Question

For a particular isomer of C8H18, the following reaction produces 5113.3 kJ of heat per mole of C8H18(g) consumed, under standard conditions.

C8H18 + 25/2(O2) -> 8CO2 + 9H2O
DeltaH= -5113.3 kJ
What is the standard enthalpy of formation of this isomer of C8H18(g)?

Answer 1

To calculate the standard enthalpy of formation for the isomer of C8H18(g), use the given reaction enthalpy and the standard enthalpies of formation for the products, assuming the standard enthalpy of formation for elemental O2 is zero. The enthalpy of formation for CO2(g) is -393.5 kJ/mol. Solve for the unknown enthalpy of the C8H18 isomer by rearranging the enthalpy of formation equation.

Step-by-step explanation:

To find the standard enthalpy of formation of the isomer of C8H18(g), we need to understand that the standard enthalpy of a compound is the heat change that occurs when one mole of the compound is formed from its elements in their standard states. For example, the standard enthalpy of formation of CO2(g) is −393.5 kJ/mol, which is the enthalpy change when carbon (in its standard state as graphite) reacts with oxygen to form one mole of CO2 at standard conditions.

The given thermochemical equation for the C8H18 isomer is:

C8H18 + 25/2 O2 → 8 CO2 + 9 H2O

ΔH = −5113.3 kJ

To calculate the enthalpy of formation of this isomer, consider the enthalpy of formation of the products (CO2 and H2O) and the enthalpy of formation of the reactants. Since we're looking for the enthalpy of formation of the isomer, assume that the enthalpy of formation of the elements (Oxygen in this case) is zero, as per standard conditions. Therefore, we sum up the standard enthalpies of the formation of the products and subtract the enthalpy of the formation of the reactant (an isomer of C8H18).

The equation representing the enthalpy of formation would be the sum of the enthalpies of all products minus the sum of the enthalpies of all reactants equals the total enthalpy change of the reaction:

(8 × −393.5 kJ/mol) + (9 × enthalpy of formation of H2O) − enthalpy of formation of C8H18 = −5113.3 kJ

Answer 2

To find the standard enthalpy of formation of the C8H18 isomer, one must use the standard enthalpies of formation for CO2 and H2O and apply Hess's Law to the given combustion reaction, using the enthalpy change of -5113.3 kJ.

Step-by-step explanation:

The question asks for the standard enthalpy of formation of a specific isomer of C8H18(g), given the heat released from its combustion. To find the enthalpy of formation, one must consider the definition of standard enthalpy of formation, which is the enthalpy change when one mole of compound is formed from its elements in their standard states.

Since we are given the enthalpy change for the combustion reaction, we can use this information alongside the known standard enthalpies of formation for CO2 and H2O (which are -393.5 kJ/mol and -241.8 kJ/mol, respectively) to calculate the enthalpy of formation for the isomer of C8H18(g).
The calculation involves using the standard heat of formation values from a reference table (such as Table T1) and applying Hess's Law, which states that the total enthalpy change in a chemical reaction is the same, regardless of the number of steps in the reaction.

The process requires summing the standard enthalpies of formation for the products and subtracting the sum of the standard enthalpies of formation for the reactants. It should be noted that for elements in their standard states, like O2, the standard enthalpy of formation is zero. The calculated value for the enthalpy of formation of the isomer of C8H18(g) will be the sum of the enthalpies of formation for the products minus the given enthalpy change for the reaction itself (-5113.3 kJ).