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Topic: Work produced by a chemical reaction  (Read 9629 times)

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Offline TheShehanigan

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Work produced by a chemical reaction
« on: September 26, 2009, 12:30:46 PM »
Hello,

Out of my First Law of Thermodynamics problem set for practice, I am unable to solve one of the problems given to me. I'd like some direction on where to proceed, since I am a bit lost with the problem. It reads:

A 15.0 g piece of Mg(s) is deposited in a container with HCl(aq). Calculate the work done by the system as a consequence of the reaction, if Patm = 1.1 atm and T = 298.15 K. Presume ideal behavior.

The reaction I calculated was: Mg(s) + 2 HCl(aq) --> H2(g) + MgCl2(aq)

Thus far, we've covered (and which are stuff I may be able to relate)

dW = -Pext dV (ext in this case, since the process is highly irreversible)
dU = dQ + dW
H = U + PV

I am assuming that the container will be open to the environment, and that thus the external pressure will be constant, making the process isobaric. However, I have doubts on how to calculate the initial volume of the system, or if my reasoning is correct at all.

I really don't want answers on how to do it, I just need tugs in the right direction to begin thinking on how to solve the problem.

Thanks.

Offline Yggdrasil

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Re: Work produced by a chemical reaction
« Reply #1 on: September 26, 2009, 12:33:36 PM »
Because solids, liquids, and solutions are much more dense than gasses, you can generally assume that their volume is negligible compared to the volume of any gas in the reaction.

Offline TheShehanigan

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Re: Work produced by a chemical reaction
« Reply #2 on: September 26, 2009, 12:40:18 PM »
Hence from said logic (which is actually a good approximation, don't know why I forgot it), I can calculate from the given quantity of Mg(s) and assuming it will be my limiting reactant the amount of H2(g) produced by the reaction, and assume the initial volume will be 0, and calculate the final volume of the gas. Then I can simply calculate from the isobaric process, using Pext given, the work by the reaction.

This actually sounds about right. I'll proceed to do that.

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