HESS'S LAW

      This law was formulated by the Swiss-born Russian chemist Germain Henri Hess (1802-1850). Hess's Law states that, if a reaction occurs in two or more steps, the enthalpy for the reaction is the sum of the enthalpies of the individual steps or it can also be stated as, the enthalpy change for any chemical reaction is independent of the intermediate stages, provided the initial and final conditions are the same for each route.

Thermal Equations Hess’s law energy level diagram ·A + B = AB, dH1 ·AB + B = AB2, dH2  then, A + 2 B = AB2, dH1 2 = dH1 + dH2     ======= A + 2 B         |   | dH1 dH1 2 | ===== AB + B        |   | dH2      ======= AB2

If both sides of the equations are multiplied by a factor to alter the number of moles, dH, dH°, or dH_f for the equation should be multiplied by the same factor, since they are quantities per equation as written. Thus, for the equation

C(graphite) + 0.5 O₂ -> CO,       dH° = -110 kJ/mol. We can write it in any of the following forms: 2 C(graphite) + O₂ -> 2 CO,       dH° = -220 kJ/mol (multiplied by 2)
6 C(graphite) + 3 O₂ -> 6 CO,       dH° = -660 kJ/mol (multiplied by 6)

For the reverse reaction, the sign of these quantities are changed (multiply by -1). The equation implies the following:

CO -> C(graphite) + 0.5 O₂,       dH° = 110 kJ/mol
2 CO -> 2 C(graphite) + O₂,       dH° = 220 kJ/mol.

Example 1

We can use Hess's law  to calculate the enthalpies of reactions that are difficult to measure. In the above example, it is very difficult to control the oxidation of graphite to give pure CO. However, enthalpy for the oxidation of graphite to CO₂ can easily be measured.  The application of Hess's law enables us to estimate the enthalpy of formation of CO. Since,

C + O₂ -> CO₂,         dH° = -393 kJ/mol
CO + ¹/₂ O₂ -> CO₂,         dH° = -283 kJ/mol. 

We should reverse the second reaction to cancel out CO₂ but remember that by doing so you should change the sign of its enthalpy.

            C + O₂ -> CO₂,         dH° = -393 kJ/mol
         CO₂ -> CO + ¹/₂ O₂      dH° = 283 kJ/mol. 


Subtracting the second equation from the first gives



C + ¹/₂ O₂ -> CO,         dH° = -393 + 283 = -110 kJ/mol* note that since oxygen gas is present on the different side of the two equation we should subtract                               O₂ - ¹/₂ O_{2   =} ¹/₂ O₂     

The equation shows the standard enthalpy of formation of CO to be -110 kJ/mol.

Example 2.
          Consider another example using the following reactions:
      

     (1) C_(s) + O_{2(g)    --->}      CO_2(g)                           dH= -393.51 kJ

     (2) H_2(g)  + ½ O_{2(g)  ----->}   H₂O_(l)                  dH= -285.83 kJ

     (3) CH_4(g) + 2 O_{2(g)  ----->} CO_2(g) + 2 H₂O_(l)     dH= -890.37 kJ

             

         C_(s) + O_{2(g)  ----->}    CO_2(g)                             dH= -393.51 kJ

      2H_2(g)  + O_2(g)  ----->  2 H₂O_(l)                     dH= -571.66 kJ

        CO_2(g) + 2 H₂O_(l)  ----->   CH_4(g) + 2 O_2(g)    dH= 890.37                                        Sum: C_(s) + 2H_{2(g) ------>}  CH_4(g)                        dH = -74.80 kJ                   

Example 3

    Determine the dH for the reaction

        2 N₂O_3(g) -----> 2 N_2(g) + 3 O_2(g)

     

      from the equations:

    

           (1)   N_2­O_3(g) ----> NO_(g) + NO_2(g)      dH = 39.7 kJ

(          (2)   ½ N_­2(g) + ½ O_2(g) ----> NO_(g)      dH = 90.4 kJ

           (3)   ½ N_2(g) + O_2(g) ----> NO_2(g)          dH = 33.8 kJ

 Answer: 

                         2 N_2­O_3(g) ----> 2NO_(g) + 2NO_2(g)      dH = 79.4 kJ

         

          2 NO_{(g) ----->}  N_­2(g) +  O_2(g)               dH= -180.8 kJ

         2 NO_{2(g) ----->}  N_2(g) + 2 O_2(g)        dH = -67.9 kJ

  SUM: 2 N_2­O_3(g) ----> 2 N_­2(g) +  3 O_2(g)   dH= -169 kJ

STANDARD HEATS OF FORMATION

EVALUATION

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