Group 15 (Nitrogen family) — N₂ inertness; NH₃ synthesis (Haber); HNO₃ (Ostwald); oxides of N (N₂O, NO, NO₂, N₂O₄, N₂O₅): Exercises

Two symbols recur, so let us fix them in plain words before any problem uses them.

Level 1 — Recognition

L1.1

State the oxidation state of nitrogen in each: $N_{2} O, N O, N O_{2}, N_{2} O_{5}$ .

Recall Solution

Rule: sum of oxidation states $= 0$ (all neutral), oxygen $= - 2$ . Let N $= x$ .

$N_{2} O$ : $2 x + (- 2) = 0 \Rightarrow 2 x = 2 \Rightarrow x = + 1$ .
$N O$ : $x + (- 2) = 0 \Rightarrow x = + 2$ .
$N O_{2}$ : $x + 2 (- 2) = 0 \Rightarrow x = + 4$ .
$N_{2} O_{5}$ : $2 x + 5 (- 2) = 0 \Rightarrow 2 x = 10 \Rightarrow x = + 5$ . WHAT we did: wrote "sum $= 0$ " and solved for the only unknown. WHY: oxygen's state is fixed and the molecule is neutral, so N is forced.

L1.2

Write the balanced Haber equation with its sign of $Δ H$ , and say how many gas moles are on each side.

Recall Solution

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g), Δ H = - 92 kJ/mol$ Left: $1 + 3 = 4$ moles of gas. Right: $2$ moles. Exothermic ( $Δ H < 0$ ).

Level 2 — Application

L2.1

In the Haber synthesis at equilibrium a vessel holds $p_{N_{2}} = 4 atm$ , $p_{H_{2}} = 2 atm$ , $p_{N H_{3}} = 8 atm$ . Compute $K_{p}$ (units atm $^{- 2}$ ).

Recall Solution

$K_{p} = \frac{p _{N H_{3}}^{2}}{p _{N_{2}} p _{H_{2}}^{3}} = \frac{8 ^{2}}{4 \cdot 2 ^{3}} = \frac{64}{4 \cdot 8} = \frac{64}{32} = 2.0 atm^{- 2}$ WHY this formula: exponents equal the balancing coefficients — $2$ for $N H_{3}$ , $1$ for $N_{2}$ , $3$ for $H_{2}$ . That is the definition, no extra physics needed.

L2.2

Ammonia is oxidised in Ostwald Step 1: $4 N H_{3} + 5 O_{2} \to 4 N O + 6 H_{2} O$ . Starting from $4.0$ mol $N H_{3}$ with excess $O_{2}$ , how many moles of $N O$ form (100% yield)?

Recall Solution

Coefficients give $4 N H_{3} : 4 N O$ , a $1 : 1$ ratio. So $4.0$ mol $N H_{3} \Rightarrow 4.0$ mol $N O$ . Change in N state: $- 3 \to + 2$ , i.e. each N loses $5$ electrons.

Level 3 — Analysis

L3.1

Ostwald Step 3 is $3 N O_{2} + H_{2} O \to 2 H N O_{3} + N O$ . Show it is a disproportionation by tracking N's oxidation state, and state how many of the three N atoms went up vs down.

Recall Solution

N in $N O_{2}$ is $+ 4$ (from L1.1). Products: N in $H N O_{3}$ is $+ 5$ ; N in $N O$ is $+ 2$ .

$+ 4 \to + 5$ : oxidation (up by 1). Two of the three N atoms do this (coefficient $2$ on $H N O_{3}$ ).
$+ 4 \to + 2$ : reduction (down by 2). One N atom does this (coefficient $1$ on $N O$ ). Same element both oxidised and reduced ⇒ disproportionation. Electron check: $2$ atoms lose $1 e^{-}$ each ( $= 2 e^{-}$ up) balances $1$ atom gaining $2 e^{-}$ ( $= 2 e^{-}$ down). ✔ See Disproportionation Reactions.

L3.2

Using the numbers from the parent note, evaluate $941 - 3 \times 159$ (kJ/mol) and explain in one line what this positive gap means physically.

Recall Solution

$3 \times 159 = 477$ ; $941 - 477 = 464 kJ/mol$ . Meaning: forming the $N \equiv N$ triple bond releases $464$ kJ/mol more than three separate N–N single bonds would — the triple bond is hugely favoured, which (via the huge energy needed to break it) underlies N₂'s kinetic inertness.

