Determine the standard enthalpy change for each of the following reactions. Report your answers to the nearest 0.1 kJ. 3NO2(g) + H2O(l) → 2HNO3(g) + NO(g)
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Step 1
2 kJ/mol) = -346.4 kJ/mol - NO(g): 90.25 kJ/mol Total for products = -346.4 kJ/mol + 90.25 kJ/mol = -256.15 kJ/mol Show more…
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A chemist measures the enthalpy change ΔH during the following reaction: H2O(l) → H2O(g) ΔH = 44 kJ Use this information to complete the table below. Round each of your answers to the nearest kJ/mol. reaction ΔH 4H2O(l) → 4H2O(g) kJ 2H2O(g) → 2H2O(l) kJ H2O(g) → H2O(l) kJ
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Calculate the enthalpy change, H, for each reaction in terms of kJ/mol of each reactant?
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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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