Question #5ecc9

1 Answer
Jun 3, 2017

Answer:

#K_c = 1.96x10^5#

Explanation:

#2N_2O_5(s) rightleftharpoons 4NO_2(g) + O_2(g)#

Use #DeltaG^o = -RTlnK_c"# => #K_c = e^(-(DeltaG^o"/"RT"))#

#DeltaG^o = DeltaH^o - TDeltaS^o#

#T = 298.15K#

#DeltaH^o = Sigman*DeltaH_f^o(Products) - Sigman*DeltaH_f^o(Reactants)#
#= [4DeltaH_f^o(NO_2(g)) + 1DeltaH_f^o(O_2(g))] - [2DeltaH_f^o(N_2O_5(s))]#
#= [4(33.18) + 1(0)]Kj - [2(11.3)] Kj#
#= +110.12Kj#

#DeltaS^o = Sigman*S^o(Products) - Sigman*S^o(Reactants)#
#= [4S^o(NO_2(g)) + 1S^o(O_2(g))] - [2S^o(N_2O_5(s))]#
#= [4(239.95) + 1(205.03)]J/K - [2(347.19)]J/K#
#= +470.45J/K = +0.47045"Kj"/K#

#DeltaG^o at (298.15K):#
#DeltaG^o = DeltaH^o - TDeltaS^o#
#= [110.12Kj] - (298.15K)[0.47045"Kj"/K]#
#= -30.14Kj#

#DeltaG^o = -RTlnK_c"#
=> #lnK_c = -(DeltaG^o"/"R*T")##= - [(-30.14)/(0.008314*298.15)]"mole"#
#= 12.18322mol"#; drop units in final value of #K_c#

#K_c = e^12.18322# =#1.96x10^5#