Calculate the standard free-energy change, ΔG° (kJ), for the formation of NO(g) from N2(g) and O2(g) at 298 K: N2(g) + O2(g) → 2 NO(g) given that ΔH° = 180.7 kJ and ΔS° = 24.7 J/K.
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7 kJ - ΔS° = 24.7 J/K - T = 298 K Show more…
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Calculate the standard free-energy change for the formation of NO(g) from N2(g) and O2(g) at 298 K: N2(g) + O2(g) → 2 NO(g), given that ΔH° = 180.7 kJ and ΔS° = 24.7 J/K. Is the reaction spontaneous under these conditions?
Madhur L.
Consider the reaction: 2N2(g) + O2(g) → 2N2O(g) Using standard thermodynamic data at 298K, calculate the free energy change when 2.360 moles of N2(g) react at standard conditions. ΔG°rxn = kJ
Susan H.
Use the following themochemical equations to calculate the standard enthalpy change of formation of $\mathrm{HNO}_{3}(0)$ \[ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}\left(\mathrm{D}+2 \mathrm{NO}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{HNO}_{2}()\right.& \Delta_{4} H^{\circ} &=-226.8 \mathrm{kJ} \mathrm{mol}^{-1} \\ \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) & \Delta_{\mathrm{r}} H^{\circ} &=+66.4 \mathrm{kJ} \mathrm{mol}^{-1} \\ \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{o}) & \Delta_{\mathrm{r}} \mathrm{H}^{\mathrm{e}} &=-187.8 \mathrm{kJ} \mathrm{mol}^{-1} \end{aligned} \] (Section $13.1)$
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