Chapter 6, Problem 79E

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:

C₆H₄(OH)₂(aq) + H₂O₂(aq) → C₆H₄O₂(aq) + 2H₂O(l)

Calculate ΔH for this reaction from the following data:

C₆H₄(OH)₂(aq) → C₆H₄O₂(aq) + H₂(g)    ΔH = 177.4 kJ

H₂(g) + O₂(g) → H₂O₂(aq)    ΔH = −191.2 kJ

H₂(g) + $\frac{1}{2}$O₂(g) → H₂O(g)    ΔH = −241.8 kJ

H₂O(g) → H₂O(l)    ΔH = −43.8 kJ

Interpretation Introduction

Interpretation:

The $\text{ΔH}$ value for the given reaction should be calculated.

Concept Introduction:

Enthalpy change: It is defined as the amount of heat released or absorbed by the chemical reaction at constant pressure conditions.

Specific heat capacity: It is the amount of heat required by the substance to increase temperature of the 1 kilogram of the substance by 1 kelvin.

Calorimeter: It is the device used to measure the heat involved in the chemical reaction.

The heat released by the reaction is measured in calorimetry by using the following formula.

Coffee cup calorimeter: It is used to measure the heat in the reaction where the atmospheric pressure remains constant. It is also regarded as constant pressure calorimeter.

Bomb calorimeter: It is used to measure the heat which the volume kept constant. Hence it enables us to find the energy changes in the reaction. It is also regarded as constant volume calorimeter.

Answer to Problem 79E

The $\text{ΔH}$ value for the given reaction is $\text{-202}\text{.6kJ}$.

To calculate: The $\text{ΔH}$ value for the given reaction.

Explanation of Solution

Examining the given process the enthalpy change value that is $\text{ΔH}$ for the given reaction calculated as follows,

Given,

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}{}_{\left(\text{aq}\right)}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}{}_{\left(\text{aq}\right)}+{\text{H}}_{\text{2}}{}_{\left(g\right)}\Delta \text{H = 177}\text{.4kJ}\\ {\text{H}}_{\text{2}}{}_{\left(g\right)}+{\text{O}}_{\text{2}}{}_{\left(g\right)}\to {\text{H}}_{\text{2}}{\text{O}}_{\text{2}\left(\text{aq}\right)}\Delta \text{H = -191}\text{.2kJ}\\ {\text{H}}_{\text{2}}{}_{\left(g\right)}+\frac{1}{2}{\text{O}}_{\text{2}}{}_{\left(g\right)}\to {\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\Delta \text{H = -241}\text{.8kJ}\\ {\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\to {\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\Delta \text{H = - 43}\text{.8kJ}\end{array}$

The final given reaction is obtained by equating all the four given reactions where second reaction is written in reverse order hence its $\text{ΔH}$ value will be then positive value.

Then third and fourth reactions are multiplied by 2 which their $\text{ΔH}$ values should also be multiplied by 2.

Hence, the final reaction ${\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}{}_{\left(\text{aq}\right)}+{\text{H}}_{\text{2}}{\text{O}}_{\text{2}\left(\text{aq}\right)}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}{}_{\left(\text{aq}\right)}+2{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$ is obtained by rearranging the given set of equations as follows,

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}{}_{\left(\text{aq}\right)}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}{}_{\left(\text{aq}\right)}+{\text{H}}_{\text{2}}{}_{\left(g\right)}\Delta \text{H = 177}\text{.4kJ}\\ {\text{H}}_{\text{2}}{\text{O}}_{\text{2}\left(\text{aq}\right)}\to {\text{H}}_{\text{2}}{}_{\left(g\right)}+{\text{O}}_{\text{2}}{}_{\left(g\right)}\Delta \text{H = 191}\text{.2kJ}\\ 2{\text{H}}_{\text{2}}{}_{\left(g\right)}+{\text{O}}_{\text{2}}{}_{\left(g\right)}\to 2{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\Delta \text{H = 2}\times \text{-241}\text{.8kJ}\\ 2{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\to 2{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\Delta \text{H = 2}\times \text{- 43}\text{.8kJ}\\ \text{ ------------------------------------------------------------------------------------}\\ {\text{ C}}_{\text{6}}{\text{H}}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}{}_{\left(\text{aq}\right)}+{\text{H}}_{\text{2}}{\text{O}}_{\text{2}\left(\text{aq}\right)}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{4}}{\text{O}}_{\text{2}}{}_{\left(\text{aq}\right)}+2{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\end{array}$

Now, addition of $\text{ΔH}$ values for the four set of arranged reactions will results to give $\text{-202}\text{.6kJ}$.

Therefore, the $\text{ΔH}$ value for the given reaction is $\text{-202}\text{.6kJ}$.

Conclusion

The $\text{ΔH}$ value for the given reaction was determined.