posted by Elizabeth on .
Calculate the work involved if a reaction with an enthalpy change of -2418 kJ is carried out in a vessel with a mobile, frictionless piston. Other details: the reaction is H2(g) + 1/2Oxygen2(g) yields H2O(g) with enthalpy change of -241.8 kJ/mol. The product is 180.16gH2O=10.000 mol H2O (which is how I got the enthalpy of reaction).
I know delta E is close to delta H, so using delta H = delta E + PdeltaV, I get PdeltaV = 0. I also know that work = -PdeltaV, but that would mean that 0 work was done. I know that this is logically impossible, because the piston would have to move. What equation(s) should I use?
I originally posted as Mark; I'm assuming that I was overlooked or no one wants to answer the question. Any help would be appreciated. Thanks!