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 $\Delta$S° at 25°C for the following reaction: 2NO + O₂ 2NO₂

Question

Quizplus · Accepted Answer

The change in standard enthalpy of the reaction is given as -56.91 kJ.

Using the equation 
ΔG° = ΔH° − TΔS° 
and rearranging it to 
ΔS° = (ΔH° − ΔG°) / T

we can calculate the change in standard entropy of the reaction.

Substituting the given values: 
ΔS° = (-56.91 kJ * 1000 J/kJ - (-1.46 * 10^3 * 8.314 J/K/mol) * 298.15 K ) / (298.15 K * 2 mol) 
ΔS° = -146 J/K

Therefore, the answer is option B.

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