Level 4 — Synthesis

L4.1

Combine all three Ostwald steps to find: how many moles of $N H_{3}$ are needed to produce $2.0$ mol of $H N O_{3}$ , if the NO from Step 3 is recycled (so it does not need fresh $N H_{3}$ )?

Recall Solution

Recycling NO makes the net process behave as: each N atom that enters as $N H_{3}$ eventually ends up as one $H N O_{3}$ (the NO is fed back to become more NO₂ and re-enters Step 3). The net overall equation is $N H_{3} + 2 O_{2} \to H N O_{3} + H_{2} O$ a $1 : 1$ N balance. So $2.0$ mol $H N O_{3}$ needs $2.0$ mol $N H_{3}$ . Cross-check by states: N goes $- 3 \to + 5$ , a loss of $8 e^{-}$ ; the $2 O_{2}$ (each O $0 \to - 2$ ) accepts $8 e^{-}$ . Balanced. ✔

L4.2

Without recycling, Step 3 ( $3 N O_{2} + H_{2} O \to 2 H N O_{3} + N O$ ) converts only a fraction of its $N O_{2}$ nitrogen to $H N O_{3}$ in one pass. What fraction of the $N O_{2}$ N atoms become $H N O_{3}$ per pass, and hence what fraction is "lost" as NO if not recycled?

Recall Solution

Of $3$ N atoms in, $2$ leave as $H N O_{3}$ and $1$ leaves as $N O$ . Fraction to $H N O_{3} = \frac{2}{3} \approx 0.667$ ; fraction "lost" as $N O = \frac{1}{3} \approx 0.333$ . This $1/3$ is exactly why industry recycles the NO — see L3.1.

Level 5 — Mastery

L5.1

The dimerisation $2 N O_{2} (g) ⇌ N_{2} O_{4} (g)$ is exothermic and reduces gas moles from $2 \to 1$ . (a) Predict the direction of shift when you cool the mixture and when you compress it. (b) $N O_{2}$ is brown and paramagnetic; $N_{2} O_{4}$ is colourless and diamagnetic. State the observed colour change on cooling and the magnetic change, justifying each from electron count.

Recall Solution

(a) Apply Le Chatelier Principle:

Cooling removes heat; the exothermic (forward) direction releases heat, so the system shifts forward → more $N_{2} O_{4}$ .
Compression raises pressure; the side with fewer gas moles ( $N_{2} O_{4}$ , $1$ mole) relieves it, so it shifts forward → more $N_{2} O_{4}$ . (b) $N O_{2}$ has an odd total valence electron count (an unpaired electron on N) ⇒ paramagnetic and coloured brown. When two $N O_{2}$ join, the two unpaired electrons pair up in a new N–N bond ⇒ $N_{2} O_{4}$ has all electrons paired ⇒ diamagnetic and colourless. So on cooling: brown fades to colourless, and the gas turns from paramagnetic to diamagnetic. See figure below.

Figure — Group 15 (Nitrogen family) — N₂ inertness; NH₃ synthesis (Haber); HNO₃ (Ostwald); oxides of N (N₂O, NO, NO₂, N₂O₄, N₂O₅)

L5.2

A student claims $N_{2} O_{5}$ is the anhydride of $H N O_{2}$ . Refute this using oxidation states, and name the correct anhydride of $H N O_{2}$ .

Recall Solution

Find N states. In $H N O_{2}$ : $+ 1 + x + 2 (- 2) = 0 \Rightarrow x - 3 = 0 \Rightarrow x = + 3$ . In $H N O_{3}$ : $+ 1 + x + 3 (- 2) = 0 \Rightarrow x - 5 = 0 \Rightarrow x = + 5$ . In $N_{2} O_{5}$ : $x = + 5$ (L1.1). An acid's anhydride must have the same N oxidation state as the acid.

$N_{2} O_{5}$ ( $+ 5$ ) matches $H N O_{3}$ ( $+ 5$ ) — so $N_{2} O_{5}$ is the anhydride of == $H N O_{3}$ ==, not $H N O_{2}$ .
The anhydride of $H N O_{2}$ ( $+ 3$ ) is the oxide with N at $+ 3$ , namely == $N_{2} O_{3}$ ==. The student's claim is refuted: states $+ 5 \neq = + 3$ .

Connections

Oxidation States and Redox — the single rule powering L1, L3, L5
Le Chatelier Principle — direction predictions in L5.1
Disproportionation Reactions — L3.1 Ostwald Step 3
N2 molecule MO diagram — origin of the 941 kJ/mol used in L3.2