For a particular isomer of C8H18, the following reaction produces 5104.1 kJ of heat per mole of C8H18(g) consumed, under standard conditions. (deltaH_rnx = -5104.1kJ)
What is the standard enthalpy of formation of this isomer of C8H18(g)?
-220.1
To find the standard enthalpy of formation of the isomer of C8H18, we need to use the given information about the reaction and apply the Hess's Law.
Hess's Law states that the overall enthalpy change of a reaction is independent of the pathway, and depends only on the initial and final states. In this case, we can use the enthalpy of the reaction and the known enthalpies of formation of other substances to calculate the standard enthalpy of formation of C8H18.
The standard enthalpy of formation (ΔHf°) is defined as the enthalpy change when one mole of a substance is formed from its elements in their standard states.
The chemical equation for the combustion of C8H18 can be written as:
C8H18(g) + 12.5O2(g) -> 8CO2(g) + 9H2O(g)
We know the enthalpy change of this reaction, ΔHrnx = -5104.1 kJ/mol.
To apply Hess's Law, we can write the enthalpy change of this reaction as the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants:
ΔHrnx = Σ(nΔHf°products) - Σ(mΔHf°reactants)
ΔHrnx = [8ΔHf°CO2 + 9ΔHf°H2O] - [ΔHf°C8H18 + 12.5ΔHf°O2]
We substitute the known enthalpy values into the equation:
-5104.1 kJ/mol = [8ΔHf°CO2 + 9ΔHf°H2O] - [ΔHf°C8H18 + 12.5ΔHf°O2]
Now, we need to look up the enthalpies of formation for CO2, H2O, and O2, which are all common compounds with known values. Let's substitute the known values:
-5104.1 kJ/mol = [8(-393.51 kJ/mol) + 9(-241.82 kJ/mol)] - [ΔHf°C8H18 + 12.5(0 kJ/mol)]
Simplifying the equation:
-5104.1 kJ/mol = [-3148.08 kJ/mol - 2176.38 kJ/mol] - ΔHf°C8H18
-5104.1 kJ/mol = -5324.46 kJ/mol - ΔHf°C8H18
Rearranging the equation to solve for ΔHf°C8H18:
ΔHf°C8H18 = -5104.1 kJ/mol + 5324.46 kJ/mol
ΔHf°C8H18 = 220.36 kJ/mol
Therefore, the standard enthalpy of formation of this isomer of C8H18(g) is 220.36 kJ/mol.
To find the standard enthalpy of formation (∆H°f) of a compound, you need to compare the energy released or absorbed during its formation with the energy changes of its constituent elements in their standard states. The balanced chemical equation for the combustion of C8H18 can be written as follows:
C8H18(g) + 12.5O2(g) → 8CO2(g) + 9H2O(l)
From the given data, we know that the enthalpy change (∆H°) for this reaction is -5104.1 kJ per mole of C8H18 consumed. The stoichiometric coefficient of C8H18 in the balanced equation is 1, so we can say that -5104.1 kJ of heat is released for the formation of 1 mole of C8H18.
Now, we need to find the standard enthalpy of formation (∆H°f) of C8H18. By definition, the standard enthalpy of formation is the enthalpy change when 1 mole of a compound is formed from its constituent elements under standard conditions.
We know that the standard enthalpy of formation of elements in their standard states is zero. In this case, we have 8 moles of carbon (C) and 18 moles of hydrogen (H2). Therefore, we need to subtract the energy released by the combustion reaction from the energy required to form the elements.
∆H°f(C8H18) = [Energy released by the combustion of C8H18] - [Energy required to form the elements]
Since combustion releases energy, we can write:
∆H°f(C8H18) = -∆H° = -5104.1 kJ/mol
So, the standard enthalpy of formation (∆H°f) of the isomer of C8H18 is -5104.1 kJ/mol. Keep in mind that the negative sign indicates an exothermic reaction, where energy is released.