Question

For each of these equations, predict the effects of temperature on the spontaneity of the reaction.

2N2O(g) es001-1.jpg 2N2(g) + O2(g) es001-2.jpgH < 0; es001-3.jpgS > 0

H2O(l) es002-1.jpg H2O(g) es002-2.jpgH > 0; es002-3.jpgS > 0

3O2(g) es003-1.jpg 2O3(g) es003-2.jpgH > 0; es003-3.jpgS < 0

C(s) + 2Cl2(g) es004-1.jpg CCl4(g) es004-2.jpgH < 0; es004-3.jpgS < 0

A 4-column table with 4 rows. Column 1 is labeled delta H. Column 2 is labeled Delta S. Column 3 is labeled Delta G. Column 4 is labeled Spontaneity. 4 rows of entries: Positive, negative, positive, nonspontaneous for all T. Negative, positive, negative, spontaneous for all T. Positive, negative, positive or negative, spontaneous at high T nonspontaneous at low T. Negative, negative, positive or negative, spontaneous at low T nonspontaneous at high T.

Answer 1

Use ΔG = ΔH − TΔS and the table you gave (signs → spontaneity).

1) 2 N2O(g) → 2 N2(g) + O2(g)
- ΔH < 0, ΔS > 0 → (−, +) case.
- ΔG is always negative → spontaneous at all T.

2) H2O(l) → H2O(g)
- ΔH > 0, ΔS > 0 → (+, +) case.
- ΔG = ΔH − TΔS becomes negative only above a threshold T = ΔH/ΔS.
- So vaporization is nonspontaneous at low T and spontaneous at high T (becomes spontaneous at T greater than the boiling point at the given pressure).

3) 3 O2(g) → 2 O3(g)
- ΔH > 0, ΔS < 0 → (+, −) case.
- ΔG is always positive → nonspontaneous at all T (requires input of energy/photons).

4) C(s) + 2 Cl2(g) → CCl4(g)
- ΔH < 0, ΔS < 0 → (−, −) case.
- ΔG = ΔH − TΔS will be negative only below a threshold T = ΔH/ΔS (note ΔH/ΔS is positive since both are negative).
- So spontaneous at low T, nonspontaneous at high T.