Using the technique of the previous problem ΔE was found to be -2,000.00 kJ/mol of an unknown liquid hydrocarbon at 298 K. In another experiment it was determined that for each mole of hydrocarbon, 4 moles of oxygen gas are consumed and 6 moles of CO2 gas and 6 moles of H2O liquid are produced. Find ΔH per mole of this hydrocarbon (in kJ) at 298 K.
To find ΔH per mole of the hydrocarbon, we will use the concept of Hess's Law, which states that the enthalpy change of a reaction is independent of the pathway between the initial and final states.
Given that ΔE for the reaction is -2,000.00 kJ/mol, we know that ΔE = ΔH - ΔnRT, where Δn is the change in the number of moles of gas between the reactants and products, R is the ideal gas constant, and T is the temperature in Kelvin.
In this case, 4 moles of oxygen gas (O2) are consumed, and 6 moles each of carbon dioxide (CO2) and liquid water (H2O) are produced. Therefore, the change in the number of moles of gas (Δn) is:
Δn = (moles of gas produced) - (moles of gas consumed)
= (6 moles of CO2 + 6 moles of H2O) - (4 moles of O2)
= 12 moles - 4 moles
= 8 moles
At 298 K, the value of R is approximately 8.3145 J/(mol·K). We need to convert the units of ΔE and R to match before plugging them into the equation.
ΔE = -2,000.00 kJ/mol = -2,000,000 J/mol (multiply by 1,000 to convert kJ to J)
Now we can rearrange the equation to solve for ΔH:
ΔH = ΔE + ΔnRT
= -2,000,000 J/mol + (8 mol × 8.3145 J/(mol·K) × 298 K)
Now, calculate the value of ΔH:
ΔH = -2,000,000 J/mol + (19,945.28 J/mol)
= 17,945.28 J/mol
Finally, convert the result to kJ/mol:
ΔH = 17,945.28 J/mol ÷ 1,000 = 17.94528 kJ/mol
Therefore, ΔH per mole of the hydrocarbon at 298 K is approximately 17.95 kJ/mol.