Name:
Date:
Heats of Reaction
In this experiment you will determine and compare the quantity of heat energy released in three exothermic
chemical reactions. Go to http://chemcollective.org/vlab/138
Reaction 1: Solid sodium hydroxide dissolved in water to form an aqueous solution of ions
NaOH (s) → Na+(aq) + OH-(aq) + x1 kJ
Reaction 2: Solid sodium hydroxide reacts with an aqueous solution of hydrogen chloride to form water and an
aqueous solution of sodium chloride.
NaOH (s) + H+(aq) + Cl-(aq) → H2O(i) + Na+(aq) + Cl-(aq) + x2 kJ
Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution of hydrogen chloride to
form water and an aqueous solution of sodium chloride.
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → H2O(l) + Na+(aq) + Cl-(aq) + x3 kJ
In order to accurately measure the heat released in each reaction, we will be using a calorimeter (a styrofoam
cup). The change in temperature that occurs for each reaction will be used to calculate the energy released in
kilojoules per mole of sodium hydroxide use. We can assume for our calculations that any heat transferred to
the styrofoam and surrounding air will be negligible. We can also assume that the specific heat of water is
4.18 J/goC.
Specific heat can be calculated using the formula
q = mcΔT
q = -ΔH
Procedure
Reaction 1
a. In the glassware menu, take out a 50 ml graduated cylinder and a styrofoam cup. From the tools menu,
take out the scale. From the chemical stockroom, move the distilled water and solid NaOH onto the
workbench.
b. Transfer 50 mL of distilled water to the styrofoam cup. To do this, drag the carboy of water onto the
graduated cylinder. (Before you release the mouse button, the cursor will show a plus sign and a
transfer text bar will appear. Enter 50.0 mL and click on pour. You will notice that the graduated
cylinder reads 50.0 mL.
c. Weigh about 1 g of solid sodium hydroxide pellets, NaOH(s), directly into the styrofoam cup and record
its mass to the nearest 0.01 g. To do this, place the styrofoam cup on the balance so it registers a
mass, then click the “Tare” button. Drag the NaOH button onto the styrofoam cup. (When you release
the mouse, the bottle will be tipped to show that it is in pour mode.) Next type “1.00’ g into the transfer
bar and then click pour. Note that the balance now reads the mass of the transferred NaOH. You may
now take the cup off of the scale and remove the scale from the workbench.
d. Click on the graduated cylinder, record its temperature and then drag it onto the styrofoam cup. (When
you release the mouse, the graduated cylinder will be tipped to show that it is in pour mode.) Enter 50.0
in the transfer bar and then click pour. Record the highest temperature.
Reaction 2
a. Take the 0.5 M HCl from the strong acids cabinet and a styrofoam cup and 50 mL graduated cylinder
from the glassware menu and place them on the workbench. The procedure for Reaction 2 is the same
as for Reaction 1 except that 50 mL of 0.5 M hydrochloric acid solution is used in place of the water.
After measuring 50.0 mL of the HCl solution into the graduated cylinder, proceed as before with steps b
through d of the procedure.
Reaction 3
a. Take out another graduated cylinder, a fresh styrofoam cup, the 1.0 M HCl and the 1.0 M NaOH. (If you
are running out of room, you may remove the previous chemicals). Measure 25.0 mL of 1.0M
hydrochloric acid solution into the styrofoam cup. Pour an equal volume (25 mL) of 1.0 M sodium
hydroxide solution in the clean graduated cylinder.
b. Record the temperature of each solution to the nearest 0.1oC. Pour the sodium hydroxide solution into
the styrofoam cup and record the high temperature obtained during the reaction.
Analysis
Observations
Reaction 1: NaOH (s) → Na+(aq) + OH-(aq) + x1 kJ
Mass of the system / Mass of the surroundings
50mL = 50g / 1g
Initial Temperature
25°C
Final Temperature
30.3°C
Heat Released (q)
Q = mcΔT
Q = (50g)(4.18J/g°C)(30.3 - 25)
Q = 1107.7J
Q = 1.1077kJ
Moles of NaOH
n = m/M
N = 1g/39.997g/mol
n = 0.025 mol
Molar Heat of Reaction (ΔH)
ΔHsys = -Qsurroundings
nΔHsol = -Q
ΔHsol = -Q/n
ΔHsol = -1107.7J/0.025
ΔHsol = -44308J
ΔHsol = -44.308kJ/mol
Reaction 2: NaOH (s) + H+(aq) + Cl-(aq) → H2O(i) + Na+(aq) + Cl-(aq) + x2 kJ
Mass of the system / Mass of the surroundings
50mL = 50g / 1g
Initial Temperature
25°C
Final Temperature
36.97°C
Heat Released (q)
Q = mcΔT
Q = (50g)(4.18 J/g°C)(36.97-25)
Q = 2501.73J
Q = 2.50173kJ
Moles of NaOH
n = m/M
n = 1g/39.997g/mol
n = 0.025mol
Molar Heat of Reaction(ΔH)
ΔHsys = -Qsurroundings
nΔHsol = -Q
ΔHsol = -Q/n
ΔHsol = -2501.73J/0.025
ΔHsol = -100069.2
ΔHsol = -100.0692kJ/mol
Reaction 3: Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → H2O(l) + Na+(aq) + Cl-(aq) + x3 kJ
Mass of the system / Mass of the surroundings
25 mL = 25g / 25mL = 25g
Initial Temperature
25°C
Final Temperature
31.67°C
Heat Released (q)
Q = mcΔT
Q = (25g)(4.18J/g°C)(31.67-25)
Q = 697.015J
Moles of NaOH
n = m/M
n = 25g/39.997g/mol
n = 0.62mol
Molar Heat of Reaction (ΔH)
ΔHsys = -Qsurroundings
nΔHneut = -Q
ΔHneut = -Q/n
ΔHneut = -697.015J/0.62mol
ΔHneut = -1,124.22
ΔHneut = -1.12422kJ/mol
2. Write the ionic equation for Reactions 1, 2 and 3. Consider the heat given off in your reactions; add
variables to your equations to represent the heat given off in kJ/mol.
Reaction 1: NaOH (s) → Na+(aq) + OH-(aq) + x1 kJ
Reaction 2: NaOH (s) + H+(aq) + Cl-(aq) → H2O(i) + Na+(aq) + Cl-(aq) + x2 kJ
Reaction 3: Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → H2O(l) + Na+(aq) + Cl-(aq) + x3 kJ
Insert overall net ionic equation here
Reaction 1:
NaOH(s) → Na^(+)(aq) + OH^(-)(aq) + 44.31 kJ/mol
Reaction 2:
NaOH(s) + H^(+)(aq) → H2O(l) + Na^(+)(aq) + 100.07 kJ/mol
Reaction 3:
OH^(-)(aq) + H^(+)(aq) → H2O(l) + 1.12 kJ/mol
3. The energy x1 in Reaction 1 represents the energy of the solution for one mole of NaOH(s). Look at the net
ionic equations for Reactions 2 and 3 and make a similar statement concerning the significance of x2 and x3.
The energy x2 in reaction 2 represents the energy of the solution for one mole of NaOH(s).
The energy x3 in reaction 3 represents the energy of the solution for one mole of NaOH(aq).
4. Find the difference between the value of x2 and the sum of x1 and x3. Account for any similarity or
difference.
= x2 - (x1 + x3)
= 100.07kJ/mol - (44.31kJ/mol + 1.12kJ/mol)
= 100.07kJ/mol - (45.43kJ/mol)
= 54.64kJ/mol
The difference between the value of x2 and the sum of x1 and x2 is 54.64kJ/mol
5. Suppose that you had used 4.00 g of NaOH(s) in Reaction 1. What would be the number of Joules released
in the reaction? What effect would this have on x1?
Q = mcΔT
Q = (50g)(4.18J/g°C)(30.3 - 25)
Q = 1107.7J
Q = 1.1077kJ
n = m/M
N = 4g/39.997g/mol
n = 0.1mol
ΔHsys = -Qsurroundings
nΔHsol = -Q
ΔHsol = -Q/n
ΔHsol = -1107.7J/0.1 mol
ΔHsol = -11077J
ΔHsol = -11.077kJ
The number of Joules released in the reaction would be -11.077kJ instead of the original -44.308 Joules that
were being released.
Discussion
1. Look up Hess’ Law and discuss how this experiment is an example of this law at work.
Hess’ law states that the total enthalpy change is equal to the sum of all stages or steps to get to that
reaction in between.
Reaction 1:
NaOH(s) → Na^(+)(aq) + OH^(-)(aq) + 44.31 kJ/mol
Reaction 2:
x 1 NaOH(s) + H^(+)(aq) + Cl^(-)(aq) → H2O(l) + Na^(+)(aq) + Cl^(-)(aq) + 100.07 kJ/mol
Reaction 3:
x-1 Na^(+)(aq) + OH^(-)(aq) + H^(+)(aq) + Cl^(-)(aq) → H2O(l) + Na^(+)(aq) + Cl^(-)(aq) + 1.12 kJ/mol
NaOH(s)→ Na^(+)(aq) ΔH = -100.07 kJ/mol
→ OH^(-)(aq) ΔH = 1.12kJ/mol
ΔH = 98.95kJ
The value of ΔH for reaction 3 is supposed to be around 50kJ which would then be able to give us the
correct value for the first equation. This experiment is an example of Hess’ law at work since Reaction 2 and
3 equations are steps to get the first reaction and adding them up is supposed to give you the total enpalthy
change of reaction 1.
2. If you were doing this experiment in the real world, what are some major sources of error that could
have occurred?
A source of error that could occur is that your equipment is not good enough quality so it might not be able to
measure the most accurate values such as your thermometer not being able to get the most accurate
temperature. Another source of error could be that your cup is not as well insulated and as a result, the
surrounding would not only include the water or solution inside the cup, but the outside air as well which
would give you the wrong result as there would be heat transfer between the air which you do not want.
Finally, a source of error could be that the equipment is not cleaned properly and has other substances on it
which could react with the substances inside of the cup and would also give you an inaccurate result.