Question

Consider the reaction, 2 NF3 (g) ' 2 N2 (g) + 3 F2 (g) ΔH° =...

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.

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Homework Answers

Answer #1

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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