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20 December, 06:21

For a particular reaction, AH = 81.95 kJ/mol and AS = 27.0 J / (mol-K) Calculate AG for this reaction at 298 K. AG = kJ/mol What can be said about the spontaneity of the reaction at 298 K? The system is spontaneous in the reverse direction O The system is at equilibrium O The system is spontaneous as written O O

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  1. 20 December, 09:01
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    The system is spontaneous in the reverse direction

    Explanation:

    According the equation of Gibb's free energy -

    ∆G = ∆H - T∆S

    ∆G = is the change in gibb's free energy

    ∆H = is the change in enthalpy

    T = temperature

    ∆S = is the change in entropy.

    And, the sign of the ΔG, determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium,

    i. e.,

    if

    ΔG < 0, the reaction is Spontaneous

    ΔG > 0, the reaction is non Spontaneous

    ΔG = 0, the reaction is at equilibrium

    from the question,

    ∆H = 81.95 kJ/mol

    (since, 1 KJ = 1000 J)

    ∆H = 81950 J/mol

    ∆S = 27.0 J / (mol-K)

    The ∆G is calculated from the above formula -

    ∆G = ∆H - T∆S

    ∆G = (81950 J/mol) - [ (298 K) x (27.0 J / (mol·K)) ]

    ∆G = (81950 J/mol) - (8046 J/mol)

    ∆G = 73904 J/mol

    ∆G = 73.904 kJ/mol

    Since ΔG > 0, the system is non spontaneous in the forward direction and hence it will be spontaneous in the reverse direction.
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