Consider the following reaction at constant P.Use the information here to determine the value of ΔS_surrat 298 K.Predict whether or not this reaction will be spontaneous at this temperature. N₂(g)+ 2 O₂(g)→ 2 NO₂(g)ΔH = +66.4 kJ

Question

Quizplus · Accepted Answer

The change in entropy of the surroundings ($\Delta S_{surr}$) can be calculated using the formula $\Delta S_{surr} = -\frac{\Delta H}{T}$, where $\Delta H$ is the change in enthalpy and $T$ is the temperature in Kelvin. Given $\Delta H = +66.4 \,kJ = +66400 \,J$ and $T = 298 \,K$, $\Delta S_{surr} = -\frac{66400}{298} = -223 \,J/K$. A negative $\Delta S_{surr}$ suggests the reaction is not spontaneous at 298 K under constant pressure.

Consider the Following Reaction at Constant P