Question

g Below which temperature does the following reaction change from being spontaneous to non-spontaneous? 2 HNO3 (aq) + NO (g) → 3 NO2 (g) + H2O (l) ΔHº = +136.5 kJ; ΔSº = +287.5 J/K Below which temperature does the following reaction change from being spontaneous to non-spontaneous? 2 HNO3 (aq) + NO (g) → 3 NO2 (g) + H2O (l) ΔHº = +136.5 kJ; ΔSº = +287.5 J/K 475 K 39.2 K 151 K This reaction is nonspontaneous at all temperatures. This reaction is spontaneous at all temperatures.

Answer 1

Answer:

Below 475 K the reaction changes from being spontaneous to non-spontaneous.

Explanation:

Let's consider the following reaction.

2 HNO₃(aq) + NO(g) → 3 NO₂(g) + H₂O(l)

The spontaneity of the reaction is given by the standard Gibbs free energy (ΔG°). When ΔG° > 0 the reaction is non-spontaneous and when ΔG° < 0 the reaction is spontaneous. ΔG° is related to the standard enthalpy (ΔHº) and the standard entropy (ΔSº) using the following expression.

ΔG° = ΔHº - T.ΔSº

The reaction will change from spontaneous to non-spontaneous when ΔG° = 0. Then,

ΔHº - T.ΔSº = 0

ΔHº = T.ΔSº

T = ΔHº/ΔSº = (136.5 × 10³ J)/(287.5 J/K) = 475 K

Answer 2

Below 475 K , this reaction is non- spontaneous.

The change in free energy is given by the formula;

ΔG = ΔH - TΔS

Where;

• ΔG = change in free energy
• ΔH = change in enthalpy
• T = temperature
• ΔS = change in entropy

Now we must try all the temperatures given;

ΔG = +136.5 × 10^3 - (475 × +287.5)

ΔG = -62.5 J (spontaneous)

ΔG = +136.5 × 10^3 - (39.2 × +287.5)

ΔG = 125230 J (non- spontaneous)

ΔG = +136.5 × 10^3 - (151 × +287.5)

ΔG = 93087.5 J (non- spontaneous)

Hence, below 475 K , this reaction is non- spontaneous.

Learn more about spontaneous reaction: https://brainly.com/question/1217654?