Measuring Heats of Reaction: Calorimetry

The amount of heat transferred between the system and the surroundings is measured experimentally by calorimetry. A calorimeter measures the temperature change accompanying a process. The temperature change of a calorimeter depends on its heat capacity, the amount of heat required to raise its temperature by 1K. Constant-volume calorimetry is carried out in a vessel of fixed volume called a bomb calorimeter. Bomb calorimeters are used to measure the heat evolved in combustion reactions.

When a reaction occurs at constant pressure inside a Styrofoam coffee-cup calorimeter, the enthalpy change involves heat, and little heat is lost to the lab (or gained from it). If the reaction evolves heat, for example, very nearly all of it stays inside the calorimeter, the amount of heat absorbed or evolved by the reaction is calculated.

Example Problem:
The reaction of an acid such as HCl with a base such as NaOH in water involves the exothermic reaction

                                      HCl(aq) + NaOH(aq) ---> NaCl(aq) + H₂O

In one experiment, a student placed 50.0 mL of 1.00 M HCl in a coffee-cup calorimeter and carefully measured its temperature to be 25.5^oC. To this was added 50.0 mL of 1.00 M NaOH solution whose temperature was also 25.5^oC. The mixture was quickly stirred, and the student noticed that the temperature of the mixture rose to 32.4^oC. What was the heat of reaction?

Assumptions:
These are solutions, not pure water. The specific heat of water is 4.184 J/g^oC. Assume that these solutions are close enough to being like water that their specific heats are also 4.1984 J/g^oC.

The density of water is 1.00 g/mL and even though these are solutions we can assume that they are close enough to water to have the same density.

Solution:
1. Calculate the heat actually evolved.

                                            q = mct

Fill in the missing info. We have mL's and we need grams.

Use density. (50 mL + 50 mL ) = 100 mL of solution.

100 mL X 1     g        =  100 grams of solution. (m = V X D)
                      mL

Find the temperature change.

delta t = t_final - t_initial = 32.4^oC - 25.5^oC = 6.9^oC

    q = mct
       = 100 grams X 4.184    J   X 6.9^oC
                                          g^oC
       = 2.9 X 10³ J

This is the heat gained by the water, but in fact it is the heat lost by the reacting HCl and NaOH, therefore q = -2.9 x 10³ J.

i.e. it is an exothermic reaction, heat was lost to the water and it got warmer.

This only gets us part way. This is the heat evolved for those specific amounts used. (Notice we used identical amounts to keep these solutions simple). We need to find the amount of heat released per mole.

How much HCl did we actually use anyways?

50.0 mL of HCl X 1.00 mol HCl = 0.0500 mol HCl
                             1000 mL HCl

The same quantity of base, 0.0500 mole NaOH, was used.

To calculate the energy per mole of acid or base, divide the number of joules by the number of moles.

i.e. molar enthalpy = J/mol = -2.9 x 10³ J / 0.0500 mol
                                         = -5.8 x 10⁴ J/mol
                                         = -58000 J/mol
                                         = -58 kJ/mol

Therefore, for the neutralization of HCl and NaOH, the enthalpy change, often called the enthalpy of reaction is DH = -58 kJ/mol

The Bomb Calorimeter
A type of calorimeter used in very precise measurements of heats of reaction is called the bomb calorimeter. It is used to measure energy changes for reactions that will not happen until they are deliberately initiated, for example, combustions which must be ignited. The reactants are put into the "bomb", which is then sealed and immersed in a large, well-insulated vat of water. When the reaction is set off, any heat that is liberated is absorbed by the bomb, the water, and any piece of the equipment sticking into the water, and the temperature of the entire contents of the vat rises. The stirrer ensures that any heat released becomes uniformly distributed before the final temperature is read. From the temperature change and the heat capacity of the calorimeter (water plus everything in the water), the heat liberated is calculated.

Example Problem:
A sample of sucrose (table sugar) with a mass of 1.32 g is burned in a bomb calorimeter. The heat capacity of this calorimeter had been previously found to be 9.43 kJ/^oC. The temperature changed from 25.00^oC to 27.31^oC. Calculate the heat of combustion of sucrose in kilojoules per mole. The formula of sucrose is C₁₂H₂₂O₁₁.

Solution:
The delta t is 2.31^OC. For each degree increase, the reaction has evolved 9.43 kJ, as we know from the heat capacity. Therefore the total heat evolved is

                          E =  2.31^oC X 9.43 kJ = 21.8 kJ
                                                                            ^oC

This heat was produced by the combustion of 1.32 g of sucrose.

               moles = g/molecular mass
                         = 1.32 g / 342.3 grams/mole
                         = 3.86 x 10^-3 mol of sucrose.

Therefore, the heat evolved per mole of sucrose is

              21.8 kJ           =   5.65 x 10³ kJ/mole
        3.86 x 10^-3 mol

Since the combustion is exothermic, this should be given a minus sign and reported as -5.65 x 10³ kJ/mol for the heat of combustion for sucrose.

Go to the Enthalpy Calorimetry Worksheet