Thermo chemistry Solved Exercise

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Solved Exercise from chapter 7 Thermo chemistry, chemistry book 1 for FSC pre engineering and pre medical for Board of Intermediate and Secondary education. Also For Entry Test Preparation for UET, NUST, PIEAS, GIKI, AIR, FAST, WAH University, UHS, other engineering Universities and Medical Colleges.
    TEXT BOOK EXERCISE

Q1.      Select the suitable answer from the given choices.
(i)                 If an endothermic reaction is allowed to take place very rapidly in the air the temperature of the surrounding air
(a)        Remains constant                    (b)        increases
(c)        dereases                                   (d)       remain unchanged
(ii)        In endothermic reactions, the heat content of the
            (a)        Products is more than that of reactants
            (b)        Reactants is more than that of reactants
            (c)        Both (a) and (b)
            (d)       Reactants and product are equal
(iii)       Calorie is equivalent to
            (a)        0.4184 J                                   (b)        41.84J
            (c)        4.184 J                                     (d)       418.4 J
(iv)       The change in heat energy of a chemical reaction at constant temperature and  pressure is called
(a)        enthalpy change                      (b)        bond energy
(c)        heat of sublimation            (d)            internal energy change
(v)        Which of the following statements is contrary to the first law of thermodynamics?
(a)        Energy can neither be created nor destroyed.
(b)        One form of energy can be transferred into an equivalent amount of of other kinds of energy.
(c)        In an adiabatic process, the work done is independent of its path.
(d)       Continuous production of mechanical work without supplying an equivalent amount of heat is possible
(vi)       For a given process, the heat changes at constant pressure (qp ) and at constant volume (qv) are related to each other as
(a)        qp =qv                                           (b)        qp < qv   
            (c)        qp > qv                                                  (d)       qp = qv/2
(vii)      For the reaction NaOH + HC1 NaC1 + H2O. The change in enthalpy is called
(a)        Heat of reaction                      (b)        heat of formation
(c)        Heat of neutralization                         (d)       heat of combustion
(viii)     The net heat change in a chemical reaction is same whether it is brought about in two or more different ways in one or several steps. It is known as
(a)        Henry’s law                             (b)        Hess’s Law
(c)        Joule’s principle                      (d)       Law of conservation of energy
(ix)       Enthalpy of neutralization of all the strong acids and strong bases has the same value because
            (a)        Neutralization leads to the formation of salts and water
            (b)        Strong acids and bases are ionic substances
            (c)        Acids always give rise to H+ ions and bases always furnish OH- ions.
            (d)       The   net chemical change involve the combination of H+ and OH- ions to form water
Ans.    i)c        ii)a       iii)c      iv)a      v)d       vi)c      vii)c     viii)b    ix)d
Q2.      Fill in the blanks with suitable words.
(i)                 The substance undergoing a physical or a chemical change forms a chemical ________.
(ii)               The change in internal energy _________be measured.
(iii)             Solids which have more than one crystalline forms possess_________values of heats of formation.
(iv)             A process is called___________if it takes place on its own without any outside assistance.
(v)               A __________is a macroscopic property of a system which is __________of the path adopted to bring about that change.
Ans.    i)system           ii)can               iii)different                 iv)spontaneous           v)state function : independent
Q3.      Indicate the true or false as the case may be.
(i)                 It is necessary that a spontaneous reaction should be exothermic.
(ii)               Amount of heat absorbed at constant volume is internal energy change.
(iii)             The work done by the system is given the positive sign.
(iv)             Enthalpy is a state function but internal energy is not.
(v)               Total heat content of a system is called enthalpy of the system.
Ans.    i)          False   ii)         True    iii)        False   iv) False    v)   True
Q4.      Define the following terms and give three examples of each
            i)          System                                                ii)         Surroundings
            iii)        State function                         iv)        Units of energy
            v)         Exothermic reaction                vi)        Endothermic reaction
            vii)       Internal energy of the system             viii)Enthalpy of the system.
