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Thermodynamics Test - 63

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Thermodynamics Test - 63
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  • Question 1
    1 / -0
    The value of G\triangle G for the process H2O(s)H2O(l)H_{2}O(s) \rightarrow H_{2}O(l) at 1 atm1\ atm and 260 K260\ K is:-
    Solution

  • Question 2
    1 / -0
    The reaction NH2CN(s)+32O2(g)N2(g)+CO2(g)+H2O(l)NH_{2}CN(s) + \dfrac {3}{2}O_{2}(g) \rightarrow N_{2}(g) + CO_{2}(g) + H_{2}O(l) was carried out at 300 k300\ k in a bomb calorimeter. The heat released was 742 kJ mol1742\ kJ\ mol^{-1}. The value of H300K\triangle H_{300K} for this reaction would be_________.
    Solution
    Enthalpy change for a reaction (ΔH)(\Delta H) is given by the expression,
    ΔH=ΔU+Δ ngRT\Delta H=\Delta U+\Delta  n_gRT
    where,
    ΔU=\Delta U=change in internal energy
    Δng=\Delta n_g=change in number of moles

    For the given reaction,

    Δng=ng(productng(reactant)\Delta n_g=\sum{n_g(product}-\sum{n_g(reactant)}
             
             =(21.5)moles=(2-1.5)moles

    Δng=0.5moles\Delta n_g=0.5 moles,

    ΔU=742.7 kJ/mol\Delta U=-742.7\ kJ/mol

    T=298 kT=298\ k

    R=8.314×103kJ/mol R=8.314\times10^{-3} kJ/mol

    Substituting the values in the expression of ΔH\Delta H

    ΔH=(742.7)+(0.5)(298)(8.314×103)\Delta H=(-742.7)+(0.5)(298)(8.314\times10^{-3})
             =742.7+1.2=-742.7+1.2
             =741.5 kJ/mol=-741.5\ kJ/mol
  • Question 3
    1 / -0
    Following reaction occurs at 25oC{25}^{o}C:
    2NO(g,1×105atm)+Cl2(g,1×102atm)2NOCl (g,1×102atm)2NO(g, 1\times { 10 }^{ -5 }atm)+{ Cl }_{ 2 }(g, 1\times { 10 }^{ -2 }atm)\rightleftharpoons 2NOCl\ \left( g, 1\times { 10 }^{ -2 }atm \right)
    ΔGo\Delta { G }^{ o } is_______________.
    Solution
    2NO1×105atm+Cl21×102atm2NOCl1×102atm\underset { 1 \times 10^{-5} atm}{2NO}+\underset {1 \times 10^{-2}atm}{Cl_2} \rightleftharpoons \underset { 1 \times 10^{-2} atm}{2NOCl}
    Kp=(PNOCl)2(PNO)2(PCl2)K_p= \cfrac {(P\quad NOCl)^2}{(P\quad NO)^2(P\quad Cl_2)}
           =(1×102)2(1×105)2×1×102=\cfrac {(1 \times 10^{-2})^2}{(1 \times 10^{-5})^2\times 1 \times 10^{-2}}
           =1×1021×1010=1×108=\cfrac { 1 \times 10^{-2}}{1 \times 10^{-10}}=1 \times 10^8
    Δu=RTlnKp=8.314×298×ln108\Delta u= -RTln K_p= -8.314 \times 298 \times ln 10^8
           =8.314×298×8×ln10= -8.314 \times 298 \times 8 \times ln10
           =8.314×298×8×2.3= -8.314 \times 298 \times 8 \times 2.3
           =45.587J/Kg= -45.587 J/Kg
           =45.6kJ=-45.6 kJ .
  • Question 4
    1 / -0
    100ml100ml of O2{O}_{2} gas diffuses in 1010 seconds. 100ml100ml of gas xx diffuses in tt seconds. Gas xx and time tt can be respectively:
    Solution
    If the volume of both the gases are same then,
    t2t1=M2M1\cfrac{t_2}{t_1}=\sqrt{\cfrac{M_2}{M_1}}
    GIven :
    t1=10st_1=10s
    M1=32gM_1=32g
    t2=tt_2=t
    M2=MM_2=M
    Now, t10=M32\cfrac{t}{10}=\sqrt{\cfrac{M}{32}}

    Since two unknowns are present we can substitute the values from the given options and then find the right answer. 
    For M=2 t=2.5secM=2\ t=2.5sec
    2.510=232\cfrac{2.5}{10}=\sqrt{\cfrac{2}{32}}
    14=116\cfrac{1}{4}=\sqrt{\cfrac{1}{16}}
    14=14\cfrac{1}{4}=\cfrac{1}{4}
    LHS=RHS

