AN ENTHALPY CHANGE OF REACTION FROM ENTHALPY CHANGE OF FORMATION VALUES

 

AN ENTHALPY CHANGE OF REACTIONFROM ENTHALPY CHANGE OF FORMATION VALUES:

 QUESTION:

             Calculate the enthalpy change for the hydration of ethane to form ethanol, by using the given values of standard enthalpy changes of formation.

             Ethane                         = + 52 KJ mole^-1

            Water              = - 286 KJ mole^-1

            Ethanol            = - 277 KJ mole^-1

           

            Reaction for hydration of ethane is;

 

            CH₂ = CH₂ + H₂O                                C₂H₅OH

 

SOLUTION:

 

                                    CH₂ = CH₂ +H₂O  C₂H₅OH

(Route 1)           + 52        -286            -277       (Route 2)

 

                                    2C + 3H₂ + ½ O

            By applying Hess’ law;

 

            (+52)+(-286) +  = - 277

            -234 +  = - 277

             = - 277 + 234

             = - 43 KJ mole^-1

 

QUESTION:

 

            Calculate enthalpy change of reaction when solid calcium oxide (Quick lime) put in water to form solid calcium hydroxide (slacked lime).

 

            The data is;

            Ca +  O₂                                 CaO  = - 635.1 KJ mole^-1

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

            Ca + H₂ + O₂                           Ca(OH)₂  = -986.1 KJ mole^-1

 

SOLUTION:

                                    Ca  + H₂O_(s) Ca  

            (Route 1)      -635.1                        -286        -986.1                  (Route 2)

 

Ca(s) +  +

 

By applying Hess’ law          

 

(-635.1) + (-286) +  = (-986.1)

 = - 986.1 + 635.1 + 286

 = - 65.2 KJ mole^-1

ENTHALPY CHANGE OF REACTION & AVERAGE BOND ENERGIES:

 

QUESTION:

 

            Calculate enthalpy change of reaction between nitrogen and hydrogen to form ammonia gas, with the help of average bond energy values from data booklet.

 

Bond                                      bond energy KJ mole^-1

                                                 994

H – H                                                   436

N – H                                                   390

 

SOLUTION:

            N N + 3H – H                         N

           

                                                            H         H         H

 

            +994           3(+436)                        6(-390)

 

                              2N + 6H

            By applying Hess’ law,

 

            = (+994) + [3(+436)] + [6(-390)]

                    = 994 + 1308 – 2340

                      = 2302 – 2340

                 = -38 KJ mole^-1

 

Note:   Bond breaking [ Endothermic + H                    

            Bond forming   [ Endothermic - H

Overall, the above reaction is an exothermic reaction.

 

QUESTION:

 

            A balanced chemical equation for the complete combustion of methane is;

 

            CH  + 2O                                     CO₂ + 2H₂O_(g) 

 

(a)       Re-write this equation to show the bonds present in each molecule.

(b)       Use the bond enthalpies (energies) from the data booklet to calculate enthalpy change of combustion reaction.

(c)       Compare the value for  with the experimentally determined value of -890 KJ mole^-1. Which value is more accurate?

 

Solution:

 

(b)        = [4x(+410)] + [2x(+496)] + [2 x (-740) + [4x(-480)]

            = 1640 + 992 – 1480 – 1840

            = 2632 – 3320

            = - 688 KJ mole^-1

 

(c)       Bond enthalpies used are base on average values and are based on breaking bonds in gaseous molecules. Water molecules are present in the gaseous state in the bond enthalpy calculations. Whereas  refers to the liquid state for water, Hence the experimental value is more accurate.

 

QUESTION:

 

            The standard enthalpy change of atomization of carbon and hydrogen are +715 KJ mole-1 and 218 KJ mole-1 respectively. Given that the enthalpy change of formation of methane is -75 KJ mole^-1 calculate, The standard enthalpy change of atomization,  of the methane and (b) the C-H-bond energy.

 

             + (-75) = 715 + 4 (218)

             = 715 + 872 + 75

             = 1662 KJ mole^-1

 

            Bond energy for C-H =

                                                 =

                                                 =  415.5 KJ mole^-1

 

SOME OTHER NUMERICAL NUMBEREXAMPLES:

 

QUESTION:

 

            Ca + 2H₂O                   Ca(OH)₂ + H₂ enthalpy change for  the reaction is -412 KJ mole^-1. Calculate the  for calcium hydroxide, Ca(OH)₂, whereas given standard enthalpy change for water is -286 KJ mole^-1.

