Question

The standard enthalpies of formation of gaseous propyne $\left({\mathrm{C}}_3{\mathrm{H}}_4\right),$ propylene $\left({\mathrm{C}}_3{\mathrm{H}}_6\right),$ and propane $\left({\mathrm{C}}_3{\mathrm{H}}_8\right)$ are +185.4 $+20.4,$ and $-103.8\mathrm{kJ}/\mathrm{mol}$, respectively. (a) Calculate the heat evolved per mole on combustion of each substance to yield ${\mathrm{CO}}_2(g)$ and ${\mathrm{H}}_2\mathrm{O}(g)$ (b) Calculate the heat evolved on combustion of $1\mathrm{kg}$ of each substance. (c) Which is the most efficient fuel in terms of heat evolved per unit mass?

Short answer

The heat evolved per mole on combustion (enthalpy change of combustion) is -2171.6 kJ/mol for propyne, -2213.0 kJ/mol for propylene, and -2219.2 kJ/mol for propane. The heat evolved on combustion of 1 kg of each substance is -54,180 kJ/kg for propyne, -52,570 kJ/kg for propylene, and -50,270 kJ/kg for propane. Propyne is the most efficient fuel in terms of heat evolved per unit mass.

Step-by-step solution

Write balanced chemical equations for the combustion reactions

For the combustion of propyne, propylene, and propane, we will write balanced chemical equations as follows: Propyne: $C_3{H}_4(g)+\frac{5}{2}{O}_2(g)\to 3C{O}_2(g)+2H_{2}O(g)$ Propylene: $C_3{H}_6(g)+\frac{9}{2}{O}_2(g)\to 3C{O}_2(g)+3H_{2}O(g)$ Propane: $C_3{H}_8(g)+5O_2(g)\to 3C{O}_2(g)+4H_{2}O(g)$

Calculate the heat evolved per mole for each combustion reaction

We will use the enthalpy change formula: $\mathrm{\Delta }{H}_{rxn}=\sum \mathrm{\Delta }{H}_{f(products)}-\sum \mathrm{\Delta }{H}_{f(reactants)}$ Propyne: $\mathrm{\Delta }{H}_{rxn}=[3\mathrm{\Delta }{H}_f(C{O}_2)+2\mathrm{\Delta }{H}_f(H_{2}O)]-[\mathrm{\Delta }{H}_f(C_3{H}_4)+\frac{5}{2}\mathrm{\Delta }{H}_f(O_2)]$ Since, $\mathrm{\Delta }{H}_f(O_2)=0$, the equation simplifies to: $\mathrm{\Delta }{H}_{rxn}=[3\mathrm{\Delta }{H}_f(C{O}_2)+2\mathrm{\Delta }{H}_f(H_{2}O)]-\mathrm{\Delta }{H}_f(C_3{H}_4)$ Propylene: $\mathrm{\Delta }{H}_{rxn}=[3\mathrm{\Delta }{H}_f(C{O}_2)+3\mathrm{\Delta }{H}_f(H_{2}O)]-[\mathrm{\Delta }{H}_f(C_3{H}_6)+\frac{9}{2}\mathrm{\Delta }{H}_f(O_2)]$ $\mathrm{\Delta }{H}_{rxn}=[3\mathrm{\Delta }{H}_f(C{O}_2)+3\mathrm{\Delta }{H}_f(H_{2}O)]-\mathrm{\Delta }{H}_f(C_3{H}_6)$ Propane: $\mathrm{\Delta }{H}_{rxn}=[3\mathrm{\Delta }{H}_f(C{O}_2)+4\mathrm{\Delta }{H}_f(H_{2}O)]-[\mathrm{\Delta }{H}_f(C_3{H}_8)+5\mathrm{\Delta }{H}_f(O_2)]$ $\mathrm{\Delta }{H}_{rxn}=[3\mathrm{\Delta }{H}_f(C{O}_2)+4\mathrm{\Delta }{H}_f(H_{2}O)]-\mathrm{\Delta }{H}_f(C_3{H}_8)$ Now, we will calculate $\mathrm{\Delta }{H}_{rxn}$ for each substance using the given values for standard enthalpies of formation: $\mathrm{\Delta }{H}_f(C{O}_2)=-393.5\mathrm{kJ}/\mathrm{mol}$, $\mathrm{\Delta }{H}_f(H_{2}O)=-241.8\mathrm{kJ}/\mathrm{mol}$, $\mathrm{\Delta }{H}_{rxn}(C_3{H}_4)=[(3\times (-393.5))+(2\times (-241.8))]-185.4$ $\mathrm{\Delta }{H}_{rxn}(C_3{H}_6)=[(3\times (-393.5))+(3\times (-241.8))]-20.4$ $\mathrm{\Delta }{H}_{rxn}(C_3{H}_8)=[(3\times (-393.5))+(4\times (-241.8))]-(-103.8)$ Calculating these values, we get: $\mathrm{\Delta }{H}_{rxn}(C_3{H}_4)=-2171.6\mathrm{kJ}/\mathrm{mol}$ $\mathrm{\Delta }{H}_{rxn}(C_3{H}_6)=-2213.0\mathrm{kJ}/\mathrm{mol}$ $\mathrm{\Delta }{H}_{rxn}(C_3{H}_8)=-2219.2\mathrm{kJ}/\mathrm{mol}$

