The reaction N $_{2}$ O $_{4}$ (g) $⇌$ 2NO $_{2}$ (g) is
carried out at 298 K and 20 bar. Five moles of each of N $_{2}$ O $_{4}$ and
NO $_{2}$ , are taken initially. Given: $Δ$ $_{f}$ G $^{\circ}$
(N $_{2}$ O $_{4}$ ) = 100 kJ mol $^{- 1}$ and $Δ$ $_{f}$ G $^{\circ}$
(NO $_{2}$ ) = 50 kJ mol $^{- 1}$ . The reaction proceeds at an initial pressure
of 20 bar. Determine the direction in which the reaction proceeds to achieve
equilibrium. Also determine the amounts of N $_{2}$ O $_{4}$ and NO $_{2}$ when the reaction attains equilibrium.

= 8.333, = 1.667

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= 1.667, = 3.333

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= 3.333, = 1.667

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= 6.667, = 1.667

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Solution

The correct option is D
= 6.667, = 1.667

Some thermodynamic data is given.

$Δ$ G = $Δ$ G $^{\circ}$ + RTlnQ $^{\circ}$ $_{p}$ - (1)

R = gas constant. T = temperature; Q $^{\circ}$ $_{p}$ = reaction quotient

Q $^{\circ}$ $_{p}$ = $\frac{(P_{N O_{2}})^{2}}{(P_{N_{2} O_{4}})}$ = $\frac{10^{2}}{10}$ =
10 bar

$Δ$ G $^{\circ}$ (reaction) = 2 $Δ$ G $^{\circ}$ (NO $_{2}$ ) -
$Δ$ G $^{\circ}$ (N $_{2}$ O $_{4}$ ) = 2 (50 kJ mol $^{- 1}$ ) - 100 kJ
mol $^{- 1}$ = 0

$Δ$ G = $Δ$ G $^{\circ}$ + RTlnQ $^{\circ}$ $_{p}$ = 0 +
(8.314 J K $^{- 1}$ mol $^{- 1}$ ) (298 K) In (10) = (8.314 x 298 x 2.303)
Jmol $^{- 1}$ = 5.7058 kJmol $^{- 1}$

$Δ$ G $^{\circ}$ = - RT InK $^{\circ}$ $_{p}$ = 0 (from calculation)

This implies that K $^{\circ}$ $_{p}$ = 1 bar; Earlier Q $^{\circ}$ $_{p}$ =
10 bar.

As Q $^{\circ}$ $_{p}$ $>$ K $^{\circ}$ $_{p}$ the reaction will proceed in
the reverse direction. Thus we choose 2x moles to disappear from NO $_{2}$ and x
moles of N $_{2}$ O $_{4}$ to appear on the other side

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$

5 mol + x 5 mol - 2x; total amount = 10 mol -x

where x is the amount of $N_{2} O_{4}$ formed at equilibrium.

$P_{N_{2} O_{4}} = \frac{5 m o l + x}{10 m o l - x}$ (20 bar) and $P_{N O_{2}} = \frac{5 m o l - 2 x}{10 m o l - x}$ (20 bar)

At equilibrium,

K $^{\circ}$ $_{p}$ = 1 = (P $_{N O 2}$ ) $^{2}$ / (P $_{N 2 O 4}$ ) \{ here we
take the equilibrium partial pressures; not the given values\}

${[\frac{5 m o l - 2 x}{10 m o l - x} (20)]}^{2} = \frac{5 m o l + x}{10 m o l - x} (20)$

20 $(5 m o l - 2 x)^{2}$ = (5 mol + x) (10 mol - x)

500 $m o l^{2}$ + 80 $x^{2}$ - 400x mol = 50 $m o l^{2}$ + 5x mol - $x^{2}$

81 $x^{2}$ - 405x mol + 450 $m o l^{2}$ = 0

$x = \frac{(405 m o l) \pm \sqrt{(405 m o l)^{2} - 4 (81) (450 m o l^{2})}}{2 \times 81}$

Solving x = 3.333 mol or x = 1.667 mol

Since (5 - 2x ) $>$ 0

X = 1.667 mol

Number of moles N $_{2}$ O $_{4}$ = 5 mol + x = 6.667 mol and Number of moles
of NO $_{2}$ = 5 mol - 2x = 1.667 mol

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Similar questions

The reaction N $_{2}$ O $_{4}$ (g) $⇌$ 2NO $_{2}$ (g) is
carried out at 298 K and 20 bar. Five moles of each of N $_{2}$ O $_{4}$ and
NO $_{2}$ , are taken initially. Given: $Δ$ $_{f}$ G $^{\circ}$
(N $_{2}$ O $_{4}$ ) = 100 kJ mol $^{- 1}$ and $Δ$ $_{f}$ G $^{\circ}$
(NO $_{2}$ ) = 50 kJ mol $^{- 1}$ .Choose the appropriate option: The values of
$Δ$ G and K $^{\circ}$ $_{p}$ at 298 K are:

Which of the following expressions is the correct equilibrium-constant expression for the equilibrium between dinitrogen tetroxide and nitrogen dioxide?

N₂O₄ (g) ⇌ 2NO₂ (g)