What is the heat of combustion of ethane, C2H6, in kilojoules per mole of ethane?
You can calculate the heat of combustion from delta Hformation.
C2H6 + 7/2 O2 ==> 2CO2 + 3H2O
dHrxn = [(2*dHf CO2)+(3*dHf H2O)]-[dHf C2H6]
The heat of combustion of ethane, C2H6, can be determined using the balanced chemical equation for its combustion.
The balanced equation for the combustion of ethane is:
C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
According to this equation, 1 mole of ethane reacts with 7/2 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.
The heat of combustion can be calculated by considering the enthalpy change of the reaction.
The standard enthalpy change (ΔH°) of combustion for ethane is -1560 kJ/mol. This represents the heat released when 1 mole of ethane is completely burned in excess oxygen under standard conditions (25°C and 1 atm).
Therefore, the heat of combustion of ethane, C2H6, is -1560 kJ/mol.
To find the heat of combustion of ethane, C2H6, in kilojoules per mole of ethane, you will need to refer to the respective thermochemical equation and use the corresponding data.
The thermochemical equation for the combustion of ethane can be written as follows:
C2H6 + (3.5 * O2) -> 2 * CO2 + 3 * H2O
In this equation, ethane (C2H6) reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The coefficient of the oxygen molecule, 3.5, ensures that the equation is balanced in terms of atoms.
To calculate the heat of combustion, you will need to refer to the molar enthalpy of formation (∆Hf) values for ethane, carbon dioxide, and water.
∆Hf values are typically given in kilojoules per mole of the substance at standard conditions (298 K and 1 atm).
The values you will need are:
∆Hf of C2H6 = -84.68 kJ/mol
∆Hf of CO2 = -393.51 kJ/mol
∆Hf of H2O = -285.83 kJ/mol
The heat of combustion (∆Hc) is equal to the sum of the molar enthalpies of formation of the products minus the sum of the molar enthalpies of formation of the reactants.
∆Hc = [2 * ∆Hf (CO2)] + [3 * ∆Hf (H2O)] - [∆Hf (C2H6)]
Substituting the given values into the equation, we get:
∆Hc = [2 * (-393.51 kJ/mol)] + [3 * (-285.83 kJ/mol)] - [-84.68 kJ/mol]
Simplifying the equation further, we have:
∆Hc = -787.02 kJ/mol + (-857.49 kJ/mol) + 84.68 kJ/mol
∆Hc = -787.02 kJ/mol - 857.49 kJ/mol + 84.68 kJ/mol
∆Hc = -1559.83 kJ/mol
Therefore, the heat of combustion of ethane, C2H6, is -1559.83 kJ/mol. Note that the negative sign indicates that the reaction is exothermic (releases heat).