Calculate the heat evolved on combustion of 1 kg of each substance

First, we need to find the molar mass for each substance: Propyne: $C_3{H}_4=3(12.01)+4(1.01)=40.07\mathrm{g}/\mathrm{mol}$ Propylene: $C_3{H}_6=3(12.01)+6(1.01)=42.09\mathrm{g}/\mathrm{mol}$ Propane: $C_3{H}_8=3(12.01)+8(1.01)=44.11\mathrm{g}/\mathrm{mol}$ Now, we will find the heat evolved per kilogram for each substance: $\mathrm{\Delta }{H}_{rxn,kg}(C_3{H}_4)=\frac{-2171.6\mathrm{kJ}/\mathrm{mol}}{40.07\mathrm{g}/\mathrm{mol}}\times 1000=-54,180\mathrm{kJ}/\mathrm{kg}$ $\mathrm{\Delta }{H}_{rxn,kg}(C_3{H}_6)=\frac{-2213.0\mathrm{kJ}/\mathrm{mol}}{42.09\mathrm{g}/\mathrm{mol}}\times 1000=-52,570\mathrm{kJ}/\mathrm{kg}$ $\mathrm{\Delta }{H}_{rxn,kg}(C_3{H}_8)=\frac{-2219.2\mathrm{kJ}/\mathrm{mol}}{44.11\mathrm{g}/\mathrm{mol}}\times 1000=-50,270\mathrm{kJ}/\mathrm{kg}$

Determine the most efficient fuel in terms of heat evolved per unit mass

Comparing the heat evolved per kilogram for each substance, we can determine the most efficient fuel: $\mathrm{\Delta }{H}_{rxn,kg}(C_3{H}_4)=-54,180\mathrm{kJ}/\mathrm{kg}$ $\mathrm{\Delta }{H}_{rxn,kg}(C_3{H}_6)=-52,570\mathrm{kJ}/\mathrm{kg}$ $\mathrm{\Delta }{H}_{rxn,kg}(C_3{H}_8)=-50,270\mathrm{kJ}/\mathrm{kg}$ As we can see, propyne has the highest heat evolved per unit mass among the three fuels, making it the most efficient fuel.

Key concepts

Enthalpy of Formation

Enthalpy of formation is the change in heat content during the formation of one mole of a compound from its constituent elements, with all substances in their standard states. It is denoted as $\mathrm{\Delta }{H}_f$ and its unit is kilojoules per mole (kJ/mol). This concept is central to understanding how much energy is released or absorbed during a chemical reaction.For example, when we consider elements such as oxygen gas, $O_2$ in its natural state has an enthalpy of formation of zero because it is already in its basic form. However, when we discuss compounds, the enthalpy of formation will vary significantly and become either negative or positive, indicating whether the reaction is exothermic (releases heat) or endothermic (absorbs heat) respectively.

Chemical Combustion Reaction

A chemical combustion reaction involves a substance (usually a hydrocarbon) reacting with oxygen gas to produce carbon dioxide, water, and heat. Combustion reactions are almost always exothermic, meaning they release heat.These reactions follow the general formula of $\text{Fuel}+O_2\to C{O}_2+H_{2}O+\text{heat}$. For hydrocarbons, the fuel can be any compound composed of carbon and hydrogen. The coefficients for each substance must be balanced to comply with the law of conservation of mass. The heat evolved in these reactions can be measured and is a crucial consideration for the application of fuels in various sectors, such as transportation and energy production.

Heat Evolved in Reaction

The heat evolved in a reaction, also known as the reaction enthalpy ($\mathrm{\Delta }{H}_{rxn}$ ), is the total amount of heat released or absorbed during a chemical reaction. In exothermic reactions, like combustion, the heat evolved is released into the surroundings, and $\mathrm{\Delta }{H}_{rxn}$ is negative. To calculate this value, we apply the formula $\mathrm{\Delta }{H}_{rxn}=\sum \mathrm{\Delta }{H}_{f(products)}-\sum \mathrm{\Delta }{H}_{f(reactants)}$ which considers the enthalpy of formation values of both the reactants and the products.In educational exercises, these values help students understand the energetics of chemical reactions and the practical energy outputs that can be expected from different fuels. An accurate calculation of the heat evolved is important for applications where energy efficiency and fuel performance are critical.

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction. It is governed by the balanced chemical equation that provides the mole ratios of the reactants and products involved. These ratios are crucial for calculating the amounts of substances required or produced in a reaction.In practical terms, stoichiometry allows us to predict the amount of product that can be formed from a given quantity of reactants. For instance, in the case of combustion reactions, stoichiometry aids in determining how much carbon dioxide and water are produced from burning a certain amount of fuel. This quantitative aspect of stoichiometry helps in making informed decisions about fuel usage and can be used to calculate the efficiency of different fuels based on their mass-to-energy conversion ratios.