Why isn't nitric acid produced by oxidizing N2 as follows?(1)N2(g)+2O2(g)→2NO2(g)(2)3NO2(g)+H2O(l)→2HNO3(aq)+NO(g)(3)2NO(g)+O2(g)→2NO2(g)3 N2(g)+6O2(g)+2H2O(l)→4HNO3(aq)+2NO(g)

Q22.70CP - Chemistry: Molecular Nature Of Matter And Change

Q22.70CP

Question

Why isn't nitric acid produced by oxidizing $N_{2}$ as follows?

$(1) N_{2} (g) + 2 O_{2} (g) \to 2 {NO}_{2} (g) (2) 3 {NO}_{2} (g) + H_{2} O (l) \to 2 {HNO}_{3} (a q) + NO (g) (3) \frac{2 NO (g) + O_{2} (g) \to 2 {NO}_{2} (g)}{3 N_{2} (g) + 6 O_{2} (g) + 2 H_{2} O (l) \to 4 {HNO}_{3} (a q) + 2 NO (g)}$

Step-by-Step Solution

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Answer

Reaction 1: represent non-spontaneous reactions at all temperatures, so use of catalyst is required.

Reaction 2: represent non-spontaneous reactions at all temperatures, so use of catalyst is required.

Reaction 3: the reaction of step (3) is spontaneous only at low temperatures

1Step 1: Gibb’s Energy for reaction 1

The thermodynamics of each step - i.e., Gibb's energy of each reaction step can be analyzed to solve the question.

Reaction:

$N_{2} (g) + 2 O_{2} (g) \to 2 {NO}_{2} (g)$

Calculate the reaction's enthalpy and entropy using Appendix for reaction (1).

$∆ H_{r x n} = \sum ∆ H_{P r o d u c t s} - \sum ∆ H_{r e a c t a n t s} ∆ H_{r x n} = [2 mol \cdot Δ H (N O_{2})] - [1 mol \cdot Δ H (N_{2}) + 2 mol \cdot Δ H (O_{2})] ∆ H_{r x n} = [2 mol \cdot 33.20 kJ / mol] - [1 mol \cdot 0.00 kJ / mol + 2 mol \cdot 0.00 kJ / mol] ∆ H_{r x n} = 66.40 kJ ∆ s_{r x n} = \sum s_{P r o d u c t s} - \sum s_{m o l} = ∆ S (N S_{2})] - [1 mol \cdot ∆ S (N_{2}) + 2 mol \cdot ∆ S (O_{2})] ∆ s_{r x n} = [2 mol \cdot 239.90 \frac{J}{mol \cdot K}] - [1 mol \cdot 191.50 \frac{J}{mol \cdot K} + 2 mol \cdot 205.00 \frac{m}{mol \cdot K}]$

$∆ s_{r x n} = - 121.70 \frac{J}{K} = - 0.1217 \frac{kJ}{K}$

The reaction 1 is nonspontaneous at all temperatures because it is endothermic $(∆ H > 0)$ and the overall entropy of the reaction drops. As Gibb's energy is positive, no spontaneous reaction occurs.

$∆ G_{25} = 66.40 kJ - (25 + 273) K \cdot - 0.1217 \frac{kJ}{K} ∆ G_{25} = 102.67 kJ$

Hence, it represents non-spontaneous reactions at all temperatures, so use of catalyst

2Step 2: Gibb’s Energy for reaction 2

2 Reaction

$3 {NO}_{2} (g) + 2 H_{2} O (l) \to 2 {HNO}_{3} (a q) + NO (g)$

Calculate the reaction's enthalpy and entropy using Appendix B for reaction (2).

