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Thermodynamics Test - 50

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Thermodynamics Test - 50
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  • Question 1
    1 / -0
    Select correct statement.
    Solution
    All the given statements are correct.
    (A) Both lattice energy and hydration energies decrease with ionic size With large size, the charge is more dispersed. hence, the extent of hydration is smaller. Also the packing is inefficient.
    (B) Lattice energy can be calculated using Born-Haber cycle The BornHaber cycle is an approach to analyze reaction energies. The cycle is concerned with the formation of an ionic compound from the reaction of a metal  (often a Group I or Group II element) with a halogen.
    (C) If the anion is larger compared to the cation, the lattice energy will remain almost constant within a particular group
  • Question 2
    1 / -0
    An ideal monoatomic gas undergoes a process in which its internal energy U and density ρ\rho vary as Uρ=constant.U\rho\, =\, constant. The ratio of change in internal energy and the work done by the gas is
    Solution
    As we know that internal energy(U) of a monoatomic gas is given as mv22\dfrac{{m}{v}^{2}}{2}
    and work done (W)=F.S
    where F=mnv23LnF=\dfrac{{mn}{v}^{2}}{3Ln}
    and S=LS=L
    so internal energywork done\dfrac{internal   energy}{work  done}=32\dfrac{3}{2}
  • Question 3
    1 / -0
    Heat of reaction for the reaction is:

    PCl5(g)+H2O(g)POCl3(g)+2HCl(g)PCl_5(g)+H_2O(g)\longrightarrow POCl_3(g)+2HCl(g)

    Given that,
    Pwhite+(3/2)Cl2(g)+12O2(g)POCl3; ΔH=135.5kcalP_{white} + (3/2)Cl_2(g) +\frac{1}{2}O_2(g)\longrightarrow POCl_3; \:  \Delta H=-135.5 \:kcal
       
    H2(g)+Cl2(g)+2HCl(g);                  ΔH=44.1kcalH_2( g ) + Cl_2( g )\longrightarrow +2HCl(g); \:                       \Delta H=-44.1 \:kcal

    P(w)+(5/2)Cl2(g)PCl5(g);                ΔH=89.6kcalP(w)+(5/2)Cl_2(g)\longrightarrow PCl_5(g); \:                                \Delta H=-89.6 \:kcal
     
     H2(g)+12O2(g)H2O(g);                   ΔH=57.8kcalH_2(g)+\frac{1}{2}O_2(g)\longrightarrow H_2O(g);\:                          \Delta H=-57.8\:kcal
    Solution
    Pwhite+32Cl2(g)+12O2(g)POCl3; ΔH1=135.5kcalP_{white} + \dfrac 32 Cl_2(g) +\dfrac{1}{2}O_2(g)\longrightarrow POCl_3; \:  \Delta H_1=-135.5 \:kcal

      H2(g)+Cl2(g)+2HCl(g);                  ΔH2=44.1kcalH_2( g ) + Cl_2( g )\longrightarrow +2HCl(g); \:                       \Delta H_2=-44.1 \:kcal

    P(w)+52Cl2(g)PCl5(g);                ΔH3=89.6kcalP(w)+ \dfrac52 Cl_2(g)\longrightarrow PCl_5(g); \:                                \Delta H_3=-89.6 \:kcal

      H2(g)+12O2(g)H2O(g);                   ΔH4=57.8kcalH_2(g)+\cfrac{1}{2}O_2(g)\longrightarrow H_2O(g);\:                          \Delta H_4=-57.8\:kcal

    The final reaction is obtained by reversing 3 and 4 and adding 1 and 2 to it:

    PCl5(g)+H2O(g)POCl3(g)+2HCl(g)PCl_5(g)+H_2O(g)\longrightarrow POCl_3(g)+2HCl(g)

               ΔH=ΔH1+ΔH2ΔH3ΔH4\Delta H = \Delta H_1 + \Delta H_2 - \Delta H_3 - \Delta H_4.

        \implies ΔH=135.544.1+89.6+57.8 \Delta H = - 135.5 - 44.1 + 89.6 + 57.8

                      =32.2kcal =- 32.2 kcal
  • Question 4
    1 / -0
    The dissolution of 1 mole of NaOH(s) in 100moleofH2O(l)100 \: mole \: of \: H_2O(l) give rise to evolution of heat as -42.34 kJ. However, if1 mole of NaOH(s) is dissolved in 1000moleofH2O(l)1000 \:mole \:of \:H_2O(l) the heat given out is 42.76 kJ. 