Q5.      (a)        Differentiate between the following:
(i)                 Internal energy and enthalpy
(ii)               Internal energy change and enthalpy change
(iii)             Exothermic and endothermic reactions
(b)       Define the following
            (i)         Standard enthalpy of reaction
            (ii)        Standard enthalpy of combustion
            (iii)       Standard enthalpy of atomization
(iv)             Standard enthalpy of solution
Q6.      (a)        What are spontaneous and non-spontaneous processes? Give examples.
(b)               Explain that burning of a candle is a spontaneous process.
(c)                Is it true that a non-spontaneous process never happens in the universe?
Explain it.
Q.7      (a)        What is the first law of thermodynamics? How does it explain that
                        (i)         qv =E           ii)         qp =H
(b)        Wow will you differentiate between E and H ? Is it true that H and E have the same values for the reaction taking place in the solution state.
Hint:    For reactions taking place in the solution state. Since the change in volume is insignificant, i.e . V=0, so
                                                H=E + PV
                                                H = E +Px0
                                                H =E
Hence, H and E have the same values for the reaction taking place in the solution state.
Q.8.     (a)        What is the difference between heat and temperature ? Write a mathematical relationship between these two parameters.
Ans.    Difference between heat and temperature
Temperature:            “ A measure of the average kinetic energy of all the particles in a system is called temperature .”
            Temperature is a state function. If we transfer energy to a system, the kinetic energy of the particles in the system increases. Therefore, the temperature rises.
Heat:   “The transfer of energy caused by a difference in temperature between a system and its surroundings or between a system and another system is called heat.”
            Heat is not a state function. It depends on the path of the system. A mathematical relationship between heat and temperature is :
                                                            q=  msT
(b)        How do you measure the heat of combustion of a substance by bomb calorimeter.
Q9.      Define heat of neutralization. When a dilute solution of a strong acid is neutralized by a dilute solution of a strong base, the heat of neutralization is found to be nearly the same in all the cases. How do you account for this?
Q10.    (a)        State the laws of thermo-chemistry and show how are they based on the firs law of thermodynameics.
Ans.    Laws of Thermochemistry
            There are two thermochmical laws. They are based on the law of conservation of energy. These law are:
1.         First Thermo chemical Law (Lavoisier and Laplace-1780)
            “The quantity of heat required to decompose a compound into its elements is equal to the heat evolved when that compound is formed from its elements, but with opposite sign.”
                        In other words the heat of decomposition of a binary compound is numerically equal to the heat of formation of a compound, but of opposite sign.
Examples:
(1)        H2(g )+  O2 (g)   H 2O(g)                H= -285.58kJ mol-1
            H2O(l )  H 2O(g) +O2 (g)                H= +285.58kJ mol-1
(2)           C(s )   O2(g) +CO2 (g)                   H= -393.7kJ mol-1
            CO(g )   C(s)     + O2 (g)                   H= +393.7kJ mol-1
(3)        H2 (g) +12 (s)  H1(g)               H= +26.48kJ mol-1
            H1(g)   H2 (g) +12 (s)              H= -26.48kJ mol-1
2.         Second Thermaochemical law (Hess’s Law-1840)
            “ The amount of heat evolved or absorbed in a chemical reaction is the same whether the reaction takes place in one step or in several steps.”
Mathematically:
                        H = H1 +H2 + ………….
Where H is the enthalpy change for the reaction when the reactants are directly converted into products in one step and H1 and H2 are the enthalpies of different steps when the reactants are converted into products in two steps.
(b)        What us a thermochmical equation? Give three examples. What information do they convey?
Ans.    Thermochmical Equation
            “A chemical equation which gives an idea about the heat evolved or absorbed during the reaction is called a thermochmical equation.”
Examples:      (i)         C(s)     +O2 (g)      CO3(g)     H= -393.7kJ mol-1
(ii)        H2(g )+  O2 (g)   H 2O(l)                 H= -285.58kJ mol-1(iii)        N2(s)     +O2 (g)    2NO(g)                    H= +180.51kJ mol-1
(iv)       N2(s)     +3H2 (g) 2NH3(g)                  H= -41.6kJ mol-1
Information conveyed by a thermochmical equation
            A thermochmical equation give the following information.
