5. ThermoChemistry

hermochemistry (Calorimetry)

Now, Let’s quickly review a few definitions we will need to use in this section.


Study of heat absorbed or evolved in chemical reactions.


Device used in Calorimetry to measure heat processes (normally thermally insulated from the surroundings).

Heat Capacity

Amount of heat to raise an object or a given amount of substance by 1℃ (1K).

Molar Heat Capacity

Amount of heat to raise one mole of a substance by 1℃ (1K).
(water has molar heat capacity of 75.4 J K-1 mol-1 .

Specific Heat

Amount of heat required to raise one gram by 1℃ (1K).
(water has specific heat of 1 cal K-1 g-1 or 4.18 J K-1 g-1.

Let’s try an example: A sample of 50. mL of a 0.20 M solution of HCl was mixed with 50. mL of 0.20 M NaOH in a coffee cup calorimeter. The initial temperature of both solutions was 22.2℃. After mixing, the temperature rose to 23.5℃. What is the enthalpy change for the neutralization reaction which occurred?

To start, we need to define the system. Since this case involves a reaction in a liquid solution where the chemicals are intimately involved with the solvent (solvation in water), we cannot really separate the water from the chemicals. We will use the chemicals and water as the system. The insulated cup is the boundary between the system and the surroundings and we will assume that the cup itself absorbs no heat and that no heat passes through the cup to the surroundings.

We have thus, a two step process to deal with.

  1. The reaction occurs, warming the system. Since no heat was evolved to the surroundings (coffee cup insulation prevented it) the value of q and of ΔH are zero since w is negligibly small (no volume change in a liquid). For that matter, ΔU is zero too (See Internal Energy at the end of this section).
  2. The system must be returned to the starting temperature by extracting heat from it or adding heat to it. Thus, the ΔT we calculate is of the step, after the reaction is over, where the system is returned to its original state.

H3O+(aq) + OH(aq) image   2 H2O (l)

Total volume = 100. mL, dilute aqueous so we assume density = 1.00 g/mL
mass of solution: msol = 100. mL × 1.00 g/mL = 100. g

For the cooling process:

ΔT = Tf – Ti = 22.2℃ – 23.5℃ = -1.3℃ (= -1.3 K)

The temperature was lowered from it’s high value back down to its starting point. Thus, we expect a negative temperature change. Another way of viewing this: We removed heat from the system to return it to its starting point. Hence, q and ΔH are negative (the reaction is exothermic).

This temperature change is happening to the system.

qp = CpΔT = mCm ΔT (Cm is specific heat, i.e., heat capacity per gram)

NOTE: to figure out which equation you need simply look at the units of the quantities you have and figure out how to cancel out the undesired units by the correct combination of C with the values you have. In this case, we have

qp = 4.18 J/K·g × 100. g × -1.3 K = -540. J

# mol H3O+ = # mol HCl = 0.50 L × 0.20 mol/L = 0.010 mol.

ΔH’ = qp = -540 J

Note that the ΔH shown here is for the whole reaction amount. It is not a molar amount; hence, the prime on the ΔH’. I use the prime here to distinguish the global quantity and no prime to show a molar quantity.

\[\Delta H = \dfrac{1}{\frac{-540 \mathrm{J}}{0.010\mathrm{ mol}}} = -54. \mathrm{kJ/mol\;H}_3\mathrm{O}^+\]

We’ll use the stoichiometry from the balanced chemical equation to convert from moles of H3O+ to moles of reaction.  According to the equation

H3O+(aq) + OH(aq) 2 H2O(l),

there is one hydronium ion for every one reaction. So, we convert to moles of equation as follows:

\[ \Delta H = \[\begin{array}{|c|c|} 540\;\mathrm{ J}  & 1\; \mathrm{H}_3\mathrm{O}^+ \\ \hline \mathrm{0.010 mol H}_3\mathrm{O}^+& \mathrm{1 equation}\end{array}  = -54.\mathrm{ kJ/mol equation}\]

The value reported for the enthalpy change must always be accompanied by the balanced chemical reaction we used in the determination.

If we had balanced the chemical equation differently, say, to show one mole of water, it might look like:
1/2 H3O+(aq) + 1/2 OH(aq) image H2O(l)
In this balanced equation, there is one half a hydronium ion for every equation thus, our conversion would be

\[ \Delta H = \[\begin{array}{|c|c|} 540\;\mathrm{J}  & 1/2\; \mathrm{H}_3\mathrm{O}^+ \\ \hline \mathrm{0.010 mol H}_3\mathrm{O}^+& \mathrm{1 equation}\end{array}  = -27.\mathrm{ kJ/mol equation}\]

Our test calorimeter is not that accurate (significant heat loss to the surroundings). The true ΔH for the first reaction is -56.02 kJ/mol and is valid for the heat of neutralization of any strong acid with a strong base in water.

Here is an example where we are able to isolate the chemicals separately as the system and directly measure the heat removed from the system.

0.510 g of ethanol is burned in a flame calorimeter containing 1200 g of water. The water is initially at 22.46℃ and is warmed up to 25.52℃ as a result of the reaction. What is the ΔH for one mole of ethanol?

In this case, we can more easily separate the gas reaction mixture from the water in the calorimeter which absorbs (all) the heat from the reaction. Therefore, we can use the chemicals themselves as the system. In this case, the insulation in the calorimeter serves to isolate a small portion of the surroundings (the water) so we can measure its temperature change directly, while ignoring the rest of the surroundings.

C2H5OH(l) + 3 O2(g) image 2CO2(g) + 3 H2O(l)

\[\Delta T_{water} = 25.52\Celsius - 22.46\Celsius = 3.06\Celsius = 3.06 K\]

We have a specified mass of water so we will start with that and look up and use a heat capacity which has grams (specific heat) to cancel out g and leave the desired units of energy.


This is the total heat absorbed by the water (surroundings) and is the negative of the heat evolved by the system. Hence,

\[q_{p,system} = -q_{surroundings} = -15.3 kJ\]

Note that the q of the surroundings does not need to have a specification of constant P or constant V.  It just doesn’t matter.  Whatever energy is absorbed by the surroundings must have been given off by the system (First Law).  Now, the system must be specified to be constant pressure if you need to equate the heat exchanged, q_p, to enthalpy change, \Delta H, of the system.

The other thing we need to do to calculate the enthalpy is to determine the heat per mole since enthalpies for a chemical reaction are always reported in units of kJ/mol of reaction.

Finally, we need to translate the enthalpy change into units of kJ/mol of equation as written, rather than kJ/mol of one particular chemical.

\[\Delta H = \begin{array}{|c|c|} -1380\mathrm{ kJ} & \mathrm{1 ethanol} \\ \hline \mathrm{1 mol ethanol} & \mathrm{1 equation} \end{array} = -1380 \mathrm{ kJ/mol}\]

It is important to note that this reaction is written for one mole of a reactant burned completely in oxygen. This type of reaction is called a  combustion reaction (test)and the ΔH is called a heat of combustion. We could simply tabulate ΔcombH (ethanol) and know that the reaction is as written above.

Generally, we cannot use or report a enthalpy change for a chemical reaction unless we have also specified the chemical reaction.  Had we written the reaction with different coefficients (still balanced) then the enthalpy change would have been different.  Consider the reaction above written as follows.

1/3 C2H5OH(l) + O2(g) image 2/3 CO2(g) + H2O(l)

This way, we have specified the reaction per mole of H2O produced.  The value for ΔH for this reaction is 1/3 that for the reaction written for one mole of ethanol.

Thus, the enthalpy change for this reaction can be calculated, starting again from the enthalpy change for one mole of ethanol

\[\Delta H = \begin{array}{|c|c|} -1380\mathrm{ kJ} & \mathrm{1/3 ethanol} \\ \hline \mathrm{1 mol ethanol} & \mathrm{1 equation} \end{array} = -460 \mathrm{ kJ/mol}\]

This is the enthalpy change for the reaction written above with only one mole of water produced.

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