Consider chemical reaction N 2 + O 2 --> 2NO Using the integrated Gibbs-Helmholt

Question

Consider chemical reaction N2 + O2 --> 2NO

Using the integrated Gibbs-Helmholtz equation, find the minimial temperature when the reaction will become spontaneous at 1 atm.

(1) Assume ?Hr(T)=const=?Hr(To)

(2) Correct the answer for the more accurate case of

?Hr(T)= ?Hr(To) + ?rCp(T - To), with all CPm = const (can use Table 4.1). Is heat released or absorbed (for the latter case).

Table 4.1:

3111 9 0 35 47 17 17 1 on ole ht 00 00 200 58 58 29 74 5 8 28 14 17 30 44 46 65 92 92 63 63 62 18 4 80 1. 220 14730 2748 18196272791 905987 223233 4929 9199 2232 2197 .9 906 551 779 112 874 0 05 185 7222 9 0 3 211 27 4429 73790 7936271 51 .9 .4 2700 131 0929 7498351173

Explanation / Answer

chemical reaction N2 + O2 --> 2NO

H° = 180.5kJ/mol and G° = 175 kJ/mol at 25°C,

first calculate the dS° at 25°C,

dG° = dH° - TdS°
175.0 = 180.5 - (298K)(dS°)

-5.5 KJ =- (298K)(dS°)

dS° = + 0.0185 kJ/K

For spontaneous calculate the temperature:

at equilirium dG = 0
dG = dH - TdS

0 = 180.5 - (T)(0.0185 kJ/K)
180.5 KJ = (T)(0.0185   kJ/K)
T = 9756.8 K

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