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15 October, 18:09

What is the ∆G for the following reaction under standard conditions (T = 298 K) for the formation of NH4NO3 (s) ?

2NH3 (g) + 2O2 (g) NH4NO3 (s) + H2O (l)

Given:

NH4NO3 (s) : ∆Hf = - 365.56 kJ ∆Sf = 151.08 J/K.

NH3 (g) : ∆Hf = - 46.11 kJ ∆Sf = 192.45 J/K.

H2O (l) : ∆Hf = - 285.830 kJ ∆Sf = 69.91 J/K.

O2 (g) : ∆Hf = 0.00 kJ ∆Sf = 205 J/K.

A. 186.6 kJ

B. 6.9 kJ

C. - 10.4 kJ

D. - 126.3 kJ

E. - 382 kJ

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Answers (1)
  1. 15 October, 21:14
    0
    Under constant temperature and pressure the change in free energy (ΔG) equals the difference between the change in enthalpy (ΔH) of the system and the product of the temperature, T, times the change in the entropy (ΔS).

    This is: ΔG = ΔH - TΔS

    The reaction is:

    2NH3 (g) + 2O2 (g) - > NH4NO3 (s) + H2O (l)

    So,

    ΔGf = ΔGf of the products - ΔGf of the reactants

    ΔGf products = ΔHf products - TΔS products

    ΔGf reactants = ΔHf reactants - TΔS reactants

    T = 298 K

    ΔHf products = ΔHf NH4NO3 (s) + ΔHf H2O (l) = - 365.56 kJ - 285.830 kJ = - 651.390 kJ

    TΔSf products = TΔSf NH4NO3 + TΔSf H2O (l) = 298K (151.08 J/K + 69.91 J/k) = 65,855.02 J = 65.855 kJ

    ΔGf products = - 651.390kJ - 65.855kJ = - 717.245 J

    ΔHf reactants = 2ΔHf NH3 (g) + 2ΔHf O2 (g) = 2 * (-46.11kJ) + 2 * (0.00kJ) = - 92.22 kJ

    TΔSf reactants = 298k*2 * (ΔSf NH3) + 298k*2 * (ΔSf O2) = 298k*2 (192.45 J/K + 205 J/K) = 236,880.2 J = 236.880 kJ

    ΔG f reactants = - 92.22 kJ - 236.880kJ = - 329.100 kJ

    ΔGf = ΔG f products - ΔG f reactants = - 717.245kJ - (-329.100kJ) = - 388.145 kJ

    Answer: - 388.145 kJ
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