Statement of Hess’s Law:

• It states that the change in enthalpy accompanying a chemical reaction is independent of the pathway between initial and final states of the chemical reaction.
• Explanation:
• Let us suppose substance ‘A’ is converted into D directly.

A   →   D,                ΔH =   Q kJ

• Suppose the same change is brought about in different steps,

i) A   →  B,               ΔH1     =   Q1 kJ

ii)  B → C,                ΔH2    =   Q2 kJ

iii)  C →D ,                ΔH3   =   Q3 kJ

• Representation : Then, according to Hess’s law,

ΔH = ΔH1 + ΔH2 + ΔH3

Q   =   Q1   +   Q2   + Q3

• Thus, Hess’s law implies that change in enthalpy of a chemical reaction depends upon the initial and final state of a chemical reaction irrespective of the number of steps involved in a chemical reaction.

#### Illustration of Hess’s Law:

• The change in enthalpy of formation of carbon dioxide can be determined in two ways.
• One step preparation :

Solid carbon is burnt in excess of oxygen.

C(s)+ O2(g)  →  CO2(g)   , Δ H  =  -395.39 kJ

• Preparation in two steps:

Step – 1: Solid carbon is burnt in limited supply of oxygen to form carbon monoxide,

C(s)+ 1/2O2(g)  →  CO(g)   , Δ H1 = -111.7 kJ

Step – 2: Carbon monoxide is then burnt in excess of oxygen to from carbon dioxide,

CO(g)+ 1/2O2(g)  → CO2(g , ΔH2=  -283.67 kJ

According to Hess’s law,

Δ H1 + ΔH2 =   -111.7 kJ  –  283.67 kJ  = -395.39 kJ   = ΔH

Thus Hess’s law is illustrated.

#### Applications of Hess’s Law:

• Thermochemical equations can be added subtracted or multiplied like ordinary algebraic equations.
• Hess’s law is useful to calculate heats of many reactions which do not take place directly.
• It is useful to find out heats of extremely slow reaction.
• It is useful to find out the heat of formation, neutralization etc.

Problems Based on Hess’s Law:

#### Example – 1:

• Calculate the enthalpy of formation of CO from given data

i) C(s)+ O2(g)  →   CO2(g)              ΔH° = -393.5 KJ

ii) CO(g)+ ½O2(g)  →   CO2(g)      ΔH° = – 283.0 KJ

• Solution:

The formation of catbon monoxide is represented by following thermochemical equation

C(s)+ ½O2(g)  →   CO(g)      ΔH° = ?

Keeping given equation (i) as it is and reversing equation (ii) we get

i) C(s)+ O2(g)  →   CO2(g)              ΔH1° = -393.5 KJ

iii) CO2(g)  →    CO(g)+ ½O2(g)    ΔH2° = + 283.0 KJ

Adding equations (i) and (iii) and by Hess’s Law we get

C(s)+ ½O2(g)  →   CO(g)      ΔH° = ΔH1° +  ΔH2° =  -393.5 KJ + 283.0 KJ

C(s)+ ½O2(g)  →   CO(g)      ΔH° =  -110.5 KJ

Hence enthalpy of formation of CO is – 110.5 kJ mol-1

#### Example – 2:

• Calculate the enthalpy of formation of CH4 from given data

i) C(s)+ O2(g)  →   CO2(g)                                   ΔH° = -393.5 KJ

ii) H2(g)+ ½O2(g)  →   H2O(l)                           ΔH° = – 285.8 KJ

iii) CH4(g)+ 2O2(g)  →   CO2(g)  +  2 H2O(l)    ΔH° = – 890.3 KJ

• Solution:

The formation of CH4 is represented by following thermochemical equation

C(s)+ 2H2(g)  →   CH4(g)      ΔH° = ?

Keeping given equation (i) as it is, multiplying equation (ii) by 2 and reversing equation (ii) we get

i) C(s)+ O2(g)  →   CO2(g)                                      ΔH1° = -393.5 KJ

iv) 2 H2(g)+  O2(g)  →   2 H2O(l)                           ΔH2° = – 571.6 KJ

v)   CO2(g)  +  2 H2O(l)   →  CH4(g)+ 2O2(g)     ΔH3° = + 890.3 KJ

Adding equations (i) (iv) and (v) and by Hess’s Law we get

C(s)+ 2H2(g)  →   CH4(g)    ΔH° = ΔH1° +  ΔH2° +  ΔH3° = -393.5 KJ – 571.6 KJ  + 890.3 kJ

C(s)+ 2H2(g)  →   CH4(g)     ΔH° =  – 74.8 KJ

Hence enthalpy of formation of CH4 is – 74.8 kJ mol-1

#### Example – 3:

• Calculate the  ΔH° for the reaction

2 ClF(g)+ O2(g)  →   Cl2O(g)    + OF2(g)

From following equations

i) F2(g)  +   ClF(g)  →   ClF3(l)                                       ΔH° = – 139.2 KJ

ii) 2 ClF3(l)  + 2 O2(g)  →   Cl2O(g)  + 3 OF2(g)         ΔH° = + 533.4 KJ

iii) F2(g)+½O2(g)  →   OF2(g)                                      ΔH° = + 24. 7  KJ

