At what temperature does the following reaction become spontaneous? You may assume that ΔHº and Sº are temperature independent.

Question

At what temperature does the following reaction become spontaneous?

You may assume that ΔHº and Sº are temperature independent.

2NO (g) + 2H₂(g) --> N₂(g) + 2H₂O (g)

1. I have the values for H and S, but I'm confused about how to utilize them, any advice is welcome!

Answer 1

So, since you did not provide the actual values, I will perform the calculation to show you how to use them. I will use approximate values and you can substitute the values that you have. Also, not sure if you have ∆Hformation and ∆Sformation or if you have ∆Hrxn and ∆Srxn. It makes a big difference as to the way we conduct the calculations.

Assuming you have ∆Hf and ∆Sf, here is the general approach. These are the approximate values I will be using. Again, you will need to sub your values.

NO(g): ∆Hf = 90 kJ/mole; ∆Sf = 211 J/molK

H₂(g): ∆Hf = 0; ∆Sf = 131 J/molK

N₂(g): ∆Hf = 0; ∆Sf = 192 J/molK

H₂O(g): ∆Hf = -242 kJ/mol; ∆Sf = 189 J/molK

Reaction: 2NO (g) + 2H₂(g) --> N₂(g) + 2H₂O (g)

∆Hrxn = ∑n∆Hf products - ∑n∆Hf products = (0 + 2*-242) - (2*90 + 2*0) = -484 - 180 = -664 kJ/mol

∆Srxn = ∑n∆Sf products - ∑n∆Sf products = (192 + 2*189) - (2*211 + 2*131) = 570 - 684 = -114 J/molK

NOTE: ∆S is in J/mol and ∆H is in kJ/mole, so we will change ∆S to -0.114 kJ/molK

∆G = ∆H - T∆S

For a spontaneous reaction, we will set ∆G = 0 and solve for T:

0 = ∆H - T∆S

∆H = T∆S

T = ∆H / ∆S

T = -664 / -0.114

T = 5825K

T = 5825K - 273 = 5552ºC

(be sure to check all of the math)