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Thermodynamics Test - 15

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Thermodynamics Test - 15
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  • Question 1
    1 / -0
    Born - Haber cycle may be used
    Solution
    Born - Haber cycle is used to calculate the lattice energy for one mole of the ionic compound formed. According to this cycle, the value of lattice energy and enthalpy of formation is equal to the summation of sublimation energy of the metal atom, Ionization energy required for the formation of the cation, enthalpy of atomization and electron affinity. 
    From this formula, we can calculate any of the above values.

  • Question 2
    1 / -0

    Gibbs energy change Δ\Delta G is related to equilibrium constant K as:

    Solution
    ΔGo=RT\Delta G^{o}=-RT  Ink

    Gibbs energy change ΔG\Delta G is related to equilibrium constant K as above
  • Question 3
    1 / -0

    Which of the following relation is incorrect?

    Solution
    ΔG=ΔHTΔS\Delta G=\Delta H-T\Delta S

    ΔS=ΔHΔGT\therefore \Delta S=\frac{\Delta H-\Delta G}{T}

    ΔG=Wnon exp.\Delta G=-W_{non\ exp.}

    Gibbs free energy is equal to useful work for a spontaneous process

    PVP-V  work =PΔV=-P\Delta V
  • Question 4
    1 / -0

    Assertion (A): The value of ΔE\Delta E is negative if the internal energy of a system increases. 

    Reason (R): The value of QQ is negative if heat is absorbed by the system.

    Solution

    The value of ΔE\Delta E is positive if the internal energy of a system increases. The value of ΔQ\Delta Q is positive, if heat is absorbed by the system.

  • Question 5
    1 / -0

    A heat engine absorbs heat Q1Q_{1} at temperature T1T_{1} and heat Q2Q_{2} at temperature T2T_2 . Work done by the engine is (Q1+Q2)J(Q_{1}+Q_{2}) J . This data:

    Solution
    According, First Law of Thermodynamics,

    Change in E= Q +E = Q  + work done

    Change in internal energy is zero for heat engines because initial and final states are same.

    Total work done == -Total heat exchange =(Q1+Q2)= -(Q_1+Q_2).
  • Question 6
    1 / -0

    Correct relation among the following.

    Solution
    ΔGsystem=TΔStotal\Delta G_{system}=T\Delta S_{total}
    ΔGsystem=ΔHsystemTΔSsystem\Delta G_{system}=\Delta H_{system}-T\Delta S_{system} 
    If an isolated system (Q == 0) is at constant pressure (Q =ΔH=\Delta H) then
    ΔH=0\Delta H=0
    Therefore, the Gibbs free energy of an isolated system is 
    ΔGsystem=TΔStotal\Delta G_{system}=-T\Delta S_{total}
  • Question 7
    1 / -0

    Gibbs -Helmholtz equation is______

    Solution
    ΔG=ΔHTΔS\Delta G=\Delta H-T\Delta S

    is the Gibbs Helmholtz equation.

    Option A is correct.
  • Question 8
    1 / -0
    Which of the following statements is incorrect?
    Solution
    Critical temperature of the substance is the temperature at a substance's critical point.
    Above TcT_c we cannot liquify gas keep pressure constant although on \uparrow pressure we can liquify.
  • Question 9
    1 / -0
    Identify the correct statement for change of Gibbs energy for a system (Δ\DeltaGsystem_{system}) at constant temperature and pressure:
    Solution
    At equilibrium, the entropy of system is maximum. Therefore change in entropy i.e; S=0\triangle S=0

     Process Sign of S\triangle S Sign of G\triangle G
     Spontaneous Positive, Stotal>0\triangle S_{total} >0Negative, G<0\triangle G <0 
     Non-SpontaneousNegative, Stotal<0\triangle S_{total} < 0 Positive, G>0\triangle G>0 
     EquilibriumStotal=0\triangle S_{total}=0  G=0\triangle G=0
  • Question 10
    1 / -0
    Cell reaction is spontaneous when : 
    Solution

    Hint: The Gibbs free energy is related to the spontaneity of a reaction.

    Explanation: If the Gibbs free energy is negative then the reaction is spontaneous.

    ΔG= nFE\Delta {G^ \circ } =  - nF{E^ \circ }

    For ΔG<0\Delta {G^ \circ } < 0, the reaction, Ecell{E^ \circ }_{cell} should be positive so the correct answer is (A)

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