What is the change of enthalpy associated with the combustion of one mole of pentane?
C5H12 + 8O2 �¨ 5CO2 + 6H2O
I'm very confused, Please help!
To determine the change in enthalpy associated with the combustion of one mole of pentane (C5H12), you need to use the balanced chemical equation and the enthalpy of formation values.
Step 1: Write the balanced chemical equation:
C5H12 + 8O2 -> 5CO2 + 6H2O
The coefficients in the balanced equation indicate the stoichiometric ratio between the reactants and products.
Step 2: Find the enthalpy of formation values:
The enthalpy of formation (ΔHf°) values represent the change in enthalpy when one mole of a compound is formed from its elements in their standard states (e.g., at 25°C and 1 atm pressure).
In this case, you need the enthalpy of formation values for pentane (C5H12), carbon dioxide (CO2), and water (H2O).
Step 3: Calculate the change in enthalpy:
The change in enthalpy (ΔH) is calculated using the following formula:
ΔH = Σ(ΔHf° products) - Σ(ΔHf° reactants)
First, multiply the enthalpy of formation values of the products by their stoichiometric coefficients and sum them up. Then, do the same for the reactants.
Finally, subtract the sum of the reactants' enthalpies from the sum of the products' enthalpies to find the change in enthalpy (ΔH).
Note: It's essential to use consistent units for the enthalpy of formation values, such as kilojoules per mole (kJ/mol), to ensure correct calculations.
I recommend using reliable reference sources like handbooks or databases to obtain accurate enthalpy of formation values.
To determine the change in enthalpy associated with the combustion of pentane, we need to use the enthalpy values of the reactants and products.
First, we need to find the enthalpy change of formation for each compound involved in the reaction. The enthalpy change of formation is the amount of energy released or absorbed when one mole of a compound is formed from its elements in their standard states at standard conditions.
Using standard enthalpy of formation values (ΔHf°), we can calculate the change in enthalpy for the reaction:
C5H12 (pentane): ΔHf° = -173.7 kJ/mol
CO2 (carbon dioxide): ΔHf° = -393.5 kJ/mol
H2O (water): ΔHf° = -285.8 kJ/mol
O2 (oxygen): O2 does not have an enthalpy of formation since it is an elemental form.
Now, we need to determine the stoichiometric coefficients (numbers in front of the compounds) for each compound in the balanced chemical equation:
C5H12 + 8O2 -> 5CO2 + 6H2O
From the balanced equation, we see that for 1 mole of pentane (C5H12), we need 8 moles of oxygen (O2) to react. This means that the reaction involves 8 moles of oxygen gas.
To calculate the change in enthalpy, we can use the following equation:
ΔH = Σ(nΔHf°(products)) - Σ(nΔHf°(reactants))
where n is the stoichiometric coefficient for each compound.
For the reactants:
Σ(nΔHf°(reactants)) = (1 mol)(ΔHf°(C5H12)) + (8 mol)(0 kJ/mol)
The 0 kJ/mol for oxygen is because it is in its elemental form and does not have an enthalpy of formation value.
For the products:
Σ(nΔHf°(products)) = (5 mol)(ΔHf°(CO2)) + (6 mol)(ΔHf°(H2O))
Now we can substitute the values:
ΔH = (1 mol)(-173.7 kJ/mol) + (5 mol)(-393.5 kJ/mol) + (6 mol)(-285.8 kJ/mol)
Calculating this expression will give you the change in enthalpy associated with the combustion of one mole of pentane.
C5H12 + 8O2 ==> 5CO2 + 6H2O
dHrxn = (n*dHf products) - (n*dHf reactants).
Look up the heat formation of the products and reactants, plug into the above and solve for dHrxn.