• Solution:

Required reaction is

2 ClF(g)+ O2(g)  →   Cl2O(g)    + OF2(g)

Multiplying equation (i) by 2, Keeping equation (ii) as it is,

and multiplying equation (iii) by 2 and reversing it.

iv) 2 F2(g)  +   2 ClF(g)  →   2 ClF3(l)                            ΔH° = – 278.4 KJ

ii) 2 ClF3(l)  + 2 O2(g)  →   Cl2O(g)  + 3 OF2(g)         ΔH° = + 533.4 KJ

v) 2 OF2(g)     →   2 F2(g)+    O2(g)                              ΔH° = – 49.4  KJ

Adding equations (iv) (ii) and (v) and by Hess’s Law we get

2 ClF(g)+ O2(g)  →   Cl2O(g)    + OF2(g)   ΔH° = ΔH1° +  ΔH2° +  ΔH3° = – 278.4 KJ  + 533.4 KJ  – 49.4 kJ

2 ClF(g)+ O2(g)  →   Cl2O(g)    + OF2(g)   ΔH° = + 205.6 KJ

Hence ΔH° for the reaction is 205.6 kJ

#### Example – 4:

• Calculate the  ΔH° for the reaction between ethene with water to form ethanol from the following data.

From following equations

i) C2H5OH(l)  +   3 O2(g)  →   2CO2(g)  + 3 H2O(l)     ΔH° = – 1368 KJ

ii) C2H4(g)  +   3 O2(g)  →   2CO2(g)  + 2 H2O(l)          ΔH° = – 1410 KJ

Is  ΔH° calculated the enthalpy of formation of liquid ethanol?

• Solution:

The reaction between ethene and water is represented by

C2H4(g)  +   H2O(l)   →   C2H5OH(l)         ΔH° = ?

Reversing equation (i), Keeping equation (ii) as it is

iii) 2CO2(g)  + 3 H2O(l)  → C2H5OH(l)  +   3 O2(g)    ΔH° = + 1368 KJ

ii) C2H4(g)  +   3 O2(g)  →   2CO2(g)  + 2 H2O(l)          ΔH° = – 1410 KJ

Adding equations (iii) and (ii) and by Hess’s Law we get

C2H4(g)  +   H2O(l)   →   C2H5OH(l)         ΔH° = ΔH1° +  ΔH2° = 1368 KJ  -1410 KJ

C2H4(g)  +   H2O(l)   →   C2H5OH(l)         ΔH° = – 42 kJ

Hence ΔH° for the reaction is – 42 kJ

• This can not be the heat of formation of ethanol, because it is not obtained from its constituent elements in their standard state.

#### Example – 5:

• Calculate the  ΔH° for the reaction       2C(graphite)  +   3 H2(g)  →   C2H6(g)

From following equations

i) C2H6(g)  +   7/2 O2(g)  →   2CO2(g)  +3 H2O(l)          ΔH° = – 1560 KJ

ii)  H2(g)  +  1/2 O2(g)  →   H2O(l)                                   ΔH° = – 285.8 KJ

ii)  C(graphite)  +  O2(g)  →   CO2(g)                                  ΔH° = – 393.5 KJ

• Solution:

The required reaction is

2C(graphite)  +   3 H2(g)  →   C2H6(g)          ΔH° =?

Reversing equation (i), Multiplying equation (ii) by 3 and multiplying equation (iii) by 2 we get

iv) 2CO2(g)  +3 H2O(l)   →   C2H6(g)  +   7/2 O2(g)          ΔH° = + 1560 KJ

v)  3H2(g)  + 3/2 O2(g)  →   3 H2O(l)                                   ΔH° = – 857.4 KJ

vi)  2C(graphite)  +  2O2(g)  →   2CO2(g)                               ΔH° = – 787.0 KJ

Adding equations (iv), (v) and (vi) and by Hess’s Law we get

2C(graphite)  +   3 H2(g)  →   C2H6(g) ,        ΔH° = ΔH1° +  ΔH2° +  ΔH3° = 1560 KJ  -857.4 KJ  – 787 kJ

2C(graphite)  +   3 H2(g)  →   C2H6(g)    ΔH° = – 84.4 kJ

Hence ΔH° for the reaction is – 84.4 kJ

#### Example – 6:

• Calculate the standard enthalpy for the reaction       2Fe(s)  +   3/2 O2(g)  →  Fe2O3(s)

From following equations

i) 2 Al(s)  +  Fe2O3(s)    →   2Fe(s)  + Al2O3(s)         ΔH° = – 847.6 KJ

ii) 2 Al(s)    +  3 /2 O2(g)  →   Al2O3(s)                       ΔH° = – 1670 KJ

• Solution:

The required reaction is

2Fe(s)  +   3/2 O2(g)  →  Fe2O3(s)           ΔH° =?

