Chapter 20, Problem 20.127QP

Interpretation Introduction

Interpretation:

• The energy released when 1 mol of $\text{He}$ generated from protons and neutrons has to be calculated.
• At $\text{25°C}$ and 725mmHg the volume of Ethane are required to release the same quantity of energy when ethane burned in Oxygen to Carbon dioxide and water has to be calculated.

Answer to Problem 20.127QP

• Energy released when 1 mol of $\text{He}$ generated from protons and neutrons is $\text{2}{\text{.732×10}}^{\text{-5}}\text{kg/mol}$.
• At $\text{25°C}$ and 725mmHg the volume of Ethane are required to release the same quantity of energy when ethane burned in Oxygen to Carbon dioxide and water is $\text{4}{\text{.907×10}}^{\text{7}}\text{= 4}{\text{.91×10}}^{\text{7}}\text{L}$

Explanation of Solution

Given

Mass of electron = $\text{0}\text{.000549}\text{amu}$

Temperature = $\text{25°C}$

$\text{2p + 2n}\stackrel{}{\to }{}_{\text{2}}^{\text{4}}\text{He}$

${\text{ Nuclear mass of }}_{\text{2}}^{\text{4}}\text{He = 4}\text{.00260 amu-(2×0}\text{.000549 amu) = 4}\text{.001502 amu}$

$\begin{array}{l}\text{On mole basis, the mass difference (}\Delta \text{m) is , }\\ \\ \Delta \text{m= }\left[\text{4}\text{.001502 -2}\left(\text{1}\text{.00728}\right)\text{-2}\left(\text{1}\text{.00867}\right)\right]\text{ g/mol = - 0}\text{.030398 g/mol}\end{array}$

$\text{ = -3}{\text{.0398×10}}^{\text{-5}}\text{kg/mol}$

$\begin{array}{l}\text{The energy released per mole is, }\\ {\text{ΔE= (Δm)c}}^{\text{2}}\text{= -3}{\text{.0398×10}}^{\text{-5}}\text{kg/mol×(2}{\text{.998×10}}^{\text{8}}{\text{m/s)}}^{\text{2}}\\ \text{= -2}{\text{.7321×10}}^{\text{12}}{\text{ kg•m}}^{\text{2}}{\text{/s}}^{\text{2}}\text{/mol }\\ \text{= -2}{\text{.7321×10}}^{\text{12}}\text{J/mol }\\ \text{= 2}{\text{.732×10}}^{\text{12}}\text{J released per mole}\\ \\ \text{Now calculate ΔH° for burning of ethane: }\\ {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{}_{\left(\text{g}\right)}{\text{ + 7/2O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{ 2CO}}_{\text{2}}{}_{\left(\text{g}\right)}{\text{ + 3H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\text{=-84}\text{.68 }\text{0 }\text{ 2(-393}\text{.5)}\text{ 3(-241}\text{.8)kJ}\end{array}$

$\begin{array}{l}\text{ΔH°=}\left[\text{2(-393}\text{.5)+3(-241}\text{.8)-(-84}\text{.68)}\right]\text{kJ}\\ \\ \text{= (-1427}\text{.72)kJ/mol etane}\end{array}$

$\begin{array}{l}\text{Now calculate the moles of ethane required to form}\\ \text{ 2}\text{.7321}\times {\text{10}}^{\text{12}}\text{J, or 2}\text{.7321}\times {\text{10}}^{\text{9}}\text{kJ heat: }\\ \text{-2}\text{.7321}\times {\text{10}}^{\text{9}}\text{kJ´ = 1}\text{.9136}\times {\text{10}}^{\text{6}}\text{mol ethane}\end{array}$

Conclusion

• Energy released when 1 mol of $\text{He}$ generated from protons and neutrons was calculated as $\text{2}{\text{.732×10}}^{\text{-5}}\text{kg/mol}$.
• At $\text{25°C}$ and 725mmHg the volume of Ethane are required to release the same quantity of energy when ethane burned in Oxygen to Carbon dioxide and water was calculated as $\text{4}{\text{.907×10}}^{\text{7}}\text{= 4}{\text{.91×10}}^{\text{7}}\text{L}$