Consider the reaction, 2 NF3 (g) ' 2 N2 (g) + 3 F2 (g)
ΔH° = + 260.8 kJ, ΔS° = + 278.5 J/K
At what temperature will the system be at equilibrium, when all gases are present at 1 atm (standard pressure conditions)?
Calculate the value of the equilibrium constant for the reaction above at 500.0°C.
1)
At equilibrium, deltaGo = 0
deltaGo = 0.0 KJ
deltaHo = 260.8 KJ
deltaSo = 278.5 J/K
= 0.2785 KJ/K
we have below equation to be used:
deltaGo = deltaHo - T*deltaSo
0.0 = 260.8 - T *0.2785
T = 936 K
Answer: 936 K
2)
T= 500.0 oC
= (500.0+273) K
= 773 K
we have below equation to be used:
deltaGo = deltaHo - T*deltaSo
deltaGo = 260.8 - 773.0 * 0.2785
deltaGo = 45.5195 KJ
deltaG = 45519.5 J
we have below equation to be used:
deltaG = -R*T*ln Kc
45519.5 = - 8.314*773.0* ln(Kc)
ln Kc = -7.0828
Kc = 8.39*10^-4
Answer: 8.39*10^-4
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