(i)         The heat evolved or absorbed in a reaction. H is always negative for an exothermic reaction where as it is positive for an endothermic reaction.
(ii)        The physical state of reaction and products because they will effect the heat of reaction.
(iii)       It is treated as a standard equation. The coefficients of reactants and products always represent their mole.
(iv)       If the conditions of the reaction are not given, then it is assumed that the reaction is taking place at 1 atm pressure and 25oC.
(v)        If the termochemical equation is reversed, the sign of the heat of reaction will be reversed but the magnitude remains the same.
(c)        Why is it necessary to mention the physical states of reactants and products in a thermochemical reaction? Apply Hess’s law to justify your answer.
Ans.    Since the heat of reaction depends upon the physical state of the reactants and products, therefore, while writing a thermochemical equation, the physical state of the reactants and products must be mentioned.
            If the reaction AB is exothermic, than BA will be endothermic.
Q11.    (a)        Sate and explain Hess’s law of constant heat summation. Explain it with examples and give its applications.
            (b)        Hess’s law helps us to calculate the heats of those reactions, which cannot be normally carried out in a laboratory. Explain it.
Q12.    (a)        What is lattice energy ? How does Bron-Haber Cycle help to calculate the lattice energy of NaC1?
            (b)        Justify that heat of formation of a compound is the sum of all the other enthalpies.
Q13.    50 cm3 of 1.0 M HC1 is mixed with 50 cm3 of 1.00 M NaOH in a glass calorimeter. The temperature of the resultant mixture increases from  21.0oC to enthalpy change mole-1 for the reactions. The density of solution to be considered is 1 g cm-3 and specific heat is 4.18 Jg -1 k-1 .
Solution:
                        Volume of 1.0 MHC1             =50cm3
                        Volume of 1.0 MNaOH          =50cm3
                        Initial temperature of HC1     =21.0oC
                        Initial temperature of NaOH  =21.0oC
            Final temperature of reaction mixture =27.5oC
            Temperature rise for reaction mixture =27.5 – 21.0 =6.5oC
Specific heat capacity of reaction mixture      =4.10 jg-1 k-1
                                                            =4.18 jg -1 oC-1
                        Density of solution     =1 gcm-3
            Volume of reaction mixture    =50 cm 3 + 50 cm3 =100 cm3
            Mass of reaction mixture        =volume x density
                                                            =100 cm3 x 1 g cm-3
                                                            =100 g
Amount of total heat evolved, q         =m x s x  T
                                                =100g x 4.18 Jg -1 oC-1 x 6.5oC
                                                                                    =2717 J , =2.717 KJ
Since the reaction is exothermic, so, q            = - 2.717 kJ
Vol. Of NaOH                                               ==0.05 dm3
No. of moles of HC1                          =Molarity x vol. of soln. In dm3
                                                            =1.0 x 0.05
                                                            0.05 mole
Vol. Of NaOH                                               ==0.05 dm3
No. of moles of HC1                          =Molarity x vol. of soln. In dm3
                                                            =1.0 x 0.05
                                                            0.05 mole


Equation of reaction: HC1(aq)       +NaOH (aq)   NaC1(aq) + H2 O(l)
                                                                            1mole                  1mole                  1mole
                                                                        0.05 mole              0.05mole                   0.05mole
Now, 0.1 mole of formation of H2 O liberates heat =-2.717 KJ
1 mole of formation of H2 O liberates heat =
                                                                                    =- 54.34 kJ
                                               
            Enthalpy of neutralization, Hn = -54.34 kJ Answer
Q14.    Hyrazine N2 H4 is a rocket fuel. It burns in o2 to N2 and H2O.
                                    N2H4(l)      +O2(g)              N2(g) + 2H2 O(g)
            1.00 g of N2H4 is burned in a bomb calorimeter. An increase of temperature 3.51oC is recorded. The heat capacity of calorimeter is 5.5 kJK-1 . Calculate the quantity of heat evolved. Also calculate the heat of combustion of 1 mole of  N2 H4.  
