[CHM 111] Thermochemistry lab -- sample calculations...

[Andrew J. Pounds]

Here is some help for doing the calculations related to the
thermochemistry lab.

*REACTION 1*

In the first reaction you did you combined 2 grams (you need to use your
actual mass) of NaOH (s) with 50.0 ml of water.  The water has a density
of 1 g/ml so the mass of the water was 50.0 g.   Therefore the TOTAL
mass of your solution was 52 g.

Remember -- this is now a solution of NaOH (aq) so you need to use the
specific heat capacity provided on the report form 3.93 J/g°C.

To calculate the heat in this case...

$q = mc_s\DeltaT = (52.00)(3.93)\Delta T$

Once you compute the heat, you need to convert that to an enthalpy by
dividing through by the moles of NaOH reacted.

Moles of NaOH = (2.00 g) / (40.00 g/mol) = 0.0500

where 40.00 is the molar mass of NaOH.  Remember to use YOUR MASS of
NaOH in your calculations.

The Enthalpy in J/mol is then $q/0.0500$.   It is exothermic, so the
sign should be NEGATIVE.


*REACTION 2*

In this reaction you are mixing 50.0 ml of 1 M NaOH with 50.0 ml of 1 M.
  This will make 100.0 ml of solution and we are told on the report form
that the density of this solution is 1.02 g/ml and that it has a
specific heat of 4.02 J/g°C.

In this case the heat is

\documentclass{article} \usepackage[utf8x]{inputenc} \pagestyle{empty}
\begin{document} q = mc_s\Delta T = (50.0 + 50.0)(1.02)(4.02) \Delta
T%this is where your LaTeX expression goes \end{document}

Where the (50.0 + 50.0)(1.02) terms convert the volume to a mass. 

As before, I need to convert the heat to an enthalpy.  In this case to
find the moles of NaOH I multiply the volume of NaOH by the molarity of
NaOH.

Moles NaOH = (1.00 M)(0.050 L) = 0.050 moles

So again, the Enthalpy in J/mol is $q/0.0500$.   It is exothermic, so
the sign should be NEGATIVE.


*REACTION 3*

In this reaction we are doing essentially what we did in REACTION 2,
except that we are using solid NaOH.   We dilute the HCl solution up to
100 ml initially because we want to have, after the addition of the
NaOH, a solution with approximately the same density and specific heat
capacity as we had in REACTION 2.   In this particular case the mass of
the solution will be 102.0 grams; this is the 100 grams of HCl solution
plus the mass of the massed NaOH solid, so make sure you use your value
for the mass of the NaOH.  To calculate the heat:

\documentclass{article} \usepackage[utf8x]{inputenc} \pagestyle{empty}
\begin{document} q = mc_s\DeltaT = (102.0)(1.02)(4.02)\Delta T%this is
where your LaTeX expression goes \end{document}

And to compute the enthalpy we need again the moles of NaOH (which
should be just like the process we followed in step one where we had
solid NaOH).

Moles of NaOH = (2.00 g) / (40.00 g/mol) = 0.0500

The Enthalpy in J/mol is then $q/0.0500$.   It is exothermic, so the
sign should be NEGATIVE.


Convert ALL of your enthalpies to kJ/mol and fill out the appropriate
places on the chart.

Let me know if you have questions!!!!



-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Director of the Computational Science Program
Mercer University,  Macon, GA 31207   (478) 301-5627

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