If the change in enthalpy is -5074.3 kj, how much work is done during the combustion?

I guess there is is a typo in your question both change in enthalpy and change in internal energy for the combustion are negative: â†H = -5074 kJ â†U = - 5084.2 kJ Work done to surrounding in a process is given by the integral: W = â« P dV from initial to final volume for a constant pressure process the integral simplifies to: W = P â™ â« dV = Pâ™â†V Enthalpy is defined as: H = U + Pâ™V So the change in enthalpy and the change in internal energy are related as: â†H = â†U + â†(Pâ™V) for a constant pressure process: â†H = â†U + Pâ™â†V Hence, â†H = â†U + W => W = â†H - â†U = -5074 kJ - (-5084.2) kJ = 10.2 kJ

Iwould sure like to answer your question but i honestly just don’t know