Chapter 6, Problem 6.75QP

A sample of ethanol, C₂H₅OH, weighing 2.84 g was burned in an excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rose from 25.00°C to 33.73°C. If the heat capacity of the calorimeter and contents was 9.63 kJ/°C, what is the value of q for burning 1.00 mol of ethanol at constant volume and 25.00°C? The reaction is

${\text{C}}_2{\text{H}}_5\text{OH}(l)+3{\text{O}}_2(g)\to 2{\text{CO}}_2(g)+3{\text{H}}_2\text{O}(l)$

Is q equal to ΔU or ΔH?

Interpretation Introduction

Interpretation:

The value of $\text{q}$ for $\text{1}\text{mol}$ of ethanol burning has to be calculated and also the value of $\text{q}$ is equal to $\text{ΔU}\text{or}\text{ΔH}$ or not has to be checked.

Concept introduction:

Enthalpy of reaction:

The enthalpy of reaction is given by

$\begin{array}{l}\text{H}\text{=}\text{U}\text{+}\text{PV}\\ \text{where,}\\ \text{H}\text{-}\text{enthalpy}\\ \text{U}\text{-}\text{internal}\text{energy}\\ \text{P}\text{-}\text{pressure}\text{and}\text{V-}\text{volume}\end{array}$

Answer to Problem 6.75QP

• The value $\text{q}$ for $\text{1}\text{mol}$ of ethanol burning is $\text{-}\text{1}\text{.36}\text{×}{\text{10}}^{\text{3}}\text{kJ}\text{.}$

The comparison of $\text{q}\text{with}\text{ΔU}\text{and}\text{ΔH}$ is given by

$\text{q}\text{=}{\text{q}}_{\text{v}}\text{=}\text{ΔU}\text{.}\text{this result}\text{also}\text{equals to ΔH}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{mass}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\text{is}\text{2}\text{.84}\text{g}\\ \text{temperature}\text{changes}\text{from}\text{25}{}^{\text{o}}\text{C}\text{to}\text{33}\text{.73}{}^{\text{o}}\text{C}\\ \text{specific}\text{heat}\text{of}\text{contents}\text{and}\text{the}\text{calorimeter}\text{is}\text{9}\text{.63}\text{kJ/}{}^{\text{o}}\text{C}\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\left(\text{l}\right)}\text{+}{\text{3O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{2CO}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}{\text{3H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\end{array}$

Calculate the value of ${\text{q}}_{\text{calorimeter}}$ by using specific heat value and temperature change

$\begin{array}{l}{\text{q}}_{\text{calorimeter}}\text{=}\left(\text{9}\text{.63}\text{kJ}\text{/}{}^{\text{o}}\text{C}\right)\left(\text{33}\text{.73}{}^{\text{o}}\text{C}\text{-}\text{25}{}^{\text{o}}\text{C}\right)\\ \text{=}\text{+}\text{84}\text{.070}\text{kJ}\text{.}\end{array}$

Thus, $\text{2}\text{.84}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ is equivalent to $\text{84}\text{.070}\text{kJ}$ of heat energy.

Molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\text{is}\text{46}\text{.07}\text{g}\text{.}$

Calculate the enthalpy change $\left(\text{ΔH}\right)$ for $\text{1}\text{mol}$ of ethanol

$\begin{array}{l}\text{1}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\text{×}\frac{\text{46}\text{.07}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}{\text{1}\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\text{×}\frac{\text{-}\text{84}\text{.070}\text{kJ}}{\text{2}\text{.84}\text{g}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\text{=}\text{-}\text{1363}\text{.8}\text{kJ}\text{.}\\ \text{=}\text{-}\text{1}\text{.36}\text{×}{\text{10}}^{\text{3}}\text{kJ}\text{.}\end{array}$

• The value $\text{q}$ for $\text{1}\text{mol}$ of ethanol burning is $\text{-}\text{1}\text{.36}\text{×}{\text{10}}^{\text{3}}\text{kJ}\text{.}$

Volume is fixed in bomb calorimeter so no pressure – volume is performed.  So the value of $\text{ΔU}$ is directly measured.

$\text{i}\text{.e}\text{.}\text{q}\text{=}{\text{q}}_{\text{v}}\text{=}\text{ΔU}$

The above result is also equal to $\text{ΔH}\text{.}$

Conclusion

The value $\text{q}$ for $\text{1}\text{mol}$ of ethanol burning was calculated and also the given condition was checked.