The standard heat of combustion of eicosane, C20H42 (s), a typical component of candle wax, is -1.332 x 104 kJ/mol (consider this number to be exact in your calculations), when it burns in pure oxygen, and the products are cooled to 25°C. The only products are CO2 (g) and H2O(l).
Calculate ΔHf° for eicosane if ΔHf°(CO2 (g)) = -393.5 kJ/mol and ΔHfo(H2O(l)) = -285.9 kJ/mol.
C20H42 + 61/2 O2 ==> 20CO2 + 21H2O
dHrxn = (n*dHf CO2) + (n*dHf H2O(l)) - (n*dHf x) where x = C20H42
You're given Hf CO2 and Hf for liquid H2O and dHrxn, solve for x.
-1.332E4 = (21*-393.5)+(20*-285.9)-(n*dHf x)
-1.332E4 = -1.39815E4-(n*dHf x)
-6.615E2 = x
Well, isn't eicosane just a fancy way of saying candle wax? No wonder candles smell so delicious when they burn!
Alright, let's break this down. We know the heat of combustion for eicosane is -1.332 x 10^4 kJ/mol. This means that when one mole of eicosane burns, it releases that amount of heat.
Now, we also know that the only products of the combustion are CO2 (g) and H2O(l). So we need to figure out how many moles of CO2 and H2O are formed when one mole of eicosane burns.
The balanced equation for the combustion of eicosane is:
C20H42 + 61O2 → 20CO2 + 21H2O
From the balanced equation, we can see that 20 moles of CO2 and 21 moles of H2O are formed when one mole of eicosane burns.
Now let's calculate the heat released when one mole of CO2 is formed. The enthalpy change of formation of CO2 is -393.5 kJ/mol. Since there are 20 moles of CO2 formed, the total heat released is 20 * -393.5 kJ/mol.
Similarly, the heat released when one mole of H2O is formed is -285.9 kJ/mol. With 21 moles of H2O formed, the total heat released is 21 * -285.9 kJ/mol.
Now, to find the ΔHf° for eicosane, we need to subtract the total heat released from the heat of combustion. So it would be:
ΔHf°(eicosane) = heat of combustion - (total heat released from CO2 + total heat released from H2O)
Plugging in the numbers, we get:
ΔHf°(eicosane) = -1.332 x 10^4 kJ/mol - (20 * -393.5 kJ/mol + 21 * -285.9 kJ/mol)
Now go ahead and crunch those numbers to find the answer. I hope this helps shed some light on the heat of formation of eicosane, while also adding a little warmth to your understanding!
To calculate ΔHf° for eicosane, we can use the equation:
ΔHf°(eicosane) = ΣΔHf°(products) - ΣΔHf°(reactants)
First, let's determine the balanced chemical equation for the combustion of eicosane:
C20H42 (s) + [x]O2 (g) → [y]CO2 (g) + [z]H2O (l)
The balanced chemical equation indicates that for every one mole of eicosane burned, we need [x] moles of oxygen, and it produces [y] moles of carbon dioxide and [z] moles of water.
From the given data, we know that:
ΔHf°(CO2 (g)) = -393.5 kJ/mol
ΔHf°(H2O(l)) = -285.9 kJ/mol
Since the reactants are eicosane (C20H42) and oxygen (O2), we need to know the enthalpy of formation for these compounds in order to determine their values. However, the problem does not provide these values explicitly.
Without the enthalpy of formation values for C20H42 and O2, we cannot directly calculate ΔHf° for eicosane. We can only determine the ΔH°, the change in enthalpy, for the combustion reaction.
ΔH° = ΣΔH°(products) - ΣΔH°(reactants)
To calculate ΔH°, we will use the standard heat of combustion of eicosane:
ΔH° = -1.332 x 10^4 kJ/mol
This information allows us to calculate the ΔHf° of eicosane indirectly but not its exact value.