    If the enthalpy change when 900moleofH2O(l)900 \:mole \:of \:H_2O(l) are added to a solution containing 1 mole of NaOH(s) in 100moleofH2O100 \:mole \:of \:H_2O is xx kJkJ:
    Solution
    ΔH1for1moleNaOHin100moleH2O(l)=42.34kJ\Delta H_1 for \: 1 \: mole \: NaOH \: in \: 100 \: mole \: H_2O(l) = -42.34 \: kJ

    NaOH(s)+100H2ONaOH(100H2O);    ΔH=42.34kJNaOH(s) + 100H_2O\longrightarrow NaOH(100H_2O); \:      \Delta H= -42.34 \: kJ    ... (i)

    NaOH(s)+1000H2ONaOH(1000H2O);ΔH=42.76kJNaOH(s) + 1000H_2O\longrightarrow NaOH(1000H_2O); \: \Delta H= -42.76 \: kJ  ... (ii)

    By eqs. [(ii) - (i)]

    NaOH(s)+900H2ONaOH(900H2O);                ΔH=0.42kJNaOH(s) + 900H_2O\longrightarrow NaOH(900H_2O); \:                 \Delta H= -0.42 \: kJ

    Hence,option A is correct.
  • Question 5
    1 / -0
    The heat measured for a reaction in bomb calorimeter is :
    Solution
    A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction.

    Hence, the heat measured for a reaction in bomb calorimeter is ΔU\Delta U.
  • Question 6
    1 / -0
    1 gram sample of NH4NO3NH_4NO_3 is decomposed in a bomb calorimeter. The temperature of the calorimeter increases by 6.12 K. The heat capacity of the system is 1.23 kJ/g-deg. What is the molar heat of decomposition for NH4NO3NH_4NO_3?
    Solution
    The expression for the heat evolved is  q=mSΔTq= mS\Delta T

    When 1g of ammonium nitrate decomposes, the heat evolved is q=1×1.23×6.12q=1\times 1.23\times 6.12

    When 1 mole of ammonium nitrate decomposes, the heat evolved is q=1×1.23×6.12×80=602.21kJ/molq=1\times 1.23\times 6.12\times 80=6 02.21 \:kJ/mol
  • Question 7
    1 / -0
    The ΔfH\Delta_f H^{\circ} for CO2(g), CO(g) and H2O(g)CO_2 (g), \: CO(g) \: and \: H_2O(g) are 393.5,110.5-393.5, -110.5 and 241.8 kJ mol1-241.8 \: kJ \: mol^{-1} respectively. The standard enthalpy change (in kJ) for the reaction is:

    CO2(g)+H2(g) CO(g)+H2O(g)CO_2 (g)+ H_2 (g)\longrightarrow CO(g)+ H_2O(g ) 
    Solution
    First write the balanced chemical reactions for the formation of carbon dioxide, carbon monoxide and water.

    C+O2CO2;   ΔH=393.5kJC+ O_2 \longrightarrow CO_2; \:    \Delta H^{\circ}= -393.5 \:kJ ... (i)   
     
    C+12O2CO;         ΔH=110.5kJC+\frac{1}{2}O_2\longrightarrow CO;\:          \Delta H^{\circ}= -110.5 \:kJ     ...(ii)    

    H2+12O2H2O; ΔH=241.8kJH_2+\frac{1}{2}O_2\longrightarrow H_2O;\:  \Delta H^{\circ}= -241.8 \:kJ     ...(iii)  
     
    Then obtain the required equation by equating (ii)+(iii)(i):(ii) + (iii) - (i):

     CO2+H2CO+H2O;         ΔH=+41.2CO_2+H_2\longrightarrow CO+H_2O; \:          \Delta H^{\circ}=+41.2    
     
    Hence, the standard enthalpy change for the reaction is 41.2 kJ/mol.
  • Question 8
    1 / -0
    The commercial production of water gas utilize the reaction under standard conditions:

                               C+H2O(g) H2(g)+COC+H_2O(g)\longrightarrow  H_2(g)+CO

    The heat required for this endothermic reaction may be supplied by adding a limited amount of air and burning some carbon to CO2CO_2. How many g of carbon must be burnt to CO2CO_2 to provide enough heat for the water-gas conversion of 100 g carbon? 