    Hence, Gas x is H2H_2 and time t is 2.5seconds2.5seconds
  • Question 5
    1 / -0
    When two moles of Hydrogen atoms join together to form a mole of hydrogen molecules in closed rigid vessel with diathermic walls:
    H(g)+H(g)H2(g)H(g)+H(g)\longrightarrow { H }_{ 2 }(g)
    Solution
    H2H_{2} has two atoms. So, a mole of H2H_{2} should have 1 mole of H2H_{2} molecules, and when broken down into atoms, 2 moles of atoms.
    H(g)+H(g)H2(g)H(g)+H(g)\rightarrow H_{2}(g).
    So, from the above reaction w<0w<0.
  • Question 6
    1 / -0
    The difference between the heat of reaction at constant pressure and constant volume for the reaction given below at 25oC{25}^{o}C in KJ is: 
    2C6H6(l)+15O2(g)12CO2(g)+6H2O(l)\quad 2{ C }_{ 6 }{ H }_{ 6(l) }+15{ O }_{ 2(g) }\longrightarrow 12{ CO }_{ 2(g) }+6{ H }_{ 2 }{ O }_{ (l) }
    Solution
    The difference between heats of reaction at constant pressure and constant volume for the following reaction at 25oC25^oC in kJ
    2C6H6(l)+15O2(g)12CO2(g)+6H2O(l)2C_6H_{6_(l)}+15O_{2_(g)}\longrightarrow12CO_{2_(g)}+6H_2O_{(l)}
    Δn=1215=3\Delta n=12-15=-3
    ΔH=ΔE+RTΔn\Delta H=\Delta E+RT\Delta n
    ΔHΔE=RTΔn\Delta H-\Delta E=RT\Delta n
                        =(3)×8.314×103×298=(-3)\times8.314\times10^{-3}\times298
                        =7.443 kJ/mole=-7.443\ kJ/mole
  • Question 7
    1 / -0
    The normal boiling point of a liquid A is 350 K. ΔHvap\Delta { H }_{ vap } at normal boiling point is 35 KJ/mole. Pick out the correct statement(s). (Assume ΔHvap\Delta { H }_{ vap } to be independent of pressure).
    Solution
    The equilibrium at the boiling point is
    A(1)A(g);ΔvapH=35 kJ/molA(1)\leftrightharpoons A(g); \Delta_{vap}H=35\ kJ/mol
    ΔvapS=ΔvapH350K=100 J/Kmol\Delta_{vap}S=\frac{\Delta_{vap}H}{350 K}=100\ J/Kmol
    At 350 K and 0.5 atm
    The boiling point (Tvap)(T_{vap}) will be lower at the decreased pressure.
    Since Tvap<35K,ΔvapS>100 J/KmolT_{vap}<35K, \Delta_{vap}S>100\ J/Kmol
  • Question 8
    1 / -0
    Given standard enthalpy of formation of COCO (-110 KJ mol1{ mol }^{ -1 }) and CO2C { O }_{ 2 } (-394 KJ mol1{ mol }^{ -1 } ). The heat of combustion when one mole of graphite burns is:
    Solution
    Enthalpy of formation of CO=110kJ/molCO=-110\, kJ/mol
    Enthalpy of formtaion of CO2=394kJ/molCO_{2}=-394\, kJ/mol
    C+O2CO2C+O_{2}\rightarrow CO_{2}              Hc=394kJ/mol\triangle H_{c}=-394\, kJ/mol

    Formation of Carbon monoxide
    C+12O2COC+\cfrac{1}{2}O_{2}\rightarrow CO           Hc=110kJ/mol\triangle H_{c}^{'}=-110\, kJ/mol

    ΔH=ΔHcΔHc\Delta H=\Delta H_{c}- \Delta H_{c}^{'}
    =394(110)=-394-(-110)
    =394+110=-394+110
    =284kJ/mol=-284\,kJ/mol

    Hence, the correct option is BB
  • Question 9
    1 / -0
    For the reaction, X2O4(l)2XO2(g){ X }_{ 2 }{ O }_{ 4 }\left( l \right) \longrightarrow 2X{ O }_{ 2 }\left( g \right)
    ΔU=2.1 k cal\Delta U=2.1\ k\ cal, ΔS=20 cal K1\Delta S=20\ cal\ { K }^{ -1 } at 300 K Hence , ΔG\Delta G is_________.
    Solution
    The change in Gibbs free energy is given byΔG=ΔHTΔS\Delta G=\Delta H -T\Delta S
    where,
    ΔH=\Delta H=enthalpy of the reaction
    ΔS=\Delta S=entropy of reaction

    Thus in order to determine ΔG\Delta G, the value of ΔH\Delta H must be known. The value of $$\Delta H$ can be calculated by the reaction,ΔH=ΔU+ΔngRT\Delta H=\Delta U+\Delta n_gRT
    where,
    ΔU=\Delta U=change in internal energy
    Δng=\Delta n_g=(number of moles of gaseous product)-(number of moles of gaseous reactant)=20=2=2-0=2
    R=R=gas constant=2 cal=2\ cal

    ΔU=2.1 kcal=2.1×103 cal\Delta U=2.1\ kcal=2.1\times10^3\ cal
    therefore,

    ΔH=(2.1×103)+(2×2×300)=3300 cal\Delta H=(2.1\times10^3)+(2\times2\times300)=3300\ cal

    Hence,
    ΔG=ΔTΔS\Delta G=\Delta -T\Delta S
    ΔG=3300(300×20)\Delta G=3300 -(300\times20)
    ΔG=2700 cal\Delta G=-2700\ cal
    ΔG=2.7 kcal\Delta G=-2.7\ kcal

























  • Question 10
    1 / -0
    The heat liberated on complete combustion of 1 mole of CH4{ CH }_{ 4 } gas to CO2(g){ CO }_{ 2 }\left( g \right) and HO2(l){ HO }_{ 2 }\left( l \right) is 890 KJ. Calculate the heat evolved by 2.4 L of CH4{ CH }_{ 4 } on complete combustion.
    Solution
    Given
    The volume of ! mol!\ mol of methane at 298 K298\ K and 1 atm1\ atm pressure is 24L24L. Let's calculate the moles of methane for it's 2.4L2.4L as
    =2.4(1 mole24 L)=2.4\left(\frac{1\ mole}{24\ L}\right)
    =0.10 mol=0.10\ mol
    Now, it says that combustion of 1 mol1\ mol of methane generates 890 J890\ J of heat. We could calculate the heat generated by the combustion of 0.10 moles0.10\ moles of methane as:
    =0.10 mol(890 kJ1 mole)=0.10\ mol\left(\frac{890\ kJ}{1\ mole}\right)
    =89kJ=89kJ
    Therefore, 89kJ of heat will be evolved by the combustion of 2.4L of methane























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