please help me? :)
topic: standard enthalpies of formation
"Calculate ΔHf° of octane, C8H18(l), given the entalpy of combustion of octane to CO2(g) and H2O(l) is -5471kJ/mol. The standard enthalpies of formation ofCO2 and H2O are given: CO2(g)ΔHf°=-393.5kJ/mol and H2O(l) ΔHf°=-285.8kJ/mol"
I'm not really sure how to go about this... do I use ΔHrxn°= (sum)nΔHf°prod - (sum)mΔHf°reactant ? Also, can you go through this with me step by step if need be? Thank you for your time!
Yes, you have the right formula. Substitute the delta H values you have listed in the problem. The equation is
2C8H18 + 25O2 ==> 16CO2 + 18H2O so delta H for THIS reaction (2 moles that is) is 2*-547 kJ.
Sure, I'll be happy to help you step by step.
To calculate the standard enthalpy of formation (ΔHf°) of octane (C8H18), you will indeed use the equation ΔHrxn° = (sum)nΔHf°prod - (sum)mΔHf°reactant.
First, you need to determine the stoichiometric coefficients (n and m) for the combustion reaction of octane (C8H18) to CO2 and H2O. From the balanced equation of the combustion reaction:
C8H18(l) + 12.5O2(g) → 8CO2(g) + 9H2O(l)
The stoichiometric coefficients for octane, CO2, and H2O are 1, 8, and 9, respectively.
Next, substitute the given standard enthalpies of formation values into the equation:
ΔHrxn° = (8 × ΔHf°CO2) + (9 × ΔHf°H2O) - ΔHf°octane
Given that:
ΔHf°CO2 = -393.5 kJ/mol (from the given data)
ΔHf°H2O = -285.8 kJ/mol (from the given data)
ΔHrxn° = -5471 kJ/mol (given enthalpy of combustion of octane)
Now, plug in the values and solve for ΔHf°octane:
-5471 kJ/mol = (8 × -393.5 kJ/mol) + (9 × -285.8 kJ/mol) - ΔHf°octane
Let's calculate it step by step:
Step 1: Multiply the stoichiometric coefficients by the respective standard enthalpies of formation:
-5471 kJ/mol = (8 × -393.5 kJ/mol) + (9 × -285.8 kJ/mol) - ΔHf°octane
-5471 kJ/mol = -3148 kJ/mol + (-2572.2 kJ/mol) - ΔHf°octane
Step 2: Combine the terms on the right-hand side:
-5471 kJ/mol = -5720.2 kJ/mol - ΔHf°octane
Step 3: Isolate ΔHf°octane by subtracting the right-hand side from both sides of the equation:
-5471 kJ/mol + 5720.2 kJ/mol = - ΔHf°octane
ΔHf°octane = -249.2 kJ/mol
Therefore, the standard enthalpy of formation of octane (C8H18) is -249.2 kJ/mol.
Sure, let's go step-by-step to calculate the standard enthalpy of formation (ΔHf°) of octane (C8H18).
Step 1: Write the balanced equation for the combustion of octane:
C8H18(l) + 12.5O2(g) -> 8CO2(g) + 9H2O(l)
Step 2: Determine the stoichiometric coefficients of the reaction. From the balanced equation, we can see that 1 mole of octane reacts with 12.5 moles of oxygen gas (O2), producing 8 moles of carbon dioxide (CO2) and 9 moles of water (H2O).
Step 3: Calculate the enthalpy change of the combustion reaction. We are given that the enthalpy change of the combustion reaction is -5471 kJ/mol. This value represents the enthalpy change when 1 mole of octane reacts with the stoichiometric amounts of oxygen gas, carbon dioxide, and water.
Step 4: Calculate the enthalpy change of formation of the products. We are given the standard enthalpies of formation (ΔHf°) for CO2 and H2O. These values represent the enthalpy changes when 1 mole of each compound is formed from their constituent elements in their standard states (usually at 25°C and 1 atm).
For CO2: ΔHf°(CO2) = -393.5 kJ/mol
For H2O: ΔHf°(H2O) = -285.8 kJ/mol
Step 5: Use the equation ΔHrxn° = Σ(nΔHf°products) - Σ(mΔHf°reactants), where n and m represent the stoichiometric coefficients from the balanced equation.
ΔHrxn° = (8 × ΔHf°(CO2)) + (9 × ΔHf°(H2O)) - ΔHcombustion
ΔHrxn° = (8 × -393.5 kJ/mol) + (9 × -285.8 kJ/mol) - (-5471 kJ/mol)
Step 6: Calculate the value of ΔHf° for octane.
ΔHf°(octane) = ΔHrxn° / 1 mol
ΔHf°(octane) = [(8 × -393.5 kJ/mol) + (9 × -285.8 kJ/mol) - (-5471 kJ/mol)] / 1 mol
Simplifying the equation will give you the numerical value of ΔHf°(octane).