Imagine you are getting relaxed and preparing yourself to take a nice bath. You pull out your butane lighter and get ready to light some candles. When you press on the wheel, instead of getting a small flame, a large, raging fire starts. This doesn't happen in real life (thankfully) because those lighters are designed so that only a tiny flame is lit. But how do these companies know how much butane should be released by the lighter? That answer is simple: constant-volume calorimetry.
In this article, we will discuss how constant-volume calorimetry is used to measure the heat released by chemical reactions.
- This article is about constant-volume calorimetry.
- First, we will define calorimetry and constant-volume calorimetry.
- Next, we will cover the math behind calorimetry.
- We will then explain the role of pressure in constant-volume calorimetry.
- Thereafter, we will walk through some examples.
- Lastly, we will compare and contrast constant-volume calorimetry and constant-pressure calorimetry.
Constant-Volume Calorimetry Definition
Calorimetry is the process of measuring the amount of heat transferred to or from a substance. It measures this by exchanging that heat with a calorimeter.
In constant-volume calorimetry, the volume is fixed during the experiment. The apparatus is typically called a bomb calorimeter since it is typically used to measure reactions that are "explosive" in nature (i.e. combustion reactions).
Now that we know the basics, let's look at what this type of calorimetry measures.
Constant-Volume Calorimetry Measures
A calorimeter's ability to measure a reaction's heat exchange hinges on the first law of thermodynamics. The first law states that the change in internal energy (ΔU) is equal to the sum of heat and pressure-volume work: $$\Delta U=q+P\Delta V$$ However, in constant-volume calorimetry the volume is being kept steady, which means that the work term (work done by the system on the external environment) is zero: $$\Delta U=q_{v}$$ Here qv means heat at constant volume.
The calorimeter is insulated, so no heat escapes. This means that: $$ q_{cal}=-q_{rxn}$$ This is because whatever heat is being released by the reaction, qrxn, will be absorbed by the calorimeter, qcal.
We calculate the heat absorbed by the calorimeter, qcal, using the temperature change and the specific heat of the calorimeter.
Specific heat (also called heat capacity) refers to the amount of heat necessary to raise the temperature of 1 g of a substance by 1°C. It is represented as Csubstance.
So the formula for the heat absorbed by the calorimeter, qcal, is: $$ q_{cal}=C_{cal} \cdot \Delta T$$ where, ΔT, is the measured temperature change. We calculate the specific heat of the calorimeter, Ccal, by adding a known amount of heat to the calorimeter and measuring the temperature change. This "calibrates" the calorimeter, so we can use it for experiments. The expression Ccal can also be expanded to $$C_{cal}=C_{bomb}+mC_{water}$$ The "bomb" refers to the chamber where the reaction takes place, which is submerged in water inside the calorimeter and, m, is the mass of the water used in the calorimeter. Using this expansion, we get another expression for qcal: $$q_{cal}=C_{bomb}\cdot \Delta T+mC_{water} \cdot \Delta T$$
It is also common for chemists to calculate the change in enthalpy (ΔH) since it tells us what would have happened if the reaction were at a constant pressure. To do this, we need to convert from q to ΔH. This conversion requires us to know the amount of work being done, which would be: $$\text{work}=\Delta n_{gases}RT$$
where, Δngases, is the change in the number of moles of gas and R is the gas constant.
Now we can do our conversion: $$ \Delta H=q+w$$ $$\Delta H=q+\Delta n_{gases}RT$$ I threw a lot of equations at you, so here is a diagram to organize them:
Equation | When to use |
$$\Delta U=q_{v}$$ | - When given "heat of combustion" or ΔU
- Typically used in combination with $$ q_{rxn}=-q_{cal} $$
|
$$ q_{cal}=-q_{rxn}$$ | - When converting from reaction to calorimeter or vice-versa
|
$$ q_{cal}=C_{cal} \cdot \Delta T$$ | - When calibrating the calorimeter
- When treating the calorimeter as a "whole" (no info about the water in the calorimeter is given)
|
$$q_{cal}=C_{bomb}\cdot \Delta T+mC_{water} \cdot \Delta T$$ | - When treating the calorimeter as its parts (info about the water is given)
|
$$\Delta H=q+\Delta n_{gases}RT$$ | - When calculating the change in enthalpy
|
So, how does a calorimeter actually measure this heat exchange? Let's look at how a bomb calorimeter is set up to learn more.Bomb Calorimeter Apparatus
Here is what a bomb calorimeter looks like:
Simplified diagram of a bomb calorimeter, where the red arrow represents the heat evolved by the system. Vaia Original.
This diagram shows a simplified version of a bomb calorimeter, but for our purposes, this is all we need to know. The sample is housed in its own chamber, and the reaction will release heat into the calorimeter (shown by the red arrow). The surrounding water absorbs this heat, and the thermometer measures this temperature change (the stirrer keeps the temperature uniform). The calorimeter is insulated so none of the heat escapes and all the heat is absorbed by the water.
Constant-Volume Calorimetry Pressure
Bomb calorimeters are designed to withstand a large change in pressure due to chemical reactions. If the volume is to be kept stable, the pressure must increase, and it will increase a lot. Here's the ideal gas equation to help you visualize: $$PV=nRT$$ Combustion reactions release a ton of heat, so the temperature value is going to be high and the pressure must compensate for that.
Let's say we have a calorimeter of 2 L, and a reaction takes place which releases 5 moles of gas and the temperature becomes 1,573 K. The calculated pressure would be:
\(PV=nRT\)
\(P=\frac{nRT}{V}\)
\(P=\frac{5\,mol \cdot 0.0821 \frac{L \, atm}{mol \, K} \cdot 1,573\,K}{2\,L}\)
\(P=322.9\,atm\)
To put that into perspective, that is 4,745 psi, which is ~95x greater than the lethal limit for humans (yikes). Bomb calorimeters are also called that since the reactions performed in them would explode other calorimeters.
Examples of Constant-Volume Calorimetry
Now that we know what constant-volume calorimetry is, let's move on to some examples!
For calibration purposes, a 0.76 g sample of benzoic acid (C6H5COOH) was ignited in a bomb calorimeter, initially at 25 °C. This produced a temperature rise of, ΔT = 1.78 K. For benzoic acid, ΔU=3,226 kJ/mol, and the molar mass is 122.12 g/mol. A sample of 1.0 mol of propane (C3H8) was ignited under identical conditions, which rose the temperature by 