$∆ H_{r x n} = [2 mol \cdot Δ H (H N O_{3}) + 1 mol \cdot Δ H (N O)] - [3 mol \cdot Δ H (N O_{2}) + 2 mol \cdot Δ H (H_{2} O)]$

$∆ H_{r x n} = [2 mol \cdot - 206.57 kJ / mol + 1 mol \cdot 90.29 kJ / mol] - [3 mol \cdot 33.20 kJ / mol + 2 mol \cdot - 285.83 kJ / mol] ∆ H_{r x n} = 149.21 kJ ∆ s_{r x n} = [2 mol \cdot Δ S (H N O_{3}) + 1 mol \cdot Δ S (N O)] - [3 mol \cdot Δ S (N O_{2}) + 2 mol \cdot Δ S (H_{2} O)]$

$∆ s_{r x n} = [2 mol \cdot 146.00 \frac{J}{mol \cdot K} + 1 mol \cdot 210.65 \frac{J}{mol \cdot K}] - [3 mol \cdot 239.90 \frac{J}{mol \cdot K} + 2 mol \cdot 69.95 \frac{J}{mol \cdot K}]$

$∆ s_{r x n} = - 356.95 \frac{J}{K} = - 0.35695 \frac{kJ}{K}$

The reaction 2 is nonspontaneous at all temperatures because it is endothermic $(D e l t a H > 0)$ and the overall entropy of the reaction drops. For example, Gibb's energy is positive at $25^{°} C$ ,indicating that the reaction is not spontaneous.

$∆ G_{25} = 149.21 kJ - \{(25 + 273) K \times - 0.35695 \frac{kJ}{K}\} ∆ G_{25} = 255.581 kJ$

Therefore, it represents non-spontaneous reactions at all temperatures, so use of catalyst is required.

3Step 3: Gibb’s Energy for reaction 3

Reaction

$2 NO (g) + O_{2} (g) \to 2 {NO}_{2} (g)$

Calculate the reaction's enthalpy and entropy using Appendix B for reaction (2).

$∆ H_{r x n} = [2 mol \cdot Δ H ({NO}_{2})] - [2 mol \cdot Δ H (N O) + 1 mol \cdot Δ H (O_{2})] ∆ H_{r x n} = [2 mol \cdot 33.20 kJ / mol] - [2 mol \cdot 90.29 kJ / mol + 1 mol \cdot 0.00 kJ / mol] ∆ S_{r x n} = [2 mol \cdot ∆ S (N H_{rxn} = - 114.18 kJ - [2 mol \cdot ∆ S (mO) + 1 mol \cdot ∆ S (O_{2})] ∆ S_{r x n} = [2 mol \cdot 239.90 \frac{J}{mol \cdot K}] - [2 mol \cdot 210.65 \frac{J}{mol \cdot K} + 1 mol \cdot 205.00 \frac{J}{mol \cdot K}] {ΔS}_{rxn} = - 146.50 \frac{J}{K} = - 0.1465 \frac{kJ}{K}$

At low temperatures, the reaction 3 is spontaneous because it is exothermic $(D e l t a H > 0)$ and the overall entropy of the process reduces. For instance, for $ $25^{°} C$ ,, Because Gibb's energy is positive, the reaction occurs naturally.

$∆ G_{25} = - 114.18 kJ - (25 + 273) K \cdot - 0.1465 \frac{kJ}{K}$

$∆ G_{25} = - 70.523 kJ$

width="145" style="max-width: none; vertical-align: -9px;" $∆ G_{25} = - 70.523 kJ$

As a result, the reaction of step (3) is spontaneous only at low temperatures.

4Step 4: Overview of the reactions

As a result, the entire reaction is implausible, because the first two reactions are not spontaneous and only the third is.

In the industrial scale, preheated air is mixed with ammonia gas and a catalyst is employed to produce nitric oxide, which is then oxidized and, in the presence of water, nitric acid is formed.

At all temperatures, the reaction steps (1) and (2) indicate non-spontaneous reactions, necessitating the employment of a catalyst. On the other hand, Step (3), is only spontaneous at low temperatures.As a result, the entire reaction is implausible

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