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Estudos Gerais07/12/2024

Dadas as seguintes equações termoquímicas: Calcule a entalpi...

Dadas as seguintes equações termoquímicas: Calcule a entalpia da reação:

2 N2+3/2O2–> 2 N2O5  2 \mathrm{~N} 2+3 / 2 \mathrm{O} 2 \text {--> } 2 \text { N2O5 }
  • 2H2( g)+O2( g)2H2O()2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell) ΔH=571,5 kJ\Delta H^{\circ}=-571,5 \mathrm{~kJ}
  • N2O5( g)+H2O()2HNO3()\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow 2 \mathrm{HNO}_{3}(\ell) ΔH=76,6 kJ\Delta H^{\circ}=-76,6 \mathrm{~kJ}
  • 12 N2( g)+32O2( g)+12H2( g)HNO3(l)\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{HNO}_{3}(\mathrm{l}) ΔH=174,1 kJ\Delta H^{\circ}=-174,1 \mathrm{~kJ}

Dadas as seguintes equações termoquímicas:
Calcule a entalpia da reação:
\[
2 \mathrm{~N} 2+3 / 2 \mathrm{O} 2 \text {--> } 2 \text { N2O5 }
\]
- \( 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell) \) \( \Delta H^{\circ}=-571,5 \mathrm{~kJ} \)
- \( \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow 2 \mathrm{HNO}_{3}(\ell) \)
\( \Delta H^{\circ}=-76,6 \mathrm{~kJ} \)
- \( \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{HNO}_{3}(\mathrm{l}) \)
\( \Delta H^{\circ}=-174,1 \mathrm{~kJ} \)
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