Question
Given the following data:$$\begin{aligned} 2 \mathrm{O}_{3}(g) \rightarrow 3 \mathrm{O}_{2}(g) & \Delta H=-427 \mathrm{kJ} \\ \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{O}(g) & \Delta H=+495 \mathrm{kJ} \end{aligned}$$$$\mathrm{NO}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\Delta H=-199 \mathrm{kJ}$$calculate $\Delta H$ for the reaction$$\mathrm{NO}(g)+\mathrm{O}(g) \rightarrow \mathrm{NO}_{2}(g)$$
Step 1
Step 1: We start with the given reaction: $$\mathrm{NO}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta H=-199 \mathrm{kJ}$$ We multiply everything in this reaction by two, including $\Delta H$, to Show more…
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Given the following data: $$S(s)+\frac{3}{2} O_{2}(g) \rightarrow S O_{3}(g) \quad \Delta H=-395.2 \mathrm{kJ}$$ $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g) \quad \Delta H=-198.2 \mathrm{kJ}$$ calculate $\Delta H$ for the reaction $$\mathrm{S}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)$$
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Given the data $$\begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) & \Delta H=+180.7 \mathrm{kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) & \Delta H=-113.1 \mathrm{kJ} \\ 2 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H=-163.2 \mathrm{kJ} \end{aligned}$$ use Hess's law to calculate $\Delta H$ for the reaction $$\mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g)$$
Calculate delta H for the reaction NO (g) + O (g) --> NO2 (g) From the following data NO (g) + O3 (g) --> NO2 (g) + O2 (g) delta H = -199 kJ NO (g) + O3 (g) --> 3/2O2 (g) delta H = -142 kJ NO (g) + (g) O3 --> 2O (g) delta H = +495 kJ
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