# Calorimetry

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 E = qv

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 (qv) 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

-qv = qwater + qbomb

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

To determine the above we will need the individual values:

qwater = mass of water * (specific heat of water) * (T) and

qbomb = heat capacity of bomb * 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 E of combustion?

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

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

So Ecombustion = - (6863 + 41171)J = -48.0 kJ

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

Ecombustion = -48.0 kJ/g

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

Ecombustion = - 5.48 x 103 kJ/mol

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

H = E + PV

H = E + (PV) = q + w + (PV) = q - pV + PV + VP

H = q + VP

and if P is constant then we would have

H = qp

So you see that if we carried out the calorimetry experiment at constant pressure we would obtain 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:

qp = mass of water * specific heat of water * T = 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 H of solution for ammonium nitrate?

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

but this is for 1.50 grams of ammonium nitrate so

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

Hsolution = -2.58 kj/mol

Relating H to E

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

qv ~ qp so that E ~ H

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

H = E + PV

The change in volume is primarily due to the gases so

V = Vgases

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 VA = nART/P and

VB = nBRT/P

so V = VB - VA = (nA - nB)RT/P

or to generalize

V = ngRT/P

where ng = # moles of gaseous products - # moles of gaseous reactants

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

H = E + RTng

 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.

Web Author: Dr. Leon L. Combs