Reversing equation (i), keeping equation (ii) as it is we get

iii) 2Fe(s)  + Al2O3(s)     2 Al(s)  +  Fe2O3(s)          ΔH° =  847.6 KJ

ii) 2 Al(s)    +  3 /2 O2(g)  →   Al2O3(s)                       ΔH° = – 1670 KJ

Adding equations (iii) and (ii) we get

2Fe(s)  +   3/2 O2(g)  →  Fe2O3(s)      ΔH° = ΔH1° +  ΔH2° = 847.6 KJ  – 1670 KJ

2Fe(s)  +   3/2 O2(g)  →  Fe2O3(s) ,     ΔH° = – 822.4 kJ

Hence standard enthalpy of the reaction is – 822.4 kJ

#### Example – 7:

• Given the following equations and ΔH° at 25 °C

i)  Si(s)  + O2(g)     →    SiO2(s)                  ΔH° = – 911 KJ

ii) 2C(graphite)  +  O2(g)  →   2CO(g)         ΔH° = – 221 KJ

iii)   Si(s)    +  C(graphite)   →   SiC(s)           ΔH° = – 65.3 KJ

Calculate ΔH° for the reaction  SiO2(s)    + 3 C(graphite)   →   SiC(s)    +    2CO(g)

• Solution:

The required reaction is

SiO2(s)    + 3 C(graphite)   →   SiC(s)    +    2CO(g)         ΔH° =?

Reversing equation (i), keeping equations (ii) and (iii)  as it is we get

iv)  SiO2(s)       →    Si(s)  + O2(g)                   ΔH° =  911 KJ

ii) 2C(graphite)  +  O2(g)  →   2CO(g)         ΔH° = – 221 KJ

iii)   Si(s)    +  C(graphite)   →   SiC(s)           ΔH° = – 65.3 KJ

Adding equations (iv), (ii) and (iii) and by Hess’s Law we get

SiO2(s)    + 3 C(graphite)   →   SiC(s)    +    2CO(g)         ΔH°  = ΔH1° +  ΔH2° +  ΔH3° = 911 kJ  -221 kJ -65.3 kJ

SiO2(s)    + 3 C(graphite)   →   SiC(s)    +    2CO(g)         ΔH°  = + 624.7 kJ

HenceΔH°   for the reaction is  624.7 kJ

#### Example – 8:

• Given the following equations and ΔH° at 25 °C

i)  2H3BO3(aq)     →    B2O3(s)    +   3 H2O(l)       ΔH° = + 14.4 KJ

ii) H3BO3(aq)    →   HBO2(aq)    +  H2O(l)            ΔH° = – 0.02 KJ

iii)   H2B4O7(s)     →   2 B2O3(s)    +  H2O(l)          ΔH° = 17.3  KJ

Calculate ΔH° for the reaction H2B4O7(s)   + H2O(l)    →   4 HBO2(aq)

• Solution:

The required reaction is

H2B4O7(s)   + H2O(l)    →   4 HBO2(aq)       ΔH° =?

Reversing equation (i) and multiplying by 2, Multiplying equations (i) by4 keeping equation (iii)  as it is we get

iv)  2B2O3(s)    +   6 H2O(l)     →    4 H3BO3(aq)        ΔH° = – 28.8 KJ

v) 4  H3BO3(aq)    →   4 HBO2(aq)    +  4 H2O(l)        ΔH° = – 0.08 KJ

iii)   H2B4O7(s)     →   2 B2O3(s)    +  H2O(l)                 ΔH° = 17.3  KJ

Adding equations (iv), (v) and (iii) and by Hess’s Law we get

H2B4O7(s)   + H2O(l)    →   4 HBO2(aq)       ΔH° = ΔH1° +  ΔH2° +  ΔH3° = – 28.8 kJ  – 0.08 kJ + 17.3kJ

H2B4O7(s)   + H2O(l)    →   4 HBO2(aq)       ΔH° = – 11.58 kJ

HenceΔH°   for the reaction is  -11.58 kJ

 Science > > You are Here