Solution:
                                    Mass of N2H4                  = 1g
                        Rise of temperature                 =3.51 oC
            Sp. Heat capacity of calorimeter         =5.5 kJK-1 =5.5 kJ oC-1
                                                                 q     =?
                                    Formula Used:                  q    = m x s x T
                                                                        =1 g x 5.5 kJ oC-1 x 3.51oC
                                                                        =19.31 kJ
            Since combustion is an exothermic reaction, so
                                                                        q =- 19.31 kJ Answer
            Now,   No of moles of N2 H4              =1 mole
                                    Mass of N2H4              =No. of moles x Molar mass
                                                                        =1 mol x 32 g mol-1
                                                                        =32 g 
            Heat of combustion of 1 g of N2 H4               = -19.31 kJ
            Heat of combustion of 32 g of N2 H4             =- 19.31 kJ x 32
            Hence, heat of combustion of N2 H4   = -618 kJ mol-1 Answer
Q15.    Octane C8 H18 is a motor fuel. 1.80 g of a sample of octane is burned in a bomb calorimeter having heat capacity 11.66 kJk-1 . The temperature of the calorimeter increases from 31.36oC to 28.78oC. Calculate the heat of combustion for 1 g of octane. Also calculate the heat for 1 mole of octane.
Solution:
            Mass of octane, C8 H18            =1.80g
Sp. Heat capacity of calorimeter         =11.66 kJK-1 =11.66 kJ oC-1
            Rise in temperature, T         =(28.78 – 21.36 ) oC =7.42 oC
            Heat of combustion, q            =?
            Formula Used:                                    q=m x s x T
                                                            =1.80 g x 11.66 kJ oC-1 x 7.42 oC
                                                            =155.73 kJ Answer
Since combustion reaction is an exothermic, so
                        Heat of combustion    = -155.73 kJ
Now, heat o combustion of 1.80 g of octane =- 155.73 kJ
Heat of combustion of 1 mole of octane        = -  x 1 g
                                                                        = - 86.52 kJ
Heat of combustion of 1 mole of octane        = -86.52 kJ
                        No, of moles of octane           =1 mole
            Molar mass of octane, C8 H18 =96+18=114 g mol-1
            Mass of octane = moles of octane x molar mass of octane
                                                                        = 1 mol x 114 g  mol-1
                                                                        =114 g
            Now, heat of combustion of 1 g of octane     = -86.52 kJ
            Heat of combustion of 114 g of octane          = x 114 g
                                                                                    = - 9863 kJ
Hence,             heat of combustion of 1 mole of octane=- 9863 kJ mol-1 Answer
Q16.    By applying  Hess’s law calculate the enthalpy change for the formation of an aqueous solution of NH4 C1 from NH3 gas and HC1. The results for the various reactions and pressure are as follows.
(i)         NH3(s)     +aq      NH3(aq)                                H= -35.16kJ mol-1              
(ii)        HC1(g)     +aq     HC1(aq)                               H= -72.41kJ mol-1
(iii)       NH3(s)     +HC1(aq)           NH4 C1(aq)            H= -35.16kJ mol-1
Solution:
            NH3(g)     +HC1(g)+ aq   NH4C1(aq)             H= ?
