Chapter 8 #101

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Michael Lonsway 3O
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Chapter 8 #101

Postby Michael Lonsway 3O » Mon Jan 23, 2017 2:15 pm

A technician carries out the reaction 2SO2(g) + O2(g) --> 2SO3(g) at 25°C and 1.00 atm in a constant-pressure cylinder fitted with a piston. Initially, 0.0300 mol SO2 and 0.0300 mol O2 are present in the cylinder. The technician then adds a catalyst to initiate the reaction. a)Calculate the volume of the cylinder containing the reactant gases before the reaction begins. b)What is the limiting reactant? c)Assuming that the reaction goes to completion and that the temperature and pressure of the reaction remain constant, what is the final volume of the cylinder (include any excess reactant)? d)How much work takes place, and is it done by the system or on the sytem? e)How much enthalpy is exchanged, and does it leave or enter the system? f)From your answers to parts d and e, calculate the change in internal energy for the reaction.

This question really confused me and I was mainly wondering how you can find the limiting reactants in the reaction. If you have all parts of the answers please post an explanation.

Alex Chen 1B
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Re: Chapter 8 #101

Postby Alex Chen 1B » Mon Jan 23, 2017 3:06 pm

The problem tells you that there are initially 0.0300 moles of SO2 and 0.0300 moles of O2 present in the cylinder. They give you the equation for the reaction, so all you have to do is figure out which reactant will completely react given the initial quantities. So in this case, if all of the O2 were to react, you would need 0.0600 moles of SO2 to react with it (because 0.0300 moles of O2 x 2 moles of SO2 per 1 mole of O2). Since we only have 0.0300 moles of SO2, you know that the amount of SO2 is limiting the reaction, so SO2 would be your limiting reactant.

Alex Dib 4H
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Re: Chapter 8 #101

Postby Alex Dib 4H » Mon Jan 23, 2017 3:09 pm


Step one, we gotta find the initial volume using PV=nRT, where n=.0300+.0300=.0600. This gives an initial volume of 1.47 L

Step two, we need to determine the limiting reactant which can be done by inspection. We're given an equal amount of moles of both SO2 and O2, but we need twice the amount of SO2 for this reaction. SO2 will be the limiting reactant, and, after reaction completes and SO2 is depleted, we will still have half the original amount of O2, which is .015 moles.

Step three, we want to find the final volume. There is a 1:1 ratio of SO2 to SO3, so we know that because we used all of our SO2, we will have an equal amount in moles of SO3 after reaction is complete (.0300), as well as the leftover O2 that was remaining (.015). Adding these gives a total of n=.045. Using PV=nRT, we get the final volume as 1.10 L

Step four, we're asked to calculate the work done. Using w=-P(Vf-Vi), we get .370 L atm. Converting to J with R/R, we get 37.5 J for work. As our value is positive, the work was done on the system. We can see this conceptually as we started with a larger volume and ended with a smaller, meaning our system compressed.

Step five, we need to find the change in enthalpy. The way we do this by first finding the enthalpy of reaction with with enthalpies of formation. Enthalpy of formation (Products) - Enthalpy of formation (Reactants) gives the enthalpy of reaction, which is -197.8 kJ/2mol SO2, or simply -98.9 kJ/mol rxn. We are only using .0300 mol SO2, so using stoichiometry, we calculate the change in enthalpy: .03mol(-98.9 kJ)/1mol = -2.98 kJ.

Step six asks for the total change in internal energy, which we know is delta U=delta H + w. Plugging in our change in enthalpy and work we previously calculated, we get -2.98 kJ + .0375 kJ = -2.94 kJ, our final answer.

I don't have the answer book, and I am not sure if the significant figures are on point, but I'm pretty sure this is the strategy used to solve it.

Good luck, may the force be with you.

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