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Wize University Chemistry Textbook > Thermodynamics: Part 1: (0th law, 1st law, Calorimetry, Enthalpy)

Relating Work and the Ideal Gas Law

The 1st Law of Thermodynamics and WorkPrevious Section
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Expansion Due to Chemical Reaction

When a chemical reaction takes place where the number of moles of gas changes, expansion can occur. Typically this would happen with a reaction in a balloon, but this also happens for a reaction happening in an open vessel.


Example:
Calcium carbonate decomposes into calcium oxide and carbon dioxide as shown below. If 12g of calcium carbonate are placed into a vessel open to the atmosphere at room temperature and the reaction takes place, how much work is done on/by the system?




The flask is open to the atmosphere so the expansion is taking place against a constant external pressure of 1 atm, however we don't know the change in volume.
w=−PexΔVw=-P_{ex}\Delta V

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Using the Ideal gas law,
V=nRTP    and        ΔV=ΔnRTPV=\frac{nRT}{P}\ \ \ \ and\ \ \ \ \ \ \ \ \Delta V=\frac{\Delta nRT}{P}

substituting we get,
w=−PexΔV=−Pex(ΔnRTP)=−ΔnRTw=-P_{ex}\Delta V=-P_{ex}\left(\frac{\Delta nRT}{P}\right)=-\Delta nRT

We can find the number of moles of gas produced by the reaction, Δn\Delta n, by examining the stoichiometry in the question.


Example:
Calcium carbonate decomposes into calcium oxide and carbon dioxide as shown below. If 12g of calcium carbonate are placed into a vessel open to the atmosphere at room temperature and the reaction takes place, how much work is done on/by the system?


nCaCO3=mMW=12 g100.09 g mol−1=0.12 molnCO2=nCaCO3×11=0.12 mol\begin{array}{c} n_{CaCO_3}=\frac{m}{MW}=\frac{12\ g}{100.09\ g\ mol^{-1}}=0.12\ mol \\ n_{CO_2}=n_{CaCO_3}\times \frac{1}{1}=0.12\ mol \\ \end{array}


We can now solve for the work done by this chemical reaction,
w=−ΔnRTw=−(0.12mol)(8.314J mol−1 K−1)(298K)=−297 J\begin{array}{c} w=-\Delta nRT \\ w = -(0.12 mol)(8.314 J\ mol^{-1}\ K^{-1})(298 K)=-297\ J \end{array}



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Example: Relating Work and the Ideal Gas Law

For the following reaction at 25oC, ΔHrxn=-405kJ/mol what is ΔUrxn?

SO2(g) + V2O5(s) → SO3(g) + 2VO2(g)SO_2\left(g\right)\ +\ V_2O_5\left(s\right)\ \rightarrow\ SO_3\left(g\right)\ +\ 2VO_2\left(g\right)

How are ΔH and ΔU related?

ΔH=ΔU+Δ(PV)\Delta H=\Delta U+\Delta (PV)

Here we do not have P or V so what should we do?

Think of ideal gas law and use the following equation instead:

ΔH=ΔU+Δ(n)RT\Delta H=\Delta U+\Delta(n)RT

Rearrange to solve for ΔU:
ΔU=ΔH - Δ(n)RT
ΔU=-405000J/mol -(1-1 moles)(8.314J/Kmol)(298K)
ΔU=-405000J/mol

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A 10.0 g piece of Mg(s) is deposited in a container with a vast excess of HCl(aq). A reaction takes place forming hydrogen gas and Magnesium Chloride.
a) Write a balanced chemical reaction for the transformation

b) Calculate the work done by the system as a consequence of the reaction, if Pex = 1.1 atm and T = 298.15 K. You may assume ideal behaviour.