(i)         NH3(g)     +aq     NH3(aq)                                H= -35.16kJ mol-1
(ii)        NH3(g)     +aq     HC1(aq)                               H= -72.41kJ mol-1
(iii)       NH3(aq)     +HC1(aq)          NH4 C1(aq)            H= -51.48kJ mol-1
Add Eq (i), Eq (ii)
(i)         NH3(g)     +aq     NH3(aq)                                H= -35.16kJ mol-1
(ii)        NH3(g)     +aq     HC1(aq)                               H= -72.41kJ mol-1
____________________________________________________________
(iv)       NH3(g)     +HC1(g)+2aq  NH3(aq)+HC1(aq) H= -107.57kJ mol-1
Now, add Eq (iii)and Eq(iv)
(iii)       NH3(aq)     +HC1(aq)          NH4 C1(aq)            H= -51.48kJ mol-1
(iv)       NH3(g)     +HC1(g)+2aq  NH3(aq)+HC1(aq) H= -107.57kJ mol-1
____________________________________________________________
(v)        NH3(g)     +HC1(g)+2aq  NH3(aq)+HC1(aq) H= -159.05kJ mol-1
Since any integer multiplied by ‘aq’ is equal to ‘q’
                        So, 2 x aq        =aq
Hence, Eq(v)becomes,
            NH3(g)     +HC1(g)+2aq  NH3(aq)+HC1(aq) H= -159.05kJ mol-1
                                                H= .159.05 kJ mol-1 Answer
Q17.    Calculate the heat of formation of ethyl alcohol from the following information
(i)                 Heat of combustion of ethyl alcohol is – 1367 kJ mol-1
(ii)               Heat of formation of ethyl alcohol is – 393.7 kJ mol-1
(iii)             Heat of formation of water is -  285.8 kJ mol-1
Solution:
            2C(s)     +3H 2(g) +O2(g)              C2 H 5OH(l)       H= ?
(i)         C2 H5 OH(l)     + 3O2(g)  2CO2(g)    +  3H2 O(g)        H= -1367kJ
(ii)        C(s)     +O2(g)       CO2(g)                                 H= -393.7kJ
(iii)       H2(s)     +O2(g)                H2 O(l)                            H= -285.8 kJ
            Multiply Eq(ii) by 2 and Eq (iii) by 3 and then add the resultant equation
(iv)       2C(s)     +2O2(g) 2CO2(g)                               H= -787.4kJ
(v)        3H2(g)     +O2(g)             3H2 O(l)              H= -857.4 kJ
____________________________________________________________
(iv)       2C(s)+ 3H2(g)     +O2(g) 2CO2(g)+3H2 O(l)H= -1644.8 kJ
Now, subtract Eq(i) from Eq(iv),
(vi)       2C(s)+ 3H2(g)     +O2(g) 2CO2(g)+3H2 O(l)H= -1644.8 kJ
(i)         C2H5OH(l)  3O2(g      2CO2(g) 3H2 O(l)H= 1367 kJ
____________________________________________________________
2C(s)+ 3H2(g)     +O2(g) – C2H5 OH(l)    0          H= -277.8 kJ
or
2C(s)+ 3H2(g)     +O2(g)  C2H5 OH(l)                        H= -277.8 kJ
            Hence, the heat of formation of ethyl alcohol is  - 277.8 kJ Answer
Q18.    If the heats of combustion of C2H2, H2 and C2 H6 are – 337.2 , - 68.3 and – 372.8 k calories respectively , then calculate the heat of the following reaction.
                        C2H2(g)+ 2H2(g)       C2H6 (g)
Solution:
                C2H2(g)+ 2H2(g)       C2H6 (g)            H= ?
(i)         C2H2(s)+      O2(g)   2CO2(g)+ C2H6(g)     H= -337.2 kJ
(ii)        H2(s)+ O2(g)  H2 O(l)                                      H= -68.3 kJ
(iii)       C2 H6(s)+ O2(g) 2CO2(g) +3H2O(l)           H= -372.8 kJ
Multiply Eq (ii) by2
(iv)       2H2(g)+ O2(g) 2H2O(l)                                        H= -136.3kJ
            Add Eq (i) , Eq (iv)
(i)         C2H2(s)+      O2(g)   2CO2(g)+ H2O(l)       H= -337.2 kJ
(ii)        2H2(g)+ O2(g) 2H2O(l)                                        H= -136.3kJ
(v)        C2 H2(s)+ 2H2(g) +O2(g) 2CO2(g) +3H2O(l)      H= -473.8 kJ
From Eq(v) subtract Eq(iii)
(v)        C2 H2(s)+ 2H2(g) +O2(g) 2CO2(g) +3H2O(l)     H= -473.8 kJ
(iii)       C2H6(g)  O2(g)      2CO2(g) 3H2 O(l)  H= 372.8 kJ
C2H2(g)+ 2H2(g)      - C2H6 (g) 0                                 H= -101.0 k cal
Or
C2H2(g)+ 2H2(g)      - C2H6 (g)                                    H= -101.0 k cal
                                    H= -101.0 k cal Answer
Q19.    Graphite and diamond are two forms of carbon. The enthalpy of combustion of graphite at 25oC is – 393.51 kJ mol-1 and that of diamond is – 395.41 kJ mol-1.
