Question

For the reaction 2NH3(g) 202(g)N20(g)+3H20(I) AH 683 kJ and AS° = -366 JK AGo would be less than zero at temperatures (above, below) K. Enter 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 = -683.0 KJ/

ΔS = -366 J/K

= -0.366 KJ/K

use:

ΔG = ΔH - T*ΔS

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 = 181.0 KJ

ΔS = 24.9 J/K

= 0.0249 KJ/K

use:

ΔG = ΔH - T*ΔS

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