Calorimetry

Most freshman chemistry books have a nice diagram of a "constant volume calorimeter". This is commonly called a "bomb calorimeter", but this terminology might be a little frightening -- there is very little danger involved in using the instrument. The devise has an outer insulated portion which is viewed as the end of the universe -- no heat or work can pass it. The contents of the outer insulated container consists of the steel "bomb", a sample dish within the "bomb", ignition wires into the "bomb" and touching the chemical sample in a sample dish, water surrounding the "bomb", a stirrer, and a thermometer. The contents of the "bomb" are the system and the other contents of the insulated container (including the walls of the "bomb") are the surroundings. The wall of the "bomb" is the boundary. The "bomb" confines the system to constant volume. Work can

W = 0 and thus delta E = q_v

The "bomb" has the chemical sample placed in the sample dish, the "bomb" is filled with oxygen, and then the sample is ignited. We then must analyze the event.

From the law of conservation of energy, we can deduce that

the heat transferred from the system = the heat transferred into the surroundings

The left term is just the heat of the reaction (q_v) and the right term is the sum of the heat absorbed by the water and the heat absorbed by the bombs' stainless steel walls so we have

-q_v = q_water + q_bomb

where the negative sign is required because heat is lost from the system (exothermic).

To determine the above we will need the individual values:

q_water = mass of water * (specific heat of water) * ( delta T) and

q_bomb = heat capacity of bomb * delta T

The heat capacity of the bomb is determined by first doing an experiment with some chemical for which you know the heat of combustion so that you can solve the equations for the heat capacity of the bomb. Then the unknown is run using the previously determined value for the heat capacity of the bomb. So let's work an example.

A one-gram sample of octane is burned in a bomb calorimeter. The mass of water used is 1.20 kg, the temperature rise is 8.20 K, and the heat capacity of the bomb is known to be 837 J/K. What is delta E of combustion?

q_bomb = 837J/K * 8.20K = 6863 J

q_water = 4.184 J/gK * (1200 g) * (8.20 K) = 41171 J

So delta E_combustion = - (6863 + 41171)J = -48.0 kJ

This result is for one gram so we can write this as

delta E_combustion = -48.0 kJ/g

and then multiply by the molecular weight of octane (114.2 g/mol) to get

delta E_combustion = - 5.48 x 10³ kJ/mol

You see that the constant volume calorimeter is going to give us delta E rather than delta H. To obtain delta H directly from experiment we can see what we would need to do:

H = E + PV

delta H = delta E + delta (PV) = q + w + delta (PV) = q - p delta V + P delta V + V delta P

delta H = q + V delta P

and if P is constant then we would have

delta H = q_p

So you see that if we carried out the calorimetry experiment at constant pressure we would obtain delta H. Many high schools and freshman chemistry labs carry out such an experiment in a styrofoam cup which is open to the atmosphere (constant P).

Since there is no bomb, the heat of the reaction can be solely determined by the rise in temperature of the water:

q_p = mass of water * specific heat of water * delta T = delta H

So do an example:

1.50 grams of ammonium nitrate is added to 35.0 g of water and dissolved. The initial temperature was 27.7 C and the final temperature was 19.4 C.

a.) is the reaction endothermic or exothermic? It is endothermic since the surroundings lost heat as measured by a decrease in temperature. b.) what is delta H of solution for ammonium nitrate?

q_water = 4.184 J/gK * (35.0 g) * (-3.3 K) = - 483.2 J

but this is for 1.50 grams of ammonium nitrate so

q_water = -322.2 J/g and if we then multiply by the molar mass of ammonium nitrate (80.0 g/mol) we then have

delta H_solution = -2.58 kj/mol

Relating H to E

If only liquids and solids are involved in a reaction then delta V ~ zero and

q_v ~ q_p so that delta E ~ delta H

But this is obviously not correct for gases which we have for combustion reactions. Now we will need to determine delta V

delta H = delta E + P delta V

The change in volume is primarily due to the gases so

delta V = delta V_gases

Let's assume that the gases behave ideally so that

V = nRT/P

and let's assume that we have

A(g) --> B(g)

So we will have V_A = n_ART/P and

V_B = n_BRT/P

so delta V = V_B - V_A = (n_A - n_B)RT/P

or to generalize

delta V = delta n_gRT/P

where delta n_g = # moles of gaseous products - # moles of gaseous reactants

So now we see that delta H = delta E + P delta n_gRT/P or

delta H = delta E + RT delta n_g

Now take a practice quiz to help you understand if you understand the basic concepts.

You must use your real name when it asks for a name.

The test will only submit when you have answers all of the questions correctly.

If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests.