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Homework Problem 8.93

Posted: Wed Jan 25, 2017 9:38 am
by Ashley Van Belle 2B
For 8.93, it says to calculate the work that must be done against the atmosphere for the expansion of the gaseous products in the combustion of 1 mol C6H6 at 25 degrees C and 1 bar. So I figured that I would use the formula w=-P'delta'V and replace it to instead use w=-'delta'nRT. Looking at the solutions, somehow they got 1.5 for the change in moles. How does this work?

Re: Homework Problem 8.93

Posted: Wed Jan 25, 2017 11:39 am
by Chem_Mod
The balanced equation for the combustion of one mole of benzene is: where oxygen, carbon dioxide, and water are all in the gas phase while benzene is in the liquid phase. Thus, there are a total of 9 moles of gas on the product side and or 7.5 moles of gas on the reactant side. When you subtract the moles of reactant in the gas phase from the moles of products in the gas phase (9 - 7.5), you get a change of 1.5 moles.