Question

Hess's Law Problems8. Calculate the ΔH for the reaction N2 (g) + 2 O2 (g) → 2 NO2 (g), from the following:N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔH = – 115 kJ2 NH3 (g) + 4 H2O (l) → 2 NO2 (g) + 7 H2 (g)ΔH = – 142.5 kJH2O (l) → H2 (g) + ½ O2 (g)ΔH = + 285.8 kJ9. Calculate the ΔH for the reaction CO2 (g) → C (s, graphite) + O2 (g), from the following:H2O (l) → H2 (g) + ½ O2 (g)ΔH = + 285.8 kJC2H6 (g) → 2 C (s, graphite) + 3 H2 (g)ΔH = + 190.6 kJ2 CO2 (g) + 3 H2O (l) → C2H6 (g) + 3 ½ O2 (g)ΔH = + 3511.1 kJ10. Calculate the ΔH for the reaction N2H4 (l) + CH4O (l) → CH2O (g) + N2 (g) + 3 H2 (g), from the following:2 NH3 (g) → N2H4 (l) + H2 (g)ΔH = + 22.5 kJ2 NH3 (g) → N2 (g) + 3 H2 (g)ΔH = + 57.5 kJCH2O (g) + H2 (g) → CH4O (l)ΔH = + 81.2 kJ11. Calculate the ΔH for the reaction H2 (g) + Cl2 (g) → 2 HCl (g), from the following:COCl2 (g) + H2O (l) → CH2Cl2 (l) + O2 (g)ΔH = + 47.5 kJ2 HCl (g) + ½ O2 (g) → H2O (l) + Cl2 (g)ΔH = + 105 kJCH2Cl2 (l) + H2 (g) + 1 ½ O2 (g) → COCl2 (g) + 2 H2O (l)ΔH = – 402.5 kJ12. Calculate the ΔH for the reaction 2 C2H2 (g) + 5 O2 → 4 CO2 (g) +2 H2O (g), from the following:C2H6 (g) → C2H2 (g) + 2 H2 (g)ΔH = + 283.5 kJH2 (g) + ½ O2 (g) → H2O (g)ΔH = – 241.8 kJ2 CO2 (g) + 3 H2O (g) → C2H6 (g) + 3 ½ O2 (g)ΔH = + 849 kJ13. Calculate the ΔH for the reaction HCl (g) + NaNO2 (s) → HNO2 (l) + NaCl (s), from the following:2 NaCl (s) + H2O (l) → 2 HCl (g) + Na2O (s)ΔH = + 507 kJNO (g) + NO2 (g) + Na2O (s) → 2 NaNO2 (s)ΔH = – 427 kJNO (g) + NO2 (g) → N2O (g) + O2 (g)ΔH = – 43 kJ2 HNO2 (l) → N2O (g) + O2 (g) + H2O (l)ΔH = +34 kJ14. Calculate the ΔH for the reaction 8 Zn (s) + 1 S8 (s) + 16 O2 (g) → 8 ZnSO4 (s), from the following:Zn (s) + ⅛ S8 (s) → ZnS (s)ΔH = – 183.9 kJ2 ZnS (s) + 3 O2 (g) → 2 ZnO (s) + 2 SO2 (g)ΔH = – 927.5 kJ2 SO2 (g) + O2 (g) → 2 SO3 (g)ΔH = – 196 kJZnO (s) + SO3 (g) → ZnSO4 (s)ΔH = – 230.3 kJ

Answer 1

-829.1kJ

Explanations:

Given the following chemical equations

`\begin{gathered} N_2(g)+3H_2(g)\rightarrow2HN_3(g)\text{ }\triangle H=-115kJ \\ 2NH_3(g)+4H_2O(l)→2NO_2(g)+7H_2(g)\text{ }ΔH=-142.5kJ \\ 2H_2O(l)→2H_2(g)+O_2(g)\text{ }ΔH=+285.8kJ\text{ \lparen multiplied through by 2\rparen} \end{gathered}`
`\begin{gathered} N_2(g)+\cancel{3H_2\left(g\right)}\rightarrow\cancel{2HN_3(g)\text{ }}\triangle H=-115kJ \\ \cancel{2NH_3\left(g\right)}+\cancel{4H_2O\left(l\right)}→2NO_2(g)+\cancel{7H_2(g)\text{ }}ΔH=-142.5kJ \\ \cancel{4H_2\left(g\right)}+2O_2(g)\rightarrow\cancel{4H_2O\left(l\right)}ΔH=-571.6kJ\text{ \lparen multiplied through by 2 and swap\rparen} \end{gathered}`

After cancelling out the compounds on different sides to generate N2 + 2O2 -> 2NO2 as shown:

`\begin{gathered} N_2(g)+2O_2(g)\rightarrow2NO_2\text{ }\triangle H=-115kJ-142.5kJ-571.6kJ \\ N_2(g)+2O_2(g)\rightarrow2NO_2\text{ }\triangle H=-829.1kJ \end{gathered}`

Hence the enthalpy change for the reaction is -829.1kJ