            What is the enthalpy change of the process? GraphiteDiamond at the same temperature?
Solution:         C(graphite) C(diamond)           H=?
(i)         C(graphite) +O2(g) CO2(g)                      H= -393.51 kJmol-1
(ii)        C(diamond ) +O2(g) CO2(g)                    H= -395.41 kJmol-1
From Eq(i) Subtract Eq(ii)
(i)         C(graphite) +O2(g) CO2(g)                      H= -393.51 kJmol-1
(ii)        C(diamond) O2(g) CO2(g)              H= 393.41 kJmol-____________________________________________________________
C(graphite) - C(diamond) 0                                           H=1.9 kJ mol-1
C(graphite) C(diamond)                                                H=1.9 kJ mol-1
                                                H=1.9 kJ mol-1 Answer
Q20.    What is the meaning of the term enthalpy ionization? If the heat of neutralization of HC1 and NaOH is – 57.3 KJ mol-1 and heat of neutralization of CH3COOH with NaOH is  - 55.2 kJ mol-1 , calculate the enthalpy of ionization of CH3COOH.
Solution:
CH3 COOH(l)  CH3 COO-(aq) + H­­-(aq)                        H=?
(i)H+(aq) + C1-(aq) + Na+(aq) +OH-(aq) Na+(aq)+H2O(l)            H=57.3 kJ mol-1
(ii)CH3COOH(l) + Na+(aq) +OH-(aq) CH3COO –(aq) + Na+(aq)+H2O(l)  H=55.2kJ mol-1       
Now, from Eq(ii), Subtract Eq(i)
(ii) CH3COOH(l) + Na+(aq) +OH-(aq) CH3COO –(aq) + Na+(aq)+H2O(l)H=-55.2kJ mol-1
(i) H+(aq)  C1-(aq) Na+(aq) OH-(aq) Na+(aq) H2O(l)                            H=57.3 kJ mol-1

CH3 COOH(l)  CH3 COO-(aq) + H­­-(aq)                        H=2.1 kJ mol-1          
CH3 COOH(l)  CH3 COO-(aq) + H­­+(aq)                       H=2.1 kJ mol-1
                                                                        H=2.1 kJ mol-1Answer
Q21.    (a)        Explain what is meant by the following terms.
(i)                 Atomization energy
(ii)               Lattice energy
(b)               Draw a complete, fully labeled Born-Haber Cycle for the formation of potassium bromide.
(c)                Using the information given in the table below , calculate the lattice energy of potassium bromide.
Reactions:                               H / kJ mol-1
K(s) + Br2(l)                -392
K(s) K(g)                                                  +90
K(g)                        +420
Br2(l)                                +112
Br(g) +                                     -342
Solution:
                                                            Hf(KBr)=-392 kJ mol-1
                                                            Hat (K) =90kJ mol-1
                                                            Hi(K)=420 kJ mol-1
                                                            HD/2 (Br2)=112 kJ mol-1
                                                            He (Br) = -342 kJ mol-1
                                                            Hl =?
                                    According to Born-Haber cycle
                                                Hf(KBr) =Hf + Hat(k) +Hi(k) +HD/2 (Br2 ) + He(Br)
                                                On putting the values,
            -392 kJ mol-1 =Hl + 90 kJ mol-1 + 420 kJ mol-1 + 112 kJ mol-1 – 342 kJ mol-1
            -392 kJ mol-1 =Hl  + 280 kJ mol-1
                        Hl = -392 kJ mol-1 – 280 kJ mol-1
                        Hl =-672 kJ mol-1
            Hence, lattice energy of KBr = - 672 kJ mol-1 Answer 

I am Mechatronics and Control Engineer from UET, Lahore. Helping Engineering students gives me pleasure.

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