consider the following reaction:
C3H8 (g)+ 5O2 (g)--> 3CO2 (g)+ 4H20 (g)
Delta H = -2440kJ
The heat of formation of CO2 is -390 kJ/mol and the heat of formation of H2O is -240 kJ/mol. What is the heat of formation of propane?
Using Hess' Law....
3*(-390)+4(-240)-Hf=-2440
solve for Hf(propane).
To calculate the heat of formation of propane (C3H8), we can use the given information about the reaction and the heat of formation values for CO2 and H2O.
The balanced equation for the combustion of propane is:
C3H8 (g) + 5O2 (g) --> 3CO2 (g) + 4H2O (g)
We know the heat of formation of CO2 is -390 kJ/mol and the heat of formation of H2O is -240 kJ/mol.
To calculate the heat of formation of propane, we need to look at the reaction in terms of the formation of the products from their elements in their standard states.
The heat of formation of a substance is defined as the enthalpy change when one mole of the substance is formed from its elements. In this case, propane (C3H8) is formed from carbon (C) and hydrogen (H2) elements.
Looking at the reaction, we can see that 1 mole of propane (C3H8) is formed from 3 moles of carbon (C) and 8 moles of hydrogen (H2). So, we need to calculate the heat of formation for 1 mole of propane (C3H8) based on these elements.
Using the given reaction, we can relate the heat of formation for propane to the heat of formation for CO2 and H2O.
The enthalpy change (ΔH) for the reaction is given as -2440 kJ. This value represents the heat released when 1 mole of propane is combusted.
We know that 3 moles of CO2 are formed in the reaction, and the heat of formation for CO2 is -390 kJ/mol. So, the enthalpy change due to the formation of 3 moles of CO2 is (-390 kJ/mol) * 3 moles = -1170 kJ.
We also know that 4 moles of H2O are formed in the reaction, and the heat of formation for H2O is -240 kJ/mol. So, the enthalpy change due to the formation of 4 moles of H2O is (-240 kJ/mol) * 4 moles = -960 kJ.
Now, we can calculate the heat of formation of propane (C3H8) using the enthalpy change of the reaction:
ΔH = Heat of formation of products - Heat of formation of reactants.
-2440 kJ = (-1170 kJ -960 kJ) - Heat of formation of propane.
Rearranging the equation, we get:
Heat of formation of propane = (-2440 kJ + 1170 kJ + 960 kJ).
Calculating this expression:
Heat of formation of propane = -2440 kJ + 1170 kJ + 960 kJ = -310 kJ/mol.
Therefore, the heat of formation of propane (C3H8) is -310 kJ/mol.
To calculate the heat of formation of propane (C3H8), we need to use the heat of formation values for carbon dioxide (CO2) and water (H2O) and apply them to the given reaction.
The heat of formation (ΔHf) represents the change in energy when one mole of a compound is formed from its constituent elements in their standard states. The standard states for elements are usually in their most stable forms at standard conditions (25°C and 1 atm).
In the given reaction: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g), the stoichiometric coefficients represent the number of moles of each compound involved in the reaction. Therefore, the heat of formation of propane can be determined by calculating the overall change in enthalpy using the heat of formation values for CO2 and H2O.
The balanced equation shows that 3 moles of CO2 are formed when 1 mole of propane is consumed. Thus, the heat of formation of propane can be calculated as follows:
ΔH = (3 × ΔHf(CO2)) + (4 × ΔHf(H2O))
Substituting the given heat of formation values:
ΔH = (3 × -390 kJ/mol) + (4 × -240 kJ/mol)
Calculating the value:
ΔH = -1170 kJ/mol - 960 kJ/mol
ΔH = -2130 kJ/mol
Therefore, the heat of formation of propane (C3H8) is -2130 kJ/mol.