In the following equilibrium N 2 O 5 ( g ) ⇌ : 2 NO 2 ( g ) , when 5 moles of each is taken and the temperature is kept at 298 K ; the total pressure was found to be 20 bar. Given Δ fG ∘ [ N 2

Question

In the following equilibrium N 2 ​ O 5 ​ ( g ) ⇌ : ​ 2 NO 2 ​ ( g ) , when 5 moles of each is taken and the temperature is kept at 298 K ; the total pressure was found to be 20 bar. Given Δ fG ∘ [ N 2 ​ O 4 ​ ( g ) ] = 100 kJ / mol ; Δ fG ∘ [ NO 2 ​ ( g ) ] = 50 kJ / mol .

TestHub · Accepted Answer

The chemical equilibrium is given by:

N₂O₄(g) $⇌$ 2NO₂(g)

First, calculate the standard Gibbs free energy change for the reaction, $Δ G^{\circ}$ .

$Δ G^{\circ} = \sum Δ_{f} G^{\circ} (p ro d u c t s) - \sum Δ_{f} G^{\circ} (re a c t an t s)$

$Δ G^{\circ} = [2 \times Δ_{f} G^{\circ} (N O_{2} (g))] - [1 \times Δ_{f} G^{\circ} (N_{2} O_{4} (g))]$

Given $Δ_{f} G^{\circ} [N_{2} O_{4} (g)] = 100 k J / m o l$ and $Δ_{f} G^{\circ} [N O_{2} (g)] = 50 k J / m o l$ .

$Δ G^{\circ} = [2 \times 50 k J / m o l] - [1 \times 100 k J / m o l]$

$Δ G^{\circ} = 100 k J / m o l - 100 k J / m o l = 0 k J / m o l$

Thus, option B is correct.

Next, relate $Δ G^{\circ}$ to the equilibrium constant $K_{p}$ using the equation:

$Δ G^{\circ} = - RT ln K_{p}$

Since $Δ G^{\circ} = 0$ , we have:

$0 = - RT ln K_{p}$

As $R$ (gas constant) and $T$ (temperature) are non-zero, $ln K_{p}$ must be zero.

$ln K_{p} = 0$

Therefore, $K_{p} = e^{0} = 1$ .

The equilibrium constant $K_{p}$ for the reaction is 1.0 bar.

Thus, option A is correct.

To check if the system is at equilibrium under the given conditions, we calculate the reaction quotient $Q_{p}$ .

Initial moles: $n_{N_{2} O_{4}} = 5 m o l$ , $n_{N O_{2}} = 5 m o l$

Total moles: $n_{t o t a l} = 5 + 5 = 10 m o l$

Total pressure: $P_{t o t a l} = 20 ba r$

Partial pressure of N₂O₄: $P_{N_{2} O_{4}} = X_{N_{2} O_{4}} \times P_{t o t a l} = \frac{5}{10} \times 20 ba r = 10 ba r$

Partial pressure of NO₂: $P_{N O_{2}} = X_{N O_{2}} \times P_{t o t a l} = \frac{5}{10} \times 20 ba r = 10 ba r$

The reaction quotient $Q_{p}$ is given by:

$Q_{p} = \frac{( P _{N O_{2}} ) ^{2}}{P _{N_{2} O_{4}}}$

$Q_{p} = \frac{( 10 ba r ) ^{2}}{10 ba r} = \frac{100 ba r ^{2}}{10 ba r} = 10 ba r$

Since $Q_{p} = 10 ba r$ and $K_{p} = 1 ba r$ , $Q_{p} > K_{p}$ .

When $Q_{p} > K_{p}$ , the reaction will proceed in the backward (reverse) direction to reach equilibrium.

Therefore, option C is correct, and option D is incorrect.

Chemistry - Thermodynamics - II Question with Solution | TestHub

Options:(select one or more)

Doubts & Discussion

JEE Advanced 2027 (Class 12)