Propane, \({{\bf{C}}_{\bf{3}}}{{\bf{H}}_{\bf{8}}}\), is a hydrocarbon that is commonly used as a fuel.

(a) Write a balanced equation for the complete combustion of propane gas.

(b) Calculate the volume of air at 25 °C and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent O₂ by volume. (Hint: we will see how to do this calculation in a later

chapter on gases—for now use the information that 1.00 L of air at 25 °C and 1.00 atm contains 0.275 g of O2 per liter.)

(c) The heat of combustion of propane is −2,219.2 kJ/mol. Calculate the heat of formation,ΔHf °of propane given thatΔHf °of H₂O(l) = −285.8 kJ/mol andΔHf °of CO₂(g) = −393.5 kJ/mol.

(d) Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00 kilograms of water, calculate the increase in temperature of the water.

(a) A combustion reaction occurs with oxygen as one reactant and propane as another reactant. This combustion process releases energy as heat or light.

A propane gas on a combustion reaction forms\({\rm{C}}{{\rm{O}}_{\rm{2}}}\)and\({{\rm{H}}_{\rm{2}}}{\rm{O}}\)as follows:

\({{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}\left( {\rm{g}} \right){\rm{ + }}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right){\rm{\;}} \to {\rm{C}}{{\rm{O}}_{\rm{2}}}\left( {\rm{g}} \right){\rm{\; + \;}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( {\rm{g}} \right).\)

To balance the equation, we have to multiply\({\rm{C}}{{\rm{O}}_{\rm{2}}}\)by 3 and\({{\rm{H}}_{\rm{2}}}{\rm{O}}\)by 4 and\({{\rm{O}}_2}\)by 5. The balanced reaction is:

\({{\bf{C}}_{\bf{3}}}{{\bf{H}}_{\bf{8}}}\left( {\bf{g}} \right){\bf{ + 5}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right){\bf{\;}} \to {\bf{3C}}{{\bf{O}}_{\bf{2}}}\left( {\bf{g}} \right){\bf{\; + \;4}}{{\bf{H}}_{\bf{2}}}{\bf{O}}\left( {\bf{g}} \right).\)

(b) The molecular weight of propane = 44.094 g/mol. The amount of propane is calculated as follows:

\(\begin{array}{c}{\rm{Amount of propane = 25g \times }}\frac{{\rm{1}}}{{{\rm{44}}{\rm{.094 g}}}}{\rm{ }}\\{\rm{ = 0}}{\rm{.566 mol}}{\rm{.}}\end{array}\)

For the combustion of 25g of propane, 2.834 moles of oxygen gas is required.

The density of the oxygen = 1.429g/L.

The molar mass of the oxygen = 31.998g/mol.

We can calculate the volume of the oxygen as follows:

\(\begin{array}{c}{{\rm{V}}_{{{\rm{O}}_{\rm{2}}}}}{\rm{ = \;\;2}}{\rm{.834 mole of }}{{\rm{O}}_{\rm{2}}}{\rm{ \times }}\frac{{{\rm{31}}{\rm{.998 g of }}{{\rm{O}}_{\rm{2}}}}}{{{\rm{1 mole of }}{{\rm{O}}_{\rm{2}}}}}{\rm{ \times \;}}\frac{{{\rm{1 L\;}}{{\rm{O}}_{\rm{2}}}}}{{{\rm{1}}{\rm{.429 g }}{{\rm{O}}_{\rm{2}}}}}\\{\rm{\; = \;\;63}}{\rm{.458 L }}{{\rm{O}}_{\rm{2}}}{\rm{.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}}\end{array}\)

By considering that oxygen is 21% by volume, we can calculate the volume of air needed to combust 25g of propane as follows:

\(\begin{array}{c}{{\rm{V}}_{{\rm{air}}}}{\rm{ = \;\;}}\frac{{{\rm{63}}{\rm{. 458 L}}}}{{{\rm{0}}{\rm{.25}}}}{\rm{\;\;}}\\{\rm{ = \;\;253}}{\rm{.832}}\\{\rm{ = \;254 L}}{\rm{.}}\end{array}\)

(c) The standard enthalpy of formation of a chemical reaction\(\left( {{\rm{\Delta H}}_{{\rm{reaction}}}^{\rm{o}}} \right)\)can be calculated byusing the expression shown below.

\({\bf{\Delta H}}_{{\bf{reaction}}}^{\bf{o}}{\bf{ = \;\;\Sigma n \times H}}_{{\bf{products}}}^{\bf{o}}{\bf{ - \Sigma n \times \Delta H}}_{{\bf{reactants}}}^{\bf{o}}.\)

By putting the values in the above equation, we get:

\(\begin{array}{*{20}{l}}{{\rm{ - 2,219}}{\rm{.2 kJ/mol\;\; = \;\;}}\left[ {{\rm{ 3 \times }}\left( {{\rm{ - 393}}{\rm{.5 kJ/mol}}} \right) + {\rm{4 \times }}\left( {{\rm{ - 285}}{\rm{.8 kJ/mol}}} \right)} \right]{\rm{ - \Delta Hf}}_{{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}^{\rm{o}}}\\{{\rm{ \Delta Hf}}_{{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}^{\rm{o}}{\rm{\; = \;\; - 2323}}{\rm{.7 kJ/mol\; + 2,219}}{\rm{.2 kJ/mol}}}\\{{\rm{\;\;\;\;\;\;\; \Delta Hf}}_{{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}^{\rm{o}}{\rm{ = \;\; - 104}}{\rm{.5 kJ/mol}}{\rm{.}}}\end{array}\)

Assuming that there is no phase transition, the amount of heat energy transferred to or from a material of mass m and specific heat capacity can be calculated by using the expression\({\rm{Q = mc\Delta T,}}\)where\({\rm{\Delta T}}\)is the temperature change of the system.

\(\begin{array}{c}{\rm{Number of moles = }}\frac{{{\rm{25g}}}}{{{\rm{44}}{\rm{.1g/mol}}}}\\{\rm{ = 0}}{\rm{.566mol}}{\rm{.}}\end{array}\)

By the given enthalpy of the combustion, the amount of heat energy released from burning 25g of propane can be calculated as follows:

\(\begin{array}{*{20}{l}}{{\rm{\Delta }}{{\rm{H}}_{{\rm{combustion}}}}{\rm{ = \;2219}}{\rm{.2 kJ/mol \times }}\left( {{\rm{0}}{\rm{.566 mol}}} \right)}\\{{\rm{\; = \;\;1256}}{\rm{.06 kJ}}{\rm{.}}}\end{array}\)

Specific heat capacity of water = 4.186kJ/ kg ˚C.

The temperature change can be evaluated as:

\(\begin{array}{*{20}{l}}{{\rm{\Delta T = \;}}\frac{{{\rm{1256}}{\rm{.06 kJ}}}}{{\left( {{\rm{ 4 kg\; \times \;4}}{\rm{.186 kJ/ kg}}{{\rm{ }}^{\rm{o}}}{\rm{C}}} \right)}}}\\{{\rm{\; = \;75}}{\rm{.01}}{{\rm{\;}}^{\rm{o}}}{\rm{C}}{\rm{.}}}\end{array}\)