Solution Found!
The chemistry of nitrogen oxides is very versatile. Given
Chapter 5, Problem 82P(choose chapter or problem)
The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes,
\(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{~g}) \quad \Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=-39.8 \mathrm{~kJ}\)
\(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=-112.5 \mathrm{~kJ}\)
\(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \quad \Delta \mathrm{H}^{\circ}{ }_{\mathrm{rxn}}=-57.2 \mathrm{~kJ}\)
\(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=-114.2 \mathrm{~kJ}\)
\(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~s}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \quad \Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=54.1 \mathrm{~kJ}\)
calculate the heat of reaction for
\(\mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{~g})+\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~s}) \rightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}(\mathrm{~g})\)
Equation Transcription:
NO(g) + NO2(g) N2O3(g)
H°rxn = -39.8 kJ
NO(g) + NO2(g) + O2(g) N2O5(g)
H°rxn = -112.5 kJ
2NO2(g) N2O4(g)
H°rxn = -57.2 kJ
2NO(g) + O2(g) 2NO2(g)
H°rxn = -114.2 kJ
N2O5(s) N2O5(g)
H°rxn = 54.1 kJ
N2O3(g) + N2O5(s) 2N2O4(g)
Text Transcription:
NO(g) + NO_2(g) rightarrow N_2O_3(g) Delta H°_rxn = -39.8 kJ
NO(g) + NO_2(g) + O_2(g) rightarrow N2O5(g) Delta H°_rxn = -112.5 kJ
2NO_2(g) rightarrow N_2O_4(g) Delta H°_rxn = -57.2 kJ
2NO(g) + O2(g) rightarrow 2NO_2(g) Delta H°_rxn = -114.2 kJ
N_2O_5(s) rightarrow N_2O_5(g) Delta H°_rxn = 54.1 kJ
N_2O_3(g) + N_2O_5(s) rightarrow 2N_2O_4(g)
Questions & Answers
QUESTION:
The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes,
\(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{~g}) \quad \Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=-39.8 \mathrm{~kJ}\)
\(\mathrm{NO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=-112.5 \mathrm{~kJ}\)
\(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \quad \Delta \mathrm{H}^{\circ}{ }_{\mathrm{rxn}}=-57.2 \mathrm{~kJ}\)
\(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g}) \quad \Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=-114.2 \mathrm{~kJ}\)
\(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~s}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \quad \Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=54.1 \mathrm{~kJ}\)
calculate the heat of reaction for
\(\mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{~g})+\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~s}) \rightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}(\mathrm{~g})\)
Equation Transcription:
NO(g) + NO2(g) N2O3(g)
H°rxn = -39.8 kJ
NO(g) + NO2(g) + O2(g) N2O5(g)
H°rxn = -112.5 kJ
2NO2(g) N2O4(g)
H°rxn = -57.2 kJ
2NO(g) + O2(g) 2NO2(g)
H°rxn = -114.2 kJ
N2O5(s) N2O5(g)
H°rxn = 54.1 kJ
N2O3(g) + N2O5(s) 2N2O4(g)
Text Transcription:
NO(g) + NO_2(g) rightarrow N_2O_3(g) Delta H°_rxn = -39.8 kJ
NO(g) + NO_2(g) + O_2(g) rightarrow N2O5(g) Delta H°_rxn = -112.5 kJ
2NO_2(g) rightarrow N_2O_4(g) Delta H°_rxn = -57.2 kJ
2NO(g) + O2(g) rightarrow 2NO_2(g) Delta H°_rxn = -114.2 kJ
N_2O_5(s) rightarrow N_2O_5(g) Delta H°_rxn = 54.1 kJ
N_2O_3(g) + N_2O_5(s) rightarrow 2N_2O_4(g)
ANSWER:
Solution 82P
Here the following reactions and their standard enthalpy changes are given,
We have to calculate the heat of reaction for
N2O3(g) + N2O5(s) → 2N2O4(g)
Among the given reactions, only reaction 3 contain N2O4(g) and reaction 1 contain N2O3(g). N2O5 appears in both reactions 2 and 5 but they have difference in their physical states i.e solid and gas.
1st we will add reaction 1, 3 and 5
We can write the equation 1 as reversed as follows,
N2O3 (g) → NO (g) + NO2(g)
Multiply equation (3) by 2, 4NO2 (g) → 2 N2O4 (g)
The equation 5 can be written as reversed as follows,
N2O5 (s) → N2O5 (g)