 

SOLUTION:

 

             of water = - 286 KJ mole^-1

             of Ca (OH)₂ = ?

                         = - 412 KJ mole^-1

            Ca + 2H₂O                   Ca(OH)₂ + H₂

                  2  (-286)                       

 

                                    Ca + 2H₂ + O₂

            By applying Hess’ Law

 

            2 + (-286) + (-412) -

            572 – 412 =

            -984 =  

             of ca(OH)₂ = - 984 KJ mole^-1

QUESTION:

 

            The yellow gas chlorine dioxide, CLO₂ has been used for may years as a flour improving agent in bread making. It can be made in the laboratory by the following reaction.

 

            2AgCLO  + Cl                              2AgCl_(s) + 2CLO   + O , , zero

 

Of  (AgCLO₃) = - 25 KJ mole^-1

            (AgCl) = - 127 KJ mole^-1

 

Solution:

 

            2AgCLO  + CL   = O KJ mole^-1     2AgCL₍₃₎ + 2CLO  + O

            2 x (-25)                                       2 x (-127)             2 x

 

 2 Ag + 2CL₂ + 3O₂

 

By applying Hess’ law

 

            2x(-127) + 2 x  = 2 x )-25) + 0

- 254 + 2  = - 50

2 = - 50 + 254

2  = +

                       

            Since the enthalpy change of formation of CLO2 is endothermic thus the gas is energetically unstable.

 

QUESTION:

 

            Hydrogen is used in large quantities in industry to convert nitrogen into ammonia, for use in fertilizers. One method of manufacturing hydrogen is to pass methane and steam over a heated nickel catalyst.

                        CH + H2O_(g)  CO_(g) + 3H ,  = + 206 KJ mole^-1.

            Use the value of  above and bond energy values from the Data Booklet, to calculate the total bond energy in the carbon monoxide molecule. Suggest why the bond energy you have calculated above is large than either  C – O or C = O and energies in the Data Booklet.

 

            Solution:

             = [ 4 x (+410)] + [ 2x(+460)] + [-B.E] + [3x(-436)]

            206 = 1640 + 920 + [ -B.E] – 1308

            206 = 2560 + [-B.E] – 1308

            206 = [-B.E] + 1252

            B.E = 1252 – 206

            B.E = 1046

            B.E of carbon monoxide = 1046 KJ mole^-1

 

            In carbon monoxide, carbon – oxygen bond is triple covalent bond i.e; C  O which stronger than C = O and C – O bond. The strength of bond order is;

 

            C  O > C = O > C – O

 

QUESTION:

 

            Dinitrogen pentoxide, N₂O₅, can be produced by the following sequence.

 

            I           N + O                             2NO_(g)  = + 180 KJ mole^-1

            II          NO(g) + ½ O                         NO  = -  57 KJ mole^-1

            III        2NO  + ½ O                          N₂O    = -  55 KJ mole^-1

 

(a)          Use the given data to calculate the standard enthalpy change of formation of dinitrogen pentoxide,  (N₂O₅).

(b)         Suggest why reaction I is endothermic.

(c)          Dinitrogen pentoxide is a covalent molecule. Draw a possible displayed formula for the molecule and suggest the bond angles in the molecule.

(d)         Suggest an equation for the reaction of dinitrogen pentoxide with water.

Solution:

                 = (+180) + [2X(-57)] + [ -55]

                = 180 – 114 – 55

                 = 180 – 169     

             = + 11 KJ MOLE^-1 

 

(b)       Reaction I is endothermic because strong N  N and O  O bonds needed to be broken and this is not compensated by the N  O bonds formed.

There are three regions of electron density around N so that N has a triangle planar geometry hence respected angle is 120⁰. Whereas in O, has a tetrahedral geometry. However two lone pairs exert repulsion on bond pair hence, expected angle is less than 109.5⁰ i.e; about 105⁰.

 

(d)       N₂O _­ + H₂O_(l)                        2HNO     

 

QUESTION:

 

            Given that  for water is – 286 KJ mole^-1. What is the standard enthalpy change of the following reaction, ?