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Here is some help for doing the calculations related to the
thermochemistry lab.<br>
<br>
*REACTION 1 (reaction 2 in the procedure document)*<br>
<br>
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<br>
of 1 g/ml so the mass of the water was 50.0 g. Therefore the TOTAL
mass of your solution was 52 g.<br>
<br>
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.<br>
<br>
To calculate the heat in this case...<br>
<br>
<img alt="$q = mc_s\DeltaT = (52.00)(3.93)\Delta T$"
style="vertical-align: -4px;"
src="cid:part1.5EDED40B.436DAA2A@mercer.edu"><br>
<br>
Once you compute the heat, you need to convert that to an enthalpy
by dividing through by the moles of NaOH reacted.<br>
<br>
Moles of NaOH = (2.00 g) / (40.00 g/mol) = 0.0500<br>
<br>
where 40.00 is the molar mass of NaOH. Remember to use YOUR MASS of
NaOH in your calculations.<br>
<br>
The Enthalpy in J/mol is then <img alt="$q/0.0500$"
style="vertical-align: 0px;"
src="cid:part2.AD9432E9.93BC1A3F@mercer.edu">. It is exothermic,
so the sign should be NEGATIVE.<br>
<br>
<br>
*REACTION 2 (reaction 3 in the procedure document) *<br>
<br>
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<br>
<p><img alt="$q = mc_s\Delta T = (50.0 + 50.0)(1.02)(4.02) \Delta
T$" style="vertical-align: -4px;"
src="cid:part3.B47CF093.61CC07B5@mercer.edu"><br>
</p>
Where the (50.0 + 50.0)(1.02) terms convert the volume to a mass. <br>
<br>
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.<br>
<br>
Moles NaOH = (1.00 M)(0.050 L) = 0.050 moles<br>
<br>
So again, the Enthalpy in J/mol is <img alt="$q/0.0500$"
style="vertical-align: 0px;"
src="cid:part2.AD9432E9.93BC1A3F@mercer.edu">. It is exothermic,
so the sign should be NEGATIVE.<br>
<br>
<br>
*REACTION 3 (Reaction 4 from the procedures) *<br>
<br>
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:<br>
<br>
<p><img alt="$q = mc_s\DeltaT = (102.0)(1.02)(4.02)\Delta T$"
style="vertical-align: -4px;"
src="cid:part5.CB38C3A0.781F82DD@mercer.edu"></p>
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).<br>
<br>
Moles of NaOH = (2.00 g) / (40.00 g/mol) = 0.0500<br>
<br>
The Enthalpy in J/mol is then <img alt="$q/0.0500$"
style="vertical-align: -4px;"
src="cid:part2.AD9432E9.93BC1A3F@mercer.edu">. It is exothermic,
so the sign should be NEGATIVE.<br>
<br>
<br>
Convert ALL of your enthalpies to kJ/mol and fill out the
appropriate places on the chart.<br>
<br>
Let me know if you have questions!!!!<br>
<br>
<br class="Apple-interchange-newline">
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Director of the Computational Science Program
Mercer University, Macon, GA 31207 (478) 301-5627
</pre>
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