Question

For the reaction 2NH3(g) 202(g)N20(g)+3H20(I) AHo 683 kJ and AS° = -366 J/K Kelvin The equilibrium constant, K, would be greater than 1 at temperatures Select above or below in the first box and enter the temperature in the second box. Assume that AH° and AS° are constant

Answer 1

1)

ΔH = -683 KJ

ΔS = -366 J/K

= -0.366 KJ/K

use:

ΔG = ΔH - T*ΔS

For K to be greater than 1, reaction should be spontaneous.

for reaction to be spontaneous, ΔG should be negative

that is ΔG<0

since ΔG = ΔH - T*ΔS

so, ΔH - T*ΔS < 0

-683.0- T *-0.366 < 0

T *0.366 < 683.0

T < 1866 K

Answer: Below 1866 K

2)

ΔH = 181 KJ

ΔS = 24.9 J/K

= 0.0249 KJ/K

use:

ΔG = ΔH - T*ΔS

For K to be greater than 1, reaction should be spontaneous.

for reaction to be spontaneous, ΔG should be negative

that is ΔG<0

since ΔG = ΔH - T*ΔS

so, ΔH - T*ΔS < 0

181.0- T *0.0249 < 0

T *0.0249 > 181.0

T > 7269 K

Answer: Above 7269 K