Q.72. For the following reaction at 50°C and 2 atm pressure,
2N₂O₃ (g) ⇌ 2N₂O₄(g) + O₂(g)
N₂O₃ is 50

Q: What is the correct answer for JEE Main Chemistry Q72?

The correct answer is 2474. The reaction is: 2N₂O₃ (g) ⇌ 2N₂O₄(g) + O₂(g) Initial moles: Let's assume 2 moles of N₂O₃ initially. Initial pressure = 2 atm. Temperature T = 50°C = 50 + 273 = 323 K. N₂O₃ is 50% dissociated, so α = 0.5. Initial: 2 moles N₂O₃, 0 moles N₂O₄, 0 moles O₂ Change: -2α moles N₂O₃, +2α moles N₂O₄, +α moles O₂ Equilibrium: 2(1-α) moles N₂O₃, 2α moles N₂O₄, α moles O₂ Substitute α = 0.5: Equilibrium mo

Q: Q.72. For the following reaction at 50°C and 2 atm pressure, 2N₂O₃ (g) ⇌ 2N₂O₄(g) + O₂(g) N₂O₃ is 50% dissociated The magnitude of standard free energy change at this temperature is x. x = ______ J…

The correct answer is 2474. The reaction is: 2N₂O₃ (g) ⇌ 2N₂O₄(g) + O₂(g) Initial moles: Let's assume 2 moles of N₂O₃ initially. Initial pressure = 2 atm. Temperature T = 50°C = 50 + 273 = 323 K. N₂O₃ is 50% dissociated, so α = 0.5. Initial: 2 moles N₂O₃, 0 moles N₂O₄, 0 moles O₂ Change: -2α moles N₂O₃, +2α moles N₂O₄, +α moles O₂ Equilibrium: 2(1-α) moles N₂O₃, 2α moles N₂O₄, α moles O₂ Substitute α = 0.5: Equilibrium mo

Question

Q.72. For the following reaction at 50°C and 2 atm pressure,
2N₂O₃ (g) ⇌ 2N₂O₄(g) + O₂(g)
N₂O₃ is 50% dissociated
The magnitude of standard free energy change at this temperature is x.
x = ______ J mol⁻¹ [Nearest integer]
Given : R = 8.314 mol⁻¹K⁻¹, log 2 = 0.30, log 3 = 0.48, ln 10 = 2.303, °C + 273 = K

FirstInTest · Accepted Answer

The reaction is: 2N₂O₃ (g) ⇌ 2N₂O₄(g) + O₂(g)
Initial moles: Let's assume 2 moles of N₂O₃ initially.
Initial pressure = 2 atm.
Temperature T = 50°C = 50 + 273 = 323 K.

N₂O₃ is 50% dissociated, so α = 0.5.

Initial: 2 moles N₂O₃, 0 moles N₂O₄, 0 moles O₂
Change: -2α moles N₂O₃, +2α moles N₂O₄, +α moles O₂
Equilibrium: 2(1-α) moles N₂O₃, 2α moles N₂O₄, α moles O₂

Substitute α = 0.5:
Equilibrium moles: N₂O₃ = 2(1-0.5) = 1 mole
N₂O₄ = 2(0.5) = 1 mole
O₂ = 0.5 mole

Total moles at equilibrium (n_total) = 1 + 1 + 0.5 = 2.5 moles.

Initial moles (n_initial) = 2 moles.

We can calculate the partial pressures at equilibrium. The total pressure at equilibrium is given as 2 atm.

P_N₂O₃ = (n_N₂O₃ / n_total) * P_total = (1 / 2.5) * 2 atm = 0.8 atm
P_N₂O₄ = (n_N₂O₄ / n_total) * P_total = (1 / 2.5) * 2 atm = 0.8 atm
P_O₂ = (n_O₂ / n_total) * P_total = (0.5 / 2.5) * 2 atm = 0.4 atm

Now, calculate the equilibrium constant K_p:
K_p = (P_N₂O₄)² * P_O₂ / (P_N₂O₃)²
K_p = (0.8)² * 0.4 / (0.8)² = 0.4

Now, calculate the standard free energy change (ΔG°):
ΔG° = -RT ln K_p
ΔG° = - (8.314 J mol⁻¹ K⁻¹) * (323 K) * ln(0.4)

We need ln(0.4). We are given ln 10 = 2.303, log 2 = 0.30, log 3 = 0.48.
ln(0.4) = ln(4/10) = ln(2/5) = ln(2) - ln(5)
ln(5) = ln(10/2) = ln(10) - ln(2) = 2.303 - 0.693 = 1.610 (using ln 2 = 0.693, which is 2.303 * log 2)
ln(0.4) = ln(2) - ln(5) = 0.693 - 1.610 = -0.917

ΔG° = - (8.314) * (323) * (-0.917)
ΔG° = 8.314 * 323 * 0.917
ΔG° = 2685.262 * 0.917
ΔG° = 2462.7 J mol⁻¹

Rounding to the nearest integer, x = 2463 J mol⁻¹.

Let's recheck the calculation with the provided log values.
ln(0.4) = 2.303 * log(0.4) = 2.303 * log(4/10) = 2.303 * (log 4 - log 10) = 2.303 * (2*log 2 - 1)
= 2.303 * (2 * 0.30 - 1) = 2.303 * (0.60 - 1) = 2.303 * (-0.40) = -0.9212

ΔG° = - (8.314) * (323) * (-0.9212)
ΔG° = 8.314 * 323 * 0.9212
ΔG° = 2685.262 * 0.9212
ΔG° = 2473.6 J mol⁻¹

Rounding to the nearest integer, x = 2474 J mol⁻¹.

The provided solution gives 2474.

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