    Neglect all heat losses to the environment. Also ΔfH\Delta _fH^{\circ} of CO, H2O(g)H_2O(g) and CO2CO_2 are -110.53, 241.81 and -393.51 kJ/mol respectively.
    Solution
    C+H2O(g) H2+COC+{ H }_{ 2 }O\left( g \right)\rightarrow { H }_{ 2 }+CO 

    ΔH=ΔHf(CO)ΔHf(H2O) \Delta { H }=\Delta { { H }_{ f } }\left( CO \right) -\Delta { { H }_{ f } }\left( { H }_{ 2 }O \right) 

    ={(110.53)(241.81) }kJ/mole=131.28kJ/mole=\left\{ \left( -110.53 \right) -\left( -241.81 \right)  \right\} kJ/mole=131.28kJ/mole

    so, total amount of energy to be provided for conversion of 100 gm of C

    (100gC)(1moleC12.0gC )(131.28kJmol )=1094kJ\displaystyle(100gC)\left( \frac { 1mole C }{ 12.0gC }  \right) \left( \frac { 131.28kJ }{ mol }  \right) =1094kJ

    To provide that energy, amount of C to be burnt

    (1094)kJ(1moleC393.51kJ )(12.0gC moleC )=33.4g\displaystyle\left( 1094 \right) kJ\left( \frac { 1mole C }{ 393.51kJ }  \right) \left( \frac { 12.0g C }{ mole C }  \right) =33.4g
  • Question 9
    1 / -0
    The enthalpies of formation of N2ON_2O and NONO are 2828 and 90 kJ mol190\ kJ mol^{-1} respectively. The enthalpy of the reaction, 2N2O(g)+O2(g) 4NO(g)2N_2O(g) +O_2(g)\longrightarrow 4NO(g) is equal to :
    Solution
    First write the balanced chemical; equations for he formation of dinitrogen oxide and nitric oxide.
     
    N2+12O2N2O;    ΔH=28kJN_ 2+\frac{1}{2}O_2\longrightarrow N_2O; \:     \Delta H=28 \:kJ...(i)
     
    12N2+12O2NO;        ΔH=90kJ\frac{1}{2}N_ 2+\frac{1}{2}O_2\longrightarrow NO; \:         \Delta H=90 \:kJ...(ii)

    Then multiply second equation with 4 and first equation with 2.
     
    2N2+O22N2O;    ΔH=2×28kJ2N_ 2+O_2\longrightarrow 2N_2O; \:     \Delta H=2 \times 28 \:kJ...(iii)
     
    2N2+2O24NO;        ΔH=4×90kJ2N_ 2+2O_2\longrightarrow 4NO; \:         \Delta H=4 \times 90 \:kJ...(iv)

    Now substract third equation from fourth equation.

    2N2O+O24NO;                       ΔH=304kJ2N_2O +O_2\longrightarrow 4NO; \:                        \Delta H=304 \:kJ.

    Thus,the enthalpy of the reaction 2N2O(g)+O2(g)4NO(g)2N_2O(g) +O_2(g)\longrightarrow 4NO(g) is 304 kJ.
  • Question 10
    1 / -0
    Which of the following has the same value as ΔfH\Delta_f H^{ \ominus } , COCO
    a. 12ΔfH(CO2)\frac { 1 }{ 2 } \Delta_f H^{ \ominus } (CO_2)
    b. 12ΔcH(graphite)\frac { 1 }{ 2 } \Delta_c H^{ \ominus } (graphite)
    c. ΔfH(CO2)ΔfH(graphite)\Delta_f H^{ \ominus } (CO_2) - \Delta_f H^{ \ominus } (graphite)
    d. ΔcH(graphite)ΔcH(CO)?\Delta_c H^{ \ominus } (graphite) - \Delta _c H^{ \ominus } (CO)?
    Solution
    As we know,
    C(graphite)+O2(g)CO2(g);ΔcH=x kJC(graphite) + O_2(g) \rightarrow CO_2(g) ;\quad \Delta _cH =x\ kJ................1

    CO+1/2O2(g)CO2(g)CO + 1/2 O_2(g) \rightarrow CO_2(g) ;   ΔcH\Delta_c H^{ \ominus }  (CO)=z kJ(CO) = z\ kJ.......................2


    Equation (1)-(2) gives (3) eqaution.

    C(graphite)+1/2O2(g)CO(g)C(graphite) + 1/2 O_2(g) \rightarrow CO(g) ;      ΔfH(CO)=y kJ\Delta_f H^{ \ominus }{(CO)}=y\ kJ.......................3


    From these equations, it can be seen that

    ΔfH\Delta_f H^{ \ominus }  (CO)(CO) = ΔcH(graphite)ΔcH(CO)\Delta_c H^{ \ominus } (graphite) - \Delta _c H^{ \ominus } (CO)
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