Calorimetry
6.03 Honors
6.03 Calorimetry honors
In this experiment, you will determine the
quantity of heat involved in two different
chemical processes. The heat given off or
absorbed will be measured using an
insulated calorimeter. Be sure to record all
measurements carefully, as you will use the
data to calculate the enthalpy change of
each process.
Procedure
Part I: The Dissolving of Solid Sodium Hydroxide in
Water
1. Measure out approximately 200 mL of distilled water and pour it
into the calorimeter. Stir carefully with a thermometer until a
constant temperature is reached. Record the volume of water and
the constant initial temperature of the water on your data table.
2. Place a plastic measuring trough on top of the digital balance, and
then zero the balance (press the tare button) so that the mass of
the trough will be “ignored” and will not be added to the total mass
measured by the balance.
3. Measure out approximately three to five scoops of solid sodium
hydroxide and record the mass to your data table.
4. Place the solid sodium hydroxide into the water in the calorimeter
and replace the lid immediately. Stir gently until the solid is
completely dissolved and record the highest temperature reached.
Procedure cont.
Part II: The Reaction of Sodium Hydroxide
Solution with Hydrochloric Acid
1. Measure out approximately 100 mL of 0.50 M HCl
solution and 100 mL of 0.50 M NaOH solution. Record
both volumes on your data table.
2. Pour the hydrochloric acid solution into the calorimeter.
Measure and record the initial temperature of each
solution and record on your data table.
3. Add the sodium hydroxide solution to the acid solution
in the calorimeter and immediately replace the lid of the
calorimeter. Stir the mixture and record the highest
temperature reached.
Data and Observations – Part I
Data and Observations – Part II
Recipe for success
Part I: The Dissolving of Solid Sodium
Hydroxide in Water
1.
2.
3.
4.
5.
Calculate mol of NaOH, n = grams/molar mass
q = m × c × Δt
m reactants = mass water + mass NaOH
Convert q from joules to kilojoules
Divide kJ by moles of NaOH to solve for ΔH
(KJ/mol) ΔH = q/n
6. Then ΔH surroundings = - ΔH system
Recipe for Success
Part II: The Reaction of Sodium Hydroxide Solution with
Hydrochloric Acid
• Calculate mol of NaOH, n = M x V,
M = 0.5 M and V = 100 mL (convert to L)
• q = m × c × Δt
m reactants = mass NaOH solution + mass HCl solution
• Convert q from Joule to kiloJoule
• Divide kJ by moles of NaOH to solve for ΔH (KJ/mol) ΔH
= q/n
• Then ΔH surroundings = - ΔH system
Conclusion
1. Use calculated values in Hess’s law to solve
for the enthalpy change of the given reaction.
ΔH [1] + ΔH [2] = ΔH [3]
NaOH (s) + H2O (l) → NaOH (aq)
[1]
NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
[2]
NaOH (s) + HCl (aq) → NaCl (aq) + H2O (l)
[3]
Conclusion
2. If the accepted enthalpy change value for the dissolving of
sodium hydroxide in water is −44.2 kilojoules per mole,
determine the percent error of the experimental value that
you calculated in Part I. Show your work.(2 points) ΔH [1]
% error = accepted enthalpy change value – experimental enthalpy change value × 100 =
accepted enthalpy change value
Conclusion
3. Give a detailed explanation, using what you know
about bonds and forces of attraction, for the
enthalpy changes you observed in parts I and II of
this lab.
http://www.docbrown.info/page03/3_51energy.htm
Conclusion
4. If the hole for the thermometer in a calorimeter
is wider than the diameter of the thermometer,
leaving a gap between the lid and the
thermometer itself, how do you think this would
this affect the temperature change observed in
the experiment? How would this affect the
calculated enthalpy change?