What is the \Delta H_{rxn} for this reaction is units of kJ/mol? NH3 + O2 ---> HNO3 + H2O
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Estimate $\Delta H_{\mathrm{rxn}}$ for the following reaction: $$4 \mathrm{NH}_{3}(g)+7 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$
Given that $4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$ $$ \Delta H_{\mathrm{rxn}}^{\circ}=-1166 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} $$ $4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)$ $$ \Delta H_{\mathrm{rxn}}^{\circ}=-1531 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} $$ calculate the value of $\Delta H_{\mathrm{rxn}}^{\circ}$ for the equation $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}(g) $$
Given the data in the table below, ΔH°rxn for the reaction 4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (l) is kJ: Substance ΔH°f (kJ/mol) H2O (l) -286 NO (g) 90 NO2 (g) 34 HNO3 (aq) -207 NH3 (g) -46 A) -1540 B) -1172 C) -1892 D) -150 E) The ΔH°f of O2 (g) is needed for the calculation.
Shaiju T.
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