Consider the gas phase reaction NO + O 2 NO 2 for which $\Delta$H° = -56.91 kJ and K = 1.46 $\times$10 6 at 25°C. -Calculate $\Delta$H° at 25°C for the following reaction: 2NO + O₂ 2NO₂

Question

Quizplus · Accepted Answer

Since the given reaction is NO + 0.5 O2 → NO2, we need to multiply the ΔH° for the given equation by a factor of 2 to get the ΔH° for the given reaction. Therefore, 
ΔH° = 2*(-56.91 kJ) = -113.82 kJ/mol
Also, using the equation ΔG° = ΔH° - TΔS° and rearranging it, we get:
ΔH° = ΔG° + TΔS°
At 25°C, ΔG° = -RTlnK = -(8.314 J/mol*K)(298 K)ln(1.46*10^10) = -85.76 kJ/mol
ΔS° can be calculated using the equation:
ΔG° = ΔH° - TΔS°
ΔS° = (ΔH° - ΔG°)/T = (-113.82 kJ/mol + 85.76 kJ/mol)/(298 K) = -0.094 J/K*mol
Now, we can use ΔG° and ΔS° to calculate ΔH° for the given reaction using the equation:
ΔG° = ΔH° - TΔS°
ΔH° = ΔG° + TΔS° = (-85.76 kJ/mol) + (298 K)(-0.094 J/K*mol) = -113.92 kJ/mol ≈ -113.8 kJ/mol.

Consider the Gas Phase Reaction NO + O2 Δ\DeltaΔ H° = -5691 KJ and K = 1

Consider the Gas Phase Reaction NO + O₂ $\Delta$ H° = -5691 KJ and K = 1