Unit 10 - Thermodynamics
Section 1 – Enthalpy and Calorimetry
1. The combustion of methane, CH4, releases 890.4 kJ/mol. That is, when one mole of methane is burned, 890.4 kJ
are given off to the surroundings. This means that the products have 890.4 kJ less than the reactants. Thus, ΔH
for the reaction = - 890.4 kJ.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l); ΔH = - 890.4 kJ
a. How much energy is given off when 2.00 mol of CH4 are burned?
2.00 mol  890.4 kJ

 1780 kJ
1
1.00 mol
b. How much energy is released when 22.4 g of CH4 are burned?
22.4 g 1.00 mol CH 4
 890.4 kJ


 1240 kJ
1
16 g CH 4
1.00 mol CH 4
c.
If you were to attempt to make 45.0 g of methane from CO2 and H2O (with O2 also being made), how much
energy would be required?
45.0 g 1.00 mol CH 4
890.4 kJ


 2504 kJ
1
16 g CH 4
1.00 mol CH 4
H+(aq) + OH−(aq) → H2O(l)
2. A student is asked to determine the molar enthalpy of neutralization, ΔHneut, for the reaction represented above.
The student combines equal volumes of 1.0 M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter. The
heat released by the reaction is determined by using the equation q = mcΔT. Assume the following. Both
solutions are at the same temperature before they are combined. The densities of all the solutions are the same
as that of water. Any heat lost to the calorimeter or to the air is negligible. The specific heat capacity of the
combined solutions is the same as that of water.
a. Give appropriate units for each of the terms in the equation q = mcΔT.
q = units of energy (joules, kilojoules, calories, etc…) m = units of mass (kg, g, etc…)
c = units of specific heat capacity (J/kg ˚C, etc…) T = units of temperature (˚C or K)
b. List the measurements that must be made in order to obtain the value of q.
Volume or mass of the NaOH or HCl solutions, initial temperature before mixing, final temperature after
mixing.
c.
Explain how to calculate each of the following.
i. The number of moles of water formed during the experiment
The number of moles of water = the same as the number of moles of HCl and NaOH because it is a 1:1
ratio.
Volume of HCl 
1.00 mol CH 4
890.4 kJ

 2504 kJ
16 g CH 4
1.00 mol CH 4
ii. The value of the molar enthalpy of neutralization, ΔHneut, for the reaction between HCl(aq) and
NaOH(aq).
Use q = mcΔT to determine q where m = the total mass of the solution. ΔHneut = q divided by the total
number of moles determined in the previous question.
d. The student repeats the experiment with the same equal volumes as before, but this time uses 2.0 M HCl
and 2.0 M NaOH.
i. Indicate whether the value of q increases, decreases, or stays the same when compared to the first
experiment. Justify your prediction.
The value for q will increase because there are more moles reacting which will increase the temperature
in the same quantity of water. (Q is extensive.)
ii. Indicate whether the value of the molar enthalpy of neutralization, ΔHneut, increases, decreases, or stays
the same when compared to the first experiment. Justify your prediction.
ΔHneut will stay the same because the number of moles also increased proportionally to the q. So the ratio
stays the same.
e.
Suppose that a significant amount of heat were lost to the air during the experiment. What effect would this
have on the calculated value of the molar enthalpy of neutralization, ΔHneut? Justify your answer.
Using the equation q = mcΔT, if there is a decrease in temperature, this will cause a decrease in the q, which
will also cause a decrease in the amount of heat lost, which will make ΔHneut a more positive number.
3. In an experiment, a sample of an unknown, pure gaseous hydrocarbon was analyzed. Results showed that the
sample contained 6.000 g of carbon and 1.344 g of hydrogen.
a. Determine the empirical formula of the hydrocarbon.
6.000 g C 1.00 mol C

 0.5000 mol C
1
12 g C
1.344 g H 1.00 mol H

 1.333 mol H
1
1.00794 g H
Because the ratio of C:H is 1:2.667, I would make the empirical formula 3:8  C3H8
b. The density of the hydrocarbon at 25 oC and 1.09 at is 1.96 g L-1.
i. Calculate the molar mass of the hydrocarbon
PV  nRT so n 
1.09 atm 1.00 L  0.04455 mol  M  1.96 g  44.0 g/mol
PV

L  atm 
RT 
0.04455 mol
 0.0821
298 K 
mol  K 

ii. Determine the molecular formula of the hydrocarbon
C3H8 = 44 g/mol so the molecular formula is C3H8.
4. In an experiment, liquid heptane, C7H16 (l), is completely combusted to produce carbon dioxide gas and liquid
water, as represented by the following equation: C7H16(l) + 11O2(g)  7CO2(g) + 8H2O(l). The heat of combustion,
ΔHocomb, for one mole of C7H16(l) is -4.85 x 103 kJ.
a. Using the information in the table below, calculate the value of ΔH of for C7H16 (l) in kJ mol-1.
Compound
CO2 (g)
H2O (l)
ΔHof (kJ mol-1)
-393.5
-285.8
ΔH f   ΔH f (reactants ) -  ΔH f (products ) 
 4,850 kJ  7- 393.5  8 285  1x   110
kJ
x  191
mol
You may want to note that all elements in their natural elemental state have a heat of formation of zero.
b. A 0.0108 mol sample of heptane is combusted in a bomb calorimeter.
i. Calculate the amount of heat released to the calorimeter.
0.0108 mol  4850 kJ

 52.4 kJ
1
1.00 mol
ii. Given the total heat capacity of the calorimeter is 9.273 kJ oC-1, calculate the temperature change of the
calorimeter.
q  C p ΔT 
ΔT 
q
 52.4 kJ

 5.65C
C p 9.273 kJ/ C