Ammonium sulphide, (NH_4)_2 S, is stable under standard conditions but decompose

Question

Ammonium sulphide, (NH_4)_2 S, is stable under standard conditions but decomposes rapidly at higher temperatures: (NH_4)_2S(s) leftarrow over rightarrow 2NH_2(g) + H_2S(g) a) Give the expression for K_p (in terms of reactants and products) for this process. b) Calculate K_p for this reaction at standard temperature, 298K. c) Calculate the equilibrium constant K_p at 450K. Assume delta H degree and delta S degree for the reaction to be independent of temperature. d) Calculate the total equilibrium pressure above solid (NH_4)_2S at 450 K.

Explanation / Answer

Q5.

a)

Kp = P-products ^p / P-reactants^r

Kp = (P-NH3)^2 * (P-H2S)

b)

Kp = from dG

dG = -RT*ln(Q)

dG = Gproducts - dGreactants --> dH - T*dS

dH = 2*-45 + -20 - (-210) = 100 kJ/mol

dS = 2*190 + 200 - 170 = 410 J/K

dG = 100 -298*410/1000

dG = -22.18 kJ/mol

dG = -RT*lnK

Kp = exp(22180/(8.314*298))

Kp = 7725.740

c)

find Kp at T = 450 K assume dH /dS independend

dG = dH - T*dS

dG = 100 -450*410/1000 = -84.5 kJ/mol

Kp = exp(84500/(8.314*450))

Kp =6439599798.9

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