Question

20) Given the following data: \[ \begin{array}{ll} \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{NO}(\mathrm{g}) & \Delta \mathrm{H}=180.6 \mathrm{~kJ} \\ \mathrm{~N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) & \Delta \mathrm{H}=-91.8 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & \Delta \mathrm{H}=-483.7 \mathrm{~kJ} \end{array} \] Calculate the \( \Delta \mathrm{H}^{\circ} \) for the reaction: \( 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \) Reverse sec Reaction \( 2 \mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{0}(\mathrm{~g}) \Delta H_{=}-91.8 \mathrm{~kJ} \) Firet Reaction \[ \begin{array}{l} {\left[\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})\right] \times 2} \\ 2 \mathrm{~N}_{2}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{HNO}(\mathrm{g}) \end{array} \] 21) Given the following data: \[ \begin{array}{l} \mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g}) \\ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \\ \mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array} \] \[ \Delta H=+177.4 \mathrm{~kJ} \] \[ \Delta H=-191.2 \mathrm{~kJ} \] \[ \Delta H=-241.8 \mathrm{~kJ} \] \[ \Delta H=-43.8 k J \] Calculate the \( \Delta \mathrm{H}^{\circ} \) for the reaction: \[ \begin{array}{l} \mathrm{H}^{\circ} \text { for the reaction: } \\ \mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{OH})_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O} \text { (1) } \end{array} \]

          20) Given the following data:
\[
\begin{array}{ll}
\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{NO}(\mathrm{g}) & \Delta \mathrm{H}=180.6 \mathrm{~kJ} \\
\mathrm{~N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) & \Delta \mathrm{H}=-91.8 \mathrm{~kJ} \\
2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & \Delta \mathrm{H}=-483.7 \mathrm{~kJ}
\end{array}
\]

Calculate the \( \Delta \mathrm{H}^{\circ} \) for the reaction: \( 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \)
Reverse sec Reaction
\( 2 \mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{0}(\mathrm{~g}) \Delta H_{=}-91.8 \mathrm{~kJ} \)
Firet Reaction
\[
\begin{array}{l}
{\left[\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})\right] \times 2} \\
2 \mathrm{~N}_{2}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{HNO}(\mathrm{g})
\end{array}
\]
21) Given the following data:
\[
\begin{array}{l}
\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{OH})_{2}(\mathrm{aq}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g}) \\
\mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \\
\mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\
\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})
\end{array}
\]
\[
\Delta H=+177.4 \mathrm{~kJ}
\]
\[
\Delta H=-191.2 \mathrm{~kJ}
\]
\[
\Delta H=-241.8 \mathrm{~kJ}
\]
\[
\Delta H=-43.8 k J
\]

Calculate the \( \Delta \mathrm{H}^{\circ} \) for the reaction:
\[
\begin{array}{l}
\mathrm{H}^{\circ} \text { for the reaction: } \\
\mathrm{C}_{6} \mathrm{H}_{4}(\mathrm{OH})_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{O}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O} \text { (1) }
\end{array}
\]

20) Given the following data:

N2( g)+O(g) → 2 NO(g) ΔH=180.6 kJ
N2( g)+3 H2( g) → 2 NH3( g) ΔH=-91.8 kJ

2 H2( g)+O2( g) → 2 H2O(g) ΔH=-483.7 kJ

Calculate the ΔH^∘ for the reaction: 4 NH3( g)+5 O2( g) → 4 NO(g)+6 H2O(g)
Reverse sec Reaction
2 NH3( g) →N2( g)+3 H0( g) Δ H=-91.8 kJ
Firet Reaction

[N2( g)+O2( g) → 2 NO(g)] × 2

2 N2( g)+2 O2( g) →HNO(g)

21) Given the following data:

C6H4(OH)2(aq) →C6H4O2(aq)+H2( g)
H2( g)+O2( g) →H2O2(aq)
H2( g)+1 / 2 O2( g) →H2O(g)
H2O(g) →H2O(l)

Δ H=+177.4 kJ

Δ H=-191.2 kJ

Δ H=-241.8 kJ

Δ H=-43.8 k J

Calculate the ΔH^∘ for the reaction:

H^∘ for the reaction:
C6H4(OH)2(aq)+H2O2(aq) →C6H4O2(aq)+2 H2O (1)

Added by Chad P.

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