APPLICATION OF HESS’ LAW, ENTHALPY CHANGE OF FORMATION FROM ENTHALPY CHANGE OF COMBUSTION VALUES

 

QUESTION:

 

When a calcium is placed in water, aqueous calcium hydroxide is formed and hydrogen is given off.

(i)        Write the equation for the reactions of calcium with water.

(ii)       When 1.00g of calcium is placed in 200g of water, the temperature increases by 12.2 ^oC when the reaction is completed. The specific heat capacity of water, C, is 4.2 Jg^-1K^-1. Calculate the heat are leased in the experiment.

(iii)      Calculate the standard enthalpy change of reaction in KJ mole^-1 for your equation in (i).

(iv)      Calculate the volume of hydrogen, measure at room temperature and pressure liberated in the experiment described in part (ii).

 

SOLUTION:

 

(i)        + 2                                    Ca(OH)  + H  

(ii)       H = ?                                    

            M= (200    20 grams

            C = 4.2 Jg^-1 K^-1

            T = 12.2 ^oC

            H = mC T               

            = 20 x 4.2 x 12.2

            = - 10, 299.24 J

            = 10.29924 KJ or – 10.3 KJ

(iii)      Number of moles of calcium;

            n =  =  = 0.025 moles, the standard volume for enthalpy change of neutralization is less than the actual value, which is -57.3 KJ mole^-1 because part of energy is absorbed by the apparatus and part of energy is lost in the surroundings.

 

QUESTION:

 

            Calculate standard enthalpy change of combustion of propanol when 0.28 grams of propanol fuel is used to heat 100 grams of water in a calorimeter. The observed temperature rise is 21.5 k. Given that specific heat of water is 4.2 Jg^-1 K^-1 and Mr of Propanol  (C₃H₇OH) is 60.

 

SOLUTION:

 

             = ?

            m = 0.28g

            c = 4.2 Jg^-1 K^-1

            T – 21.5 k    

             = mC T              

            = 0.28x4.2x21.5

            = 9030 J

 

            Number of moles of propanol,

            n =           

            =

            = 0.00467 moles.

            0.00467 moles = 9030J

            1 mole =                                   

            = 1933618.84 J mole^-1

 

            Combustion is an exothermic reaction mass, C is specific heat of water which os 4.2 Jg^-1K^-1 and  is the change in temperature.

 

 

QUESTION:

 

            When 50 cm³ of hydrochloric acid and 50 cm³ of sodium hydroxide each of concentration 1.0 mole dm^-3 are mixed in  a polystyrene cup, there is a temperature rise of 6.5 ^oC. Calculate the molar heat of neutralization?

 

SOLUTION:

 

            Assume 1 cm^-3 = 1 gram, as density of solution is 1 g cm^-3

            m = (50+50) = 100 cm³ or 100 grams.

            C = 4.2 Jg^-1 K^-1

            T = 6.5 ^oC

 

            Enthalpy change = mcT

                                          = 100x4.2x6.5

                                          = 2730 J

            Number of moles of HCL or NaOH, n =                       

                                                                            

   =  

                 = 0.05 moles.

                 0.05 mole = 2730 J

                 1 mole =  x 1

                                                                            = 54600 J mole^-1   

                                                                           

            Temperature rises shows is an exothermic reaction hence, enthalpy change is negative. Through this experiment of conservation of energy or universal first law of thermodynamics.

 

APPLICATION OF HESS’ LAW:

 

            Different routes of a chemical reaction are called enthalpy cycles. By using these cycles enthalpy for a chemical reaction is calculated, which directly can not be determined easily.  By applying Hess’ law and calculations involving such cycles following enthalpy changes can be calculated that can not  be found by direct experiment e.g; an Enthalpy change of formation from enthalpy changes of  combustion. An enthalpy change of reaction from enthalpy changes of formation, average bond energies (enthalpies) and an enthalpy change of reaction from enthalpy change of combustion and formation values.

 

ENTHALPY CHANGE OF FORMATION FROM ENTHALPY CHANGE OF COMBUSTION  VALUES:

 

QUESTION

 

            Calculate enthalpy change of formation of methane from carbon and hydrogen by using the following data.

 

             + O                                CO     = - 394 KJ mole^-1                                       

 

            H  + ½ O                           H₂O_(l)   = - 286 KJ mole^-1

           

            CH  + 2O₂                            CO  + 2H₂     = - 890 KJ mole^-1

 

 

 

SOLUTION:

 

            According to the Hess’ law enthalpy changes for different pathways is same by applying Hess’ law,

 

            Enthalpy change                    Enthalpy change

            For route 1                =          for route 2

             + (-890) = (-394) + 2 (-286)              

 

             - 890 = - 394 – 572

 

             - 890 = - 966

 

             - 890 = -  966 + 890

 

                         - 890 = - 76 KJ mole^-1

 

QUESTION:

 

            Calculate standard enthalpy change of formation of ethanol from the following experimental data.

            C + O₂                           CO₂  = - 394 KJ mole^-1

            H₂ + ½ O₂                   H₂O  = - 286 KJ mole^-1

            C₂H₅H + 3O₂               2CO₂ + 3H₂O  = - 1367 KJ mole^-1