Question

Question : The bombardier beetle uses an explosive discharge as a defensive measure. The chemical reaction involved is the oxidation of hydroquinone by hydrogen peroxide to produce quinone and water:

${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{4}}{\left(\mathrm{OH}\right)}_{\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{}\mathbf{+}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{}\stackrel{}{\mathbf{\to }}\mathbf{}\mathbf{}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)$

Calculate ∆H for this reaction from the following data:

$$\begin{gathered} {\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{\left(}\mathbf{OH}\mathbf{\right)}}_{\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{}\stackrel{}{\mathbf{\to }\mathbf{}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{4}}{\mathbf{O}}_{\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{+}\mathbf{}\mathbf{H}\left(g\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{∆}\mathbf{H}\mathbf{=}\mathbf{177}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{kJ}} \\ {\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{}\left(g\right)\stackrel{}{\mathbf{\to }}\mathbf{}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{O}}_{\mathbf{2}}\mathbf{}\left(\mathrm{aq}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{∆}\mathbf{H}\mathbf{=}\mathbf{131}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{kJ} \\ \mathbf{2}{\mathbf{H}}_{\mathbf{2}}\left(g\right)\mathbf{}\mathbf{+}\mathbf{}{\mathbf{O}}_{\mathbf{2}\mathbf{}}\stackrel{}{\left(g\right)\mathbf{\to }}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{∆}\mathbf{H}\mathbf{=}\mathbf{-}\mathbf{483}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{kJ} \\ \mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)\mathbf{}\stackrel{}{\mathbf{\to }}\mathbf{}\mathbf{}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{∆}\mathbf{H}\mathbf{=}\mathbf{-}\mathbf{87}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{kJ} \end{gathered}$$

Short answer

∆H for the reaction = -202.6 kJ

Step-by-step solution

Numbering of equations

$$\begin{gathered} {\mathrm{C}}_6{\mathrm{H}}_4{\left(\mathrm{OH}\right)}_2\stackrel{}{\to {\mathrm{C}}_6{\mathrm{H}}_4{\mathrm{O}}_2+\mathrm{H}..eq.1∆\mathrm{H}=177.4\mathrm{kJ}} \\ {\mathrm{H}}_2{\mathrm{O}}_2\stackrel{}{\to }{\mathrm{H}}_2+{\mathrm{O}}_2..eq.2∆\mathrm{H}=131.2\mathrm{kJ} \\ 2{\mathrm{H}}_2+{\mathrm{O}}_2\stackrel{}{\to }2{\mathrm{H}}_2\mathrm{O}..eq.3∆\mathrm{H}=-483.6\mathrm{kJ} \\ 2{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }2{\mathrm{H}}_2\mathrm{O}..eq.4∆\mathrm{H}=-87.6\mathrm{kJ} \end{gathered}$$

Reverse equation 2

${\mathrm{H}}_2{\mathrm{O}}_2\stackrel{}{\to }{\mathrm{H}}_2+{\mathrm{O}}_2∆\mathrm{H}=131.2\mathrm{kJ}$

Multiply equation 3 by 2

$2{\mathrm{H}}_2+{\mathrm{O}}_2\stackrel{}{\to }2{\mathrm{H}}_2\mathrm{O}∆\mathrm{H}=-483.6\mathrm{kJ}$

Multiply equation 4 by 2 

$2{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }2{\mathrm{H}}_2\mathrm{O}∆\mathrm{H}=-87.6\mathrm{kJ}$

Calculation of ∆H by Hess Law

$$\begin{gathered} {\mathrm{C}}_6{\mathrm{H}}_4{\left(\mathrm{OH}\right)}_2\stackrel{}{\to {\mathrm{C}}_6{\mathrm{H}}_4{\mathrm{O}}_2+\mathrm{H}∆\mathrm{H}=177.4\mathrm{kJ}} \\ {\mathrm{H}}_2{\mathrm{O}}_2\stackrel{}{\to }{\mathrm{H}}_2+{\mathrm{O}}_2∆\mathrm{H}=131.2\mathrm{kJ} \\ 2{\mathrm{H}}_2+{\mathrm{O}}_2\stackrel{}{\to }2{\mathrm{H}}_2\mathrm{O}∆\mathrm{H}=-483.6\mathrm{kJ} \\ 2{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }2{\mathrm{H}}_2\mathrm{O}∆\mathrm{H}=-87.6\mathrm{kJ} \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \end{gathered}$$

${\mathrm{C}}_6{\mathrm{H}}_4{\left(\mathrm{OH}\right)}_2\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{aq}\right)\stackrel{}{\to }{\mathrm{C}}_6{\mathrm{H}}_4{\mathrm{O}}_2\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\mathrm{J}∆\mathrm{H}=-202.6\mathrm{kJ}$