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NEET-XI-Chemistry

06: Chemical Thermodynamics

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  • Qstn #15
    Calculate the enthalpy change for the process

    CCl4(g) → C(g) + 4Cl(g)

    and calculate bond enthalpy of C-Cl in CCl4(g).

    ΔvapHθ (CCl4) = 30.5 kJ mol-1.

    ΔfHθ (CCl4) = -135.5 kJ mol-1.

    ΔaHθ (C) = 715.0 kJ mol-1, where ΔaHθ is enthalpy of atomisation

    ΔaHθ (Cl2) = 242 kJ mol-1
    Ans : The chemical equations implying to the given values of enthalpies are:

    ΔvapHθ = 30.5 kJ mol-1

    ΔaHθ = 715.0 kJ mol-1

    ΔaHθ = 242 kJ mol-1

    ΔfH = -135.5 kJ mol-1

    Enthalpy change for the given process can be calculated using the following algebraic calculations as:

    Equation (ii) + 2 × Equation (iii) - Equation (i) - Equation (iv)

    ΔH = ΔaHθ(C) + 2ΔaHθ (Cl2) - ΔvapHθ - ΔfH

    = (715.0 kJ mol-1) + 2(242 kJ mol-1) - (30.5 kJ mol-1) - (-135.5 kJ mol-1)

    ΔH = 1304 kJ mol-1

    Bond enthalpy of C-Cl bond in CCl4 (g)



    = 326 kJ mol-1
  • Qstn #16
    For an isolated system, ΔU = 0, what will be ΔS?
    Ans : ΔS will be positive i.e., greater than zero

    Since ΔU = 0, ΔS will be positive and the reaction will be spontaneous.
  • Qstn #17
    For the reaction at 298 K,

    2A + B → C

    ΔH = 400 kJ mol-1 and ΔS = 0.2 kJ K-1 mol-1

    At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
    Ans : From the expression,

    ΔG = ΔH - TΔS

    Assuming the reaction at equilibrium, ΔT for the reaction would be:



    (ΔG = 0 at equilibrium)



    T = 2000 K

    For the reaction to be spontaneous, ΔG must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.
  • Qstn #18
    For the reaction,

    2Cl(g) → Cl2(g), what are the signs of ΔH and ΔS ?
    Ans : ΔH and ΔS are negative

    The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, ΔH is negative.

    Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ΔS is negative for the given reaction.
  • Qstn #19
    For the reaction

    2A(g) + B(g) → 2D(g)

    ΔUθ = -10.5 kJ and ΔSθ= -44.1 JK-1.

    Calculate ΔGθ for the reaction, and predict whether the reaction may occur spontaneously.
    Ans : For the given reaction,

    2 A(g) + B(g) → 2D(g)

    Δng = 2 - (3)

    = -1 mole

    Substituting the value of ΔUθ in the expression of ΔH:

    ΔHθ = ΔUθ + ΔngRT

    = (-10.5 kJ) - (-1) (8.314 × 10-3 kJ K-1 mol-1) (298 K)

    = -10.5 kJ - 2.48 kJ

    ΔHθ = -12.98 kJ

    Substituting the values of ΔHθ and ΔSθ in the expression of ΔGθ:

    ΔGθ = ΔHθ - TΔSθ

    = -12.98 kJ - (298 K) (-44.1 J K-1)

    = -12.98 kJ + 13.14 kJ

    ΔGθ = + 0.16 kJ

    Since ΔGθ for the reaction is positive, the reaction will not occur spontaneously.
  • Qstn #20
    The equilibrium constant for a reaction is 10. What will be the value of ΔGθ? R = 8.314 JK-1 mol-1, T = 300 K.
    Ans : From the expression,

    ΔGθ = -2.303 RT logKeq

    ΔGθ for the reaction,

    = (2.303) (8.314 JK-1 mol-1) (300 K) log10

    = -5744.14 Jmol-1

    = -5.744 kJ mol-1
  • Qstn #21
    Comment on the thermodynamic stability of NO(g), given

    N2(g) + O2(g) → NO(g) ; ΔrHθ = 90 kJ mol-1

    NO(g) +O2(g) → NO2(g) : ΔrHθ= -74 kJ mol-1
    Ans : The positive value of ΔrH indicates that heat is absorbed during the formation of NO(g). This means that NO(g) has higher energy than the reactants (N2 and O2). Hence, NO(g) is unstable.

    The negative value of ΔrH indicates that heat is evolved during the formation of NO2(g) from NO(g) and O2(g). The product, NO2(g) is stabilized with minimum energy.

    Hence, unstable NO(g) changes to stable NO2(g).
  • Qstn #22
    Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ΔfHθ = -286 kJ mol-1.
    Ans : It is given that 286 kJ mol-1 of heat is evolved on the formation of 1 mol of H2O(l). Thus, an equal amount of heat will be absorbed by the surroundings.

    qsurr = +286 kJ mol-1

    Entropy change (ΔSsurr) for the surroundings =



    ΔSsurr = 959.73 J mol-1 K-1
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