Question

In: Chemistry

Classify each of the following reactions as one of the four possible types summarized in Table...

Classify each of the following reactions as one of the four possible types summarized in Table 19.3: (i) spontanous at all temperatures; (ii) not spontaneous at any temperature; (iii) spontaneous at low T but not spontaneous at high T ; (iv) spontaneous at high T but not spontaneous at low T . (a) N 2 1 g 2 + 3 F 2 1 g 2 ¡ 2 NF 3 1 g 2 ∆ H ° = - 249 kJ; ∆ S ° = - 278 J > K

(b) N 2 1 g 2 + 3 Cl 2 1 g 2 ¡ 2 NCl 3 1 g 2 ∆ H ° = 460 kJ; ∆ S ° = - 275 J > K

(c) N 2 F 4 1 g 2 ¡ 2 NF 2 1 g 2 ∆ H ° = 85 kJ; ∆ S ° = 198 J > K

Solutions

Expert Solution

a)   answer : (iii) spontaneous at low T but not spontaneous at high T

N2 + 3F2 --------------------> 2NF3 ∆ H ° = - 249 kJ; ∆ S ° = - 278 J / K = -0.278 kJ / K

∆ G ° = ∆ H ° - T ∆ S °

          = -249 - T (-0.278)

          = -249 + 0.278 T

for spontaneous ∆ G ° should be negative. for that in this reaction T (temperature should be low)

spontaneous at low T

b)   answer : (ii) not spontaneous at any temperature

N2 + 3 Cl2 -------------------> 2 NCl3 , ∆ H ° = 460 kJ; ∆ S ° = - 275 J / K = - 0.275 kJ/K

∆ G ° = ∆ H ° - T ∆ S °

           = 460   - T (-0.275)

           = 460 + 0.275 T

this reaction never spontaneous because we get ∆ G ° positive at any temperature.

c) answer : (iv) spontaneous at high T but not spontaneous at low T

N2F4 ----------------------> 2 NF2 , ∆ H ° = 85 kJ; ∆ S ° = 198 J > K = 0.198 kJ /k

∆ G ° = ∆ H ° - T ∆ S °

          = 85 - T (0.198)

at high temperature ∆ G ° negative and reaction is spontaneous.


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