Question

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.

Short answer

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

(b) The volume of air = 254L.

(c)\({\rm{\Delta Hf}}_{{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}^{\rm{o}}\)= -104. 5 kJ/mol.

(d) T = 75.01˚C.

Step-by-step solution

Combustion reaction of propane gas

(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:

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

Balanced reaction

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:

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

Calculation of volume of oxygen

(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:

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

Calculation of the volume of air

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

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

Heat of formation

(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:

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

Heat of combustion

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:

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

Change in temperature of water

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

The temperature change can be evaluated as:

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