Delta H is given and is-6186.0
Calculate the difference, deltaH-deltaE=delta(PV) for the combustion reaction of 1 mole of methyl octane.
(Assume standard state conditions and 298 K for all reactants and products.)
To calculate the difference between ΔH and ΔE (deltaH - deltaE), we need to know the change in thermal energy (ΔE) for the combustion reaction of 1 mole of methyl octane.
To calculate ΔE, we can use the equation:
ΔE = ΔH - Δ(PV)
Where:
ΔH is the change in enthalpy for the reaction (given as -6186.0)
Δ(PV) is the change in pressure-volume work
In this case, we assume standard state conditions and 298 K for all reactants and products. This means that we can ignore the Δ(PV) term since standard thermodynamic tables usually report enthalpy changes under constant pressure conditions. Therefore, the change in pressure-volume work (Δ(PV)) is assumed to be negligible.
So, to calculate the difference ΔH - ΔE, we simply have ΔH = ΔE:
ΔH - ΔE = 0
Therefore, the difference between ΔH and ΔE for the combustion reaction of 1 mole of methyl octane under standard state conditions and 298 K is zero.