Question

Calculate the mass of octane, ${\mathrm{C}}_8{\mathrm{H}}_{18}(l),$ that must be burned in air to evolve the same quantity of energy as produced by the fusion of $1.0\mathrm{g}$ of hydrogen in the following fusion reaction: $$4_1^1\mathrm{H}⟶{}_2^4\mathrm{He}+2{}_1^0\mathrm{e}$$ Assume that all the products of the combustion of octane are in their gas phases. Use data from Exercise $21.50,$ Appendix $\mathrm{C}$, and the inside covers of the text. The standard enthalpy of formation of octane is $-250.1\mathrm{kJ}/\mathrm{mol}$

Short answer

To evolve the same quantity of energy as produced by the fusion of 1 g of hydrogen, $5.914\times {10}^{6}g$ of octane must be burned in air.

Step-by-step solution

Calculate energy released in fusion of hydrogen

We have 1 g of hydrogen, and we need to calculate the energy produced when it undergoes fusion. First, we have to convert the amount of hydrogen in grams to moles using the formula: $moles=\frac{mass}{molarmass}$ The molar mass of hydrogen is 1 g/mol, so we have: $molesofhydrogen=\frac{1g}{1g/\mathrm{mol}}=1mol$ The fusion reaction can be written as: $4{}_1^1\mathrm{H}⟶{}_2^4\mathrm{He}+2{}_1^0\mathrm{e}$ The energy released is equal to the difference between the products' and reactants' binding energies. Using data from Appendix C, we find: Binding energy of ${}_2^4\mathrm{He}:28.3MeV$ Binding energy of ${}_1^1\mathrm{H}:0MeV$ The energy released per atom during fusion: $Q=(4\times 0-28.3)\times -1MeV=28.3MeV$ Energy released for 1 mol of hydrogen: $Q=N_A\times 28.3MeV=6.022\times {10}^{23}\times 28.3MeV$ Now, we convert the energy from MeV to Joules: 1 MeV = 1.6 × 10⁻¹³ J $Energyreleased=6.022\times {10}^{23}\times 28.3\times 1.6\times {10}^{-13}J=2.71\times {10}^{11}J$

Calculate energy released in combustion of octane

The balanced combustion reaction of octane is: $2{\mathrm{C}}_8{\mathrm{H}}_{18}(l)+25{\mathrm{O}}_2(g)⟶16{\mathrm{CO}}_2(g)+18{\mathrm{H}}_2\mathrm{O}(g)$ The standard enthalpy of formation of octane is given as -250.1 kJ/mol. Using the data from Exercise 21.50 and Appendix C: $\mathrm{\Delta }{H}_{f}C{O}_2(g)=-393.5kJ/mol$ $\mathrm{\Delta }{H}_f{H}_{2}O(g)=-241.8kJ/mol$ Then, we can calculate the enthalpy of the reaction: $\mathrm{\Delta }{H}_{rxn}=(16\times -393.5)+(18\times -241.8)-(2\times -250.1)$ $\Delta H_{rxn} = -10467\,kJ\) per 2 moles of octane burned. Now, we need to find the energy released per 1 mole of octane: $\Delta H_{rxn}^{1\,mole} = \frac{-10467\,kJ}{2} = -5233.5\,kJ/mol\)

Calculate mass of octane needed to evolve same quantity of energy

We have the energy released by both the fusion of hydrogen and combustion of octane. Now, we can find the required moles of octane to release the same amount of energy: $Molesofoctane=\frac{Energyfromfusion}{Energypermoleofcombustion}$ $Molesofoctane=\frac{2.71\times {10}^{11}J}{(-5233.5kJ/mol)\times {10}^{3}J/\mathrm{kJ}}$ $Molesofoctane=5.1776\times {10}^{4}mol$ Now, we can find the mass of the required octane. Use the molar mass of octane (C₈H₁₈) = 114.22 g/mol. Mass of octane needed = moles × molar mass $Massofoctaneneeded=5.1776\times {10}^{4}mol\times 114.22g/\mathrm{mol}$ $Massofoctaneneeded=5.914\times {10}^{6}g$ Therefore, $5.914\times {10}^{6}g$ of octane must be burned in air to evolve the same quantity of energy as produced by the fusion of 1 g of hydrogen.

Key concepts

Fusion Reaction

A fusion reaction is a process where two lighter atomic nuclei combine to form a heavier nucleus. In our exercise, we observe the fusion of hydrogen atoms. During this fusion, four hydrogen nuclei, which are essentially protons, merge to create a helium nucleus and two positron particles.
The equation is simplified as:

• 4 ${}_1^1\mathrm{H}\to {}_2^4\mathrm{He}+2{}_1^0\mathrm{e}$
  

Fusion reactions release incredible amounts of energy. This energy comes from the difference in binding energy between reactants and products. In this case, helium's binding energy is much greater than that of the individual hydrogen nuclei.
This large energy release is due to the principle of mass-energy equivalence, as described by Einstein's famous equation $E=m{c}^2$. Such reactions power the sun and stars, making them immensely powerful sources of energy.

Combustion Reaction

A combustion reaction occurs when a substance, typically a hydrocarbon, reacts with oxygen to produce carbon dioxide, water, and energy. In the given exercise, octane undergoes combustion. Octane, with its chemical formula ${\mathrm{C}}_8{\mathrm{H}}_{18}$, reacts with oxygen in a highly exothermic process.
The balanced equation for this reaction is:

• 2 ${\mathrm{C}}_8{\mathrm{H}}_{18}(l)+25{\mathrm{O}}_2(g)\to 16{\mathrm{CO}}_2(g)+18{\mathrm{H}}_2\mathrm{O}(g)$
  

This type of reaction is what allows cars and other machines to operate, as the energy released is converted into useful work. Combustion reactions are characterized by the release of energy in the form of heat and light.
Understanding the energy released in combustion reactions, often measured in kilojoules per mole, is crucial in calculations involving fuels and energy balances.

Enthalpy of Formation

The standard enthalpy of formation, $\mathrm{\Delta }{H}_f^{°}$, represents the change in enthalpy when one mole of a compound forms from its elements in their standard states.
In our scenario, it's crucial to determine the energy involved when octane combusts. Given:

• The $\mathrm{\Delta }{H}_f^{°}$ for octane is -250.1 kJ/mol
  
• For ${\mathrm{CO}}_2(g)$ is -393.5 kJ/mol, and for ${\mathrm{H}}_2\mathrm{O}(\mathrm{g})$ is -241.8 kJ/mol

To find the total reaction enthalpy, use the formula:$$\mathrm{\Delta }{H}_{rxn}=\sum (\mathrm{\Delta }{H}_f^{\circ }\text{{ of products}})-\sum (\mathrm{\Delta }{H}_f^{\circ }\text{{ of reactants}})$$ This calculation tells us how much energy is evolved or absorbed during reactions.
Understanding $\mathrm{\Delta }{H}_f^{°}$ values is important for predicting the energetics of reactions, aiding in the comparison of different chemical processes.

Molar Mass Calculation

Molar mass is a fundamental concept that links mass with the amount of substance. It's the mass of one mole of a substance, normally expressed in grams per mole.
Calculating molar mass involves adding up the atomic masses of the elements present in the compound. For octane (${\mathrm{C}}_8{\mathrm{H}}_{18}$), calculate as follows:

• Carbon: 8 atoms $\times 12.01g/mol=96.08g/mol$
• Hydrogen: 18 atoms $\times 1.008g/mol=18.144g/mol$

Hence, the molar mass of octane is roughly 114.22 g/mol. Having this value is essential when converting between grams and moles, especially in stoichiometric calculations.
It's important because it allows us to relate the microscopic molecular scale to macroscopic quantities, making it possible to measure and react suitable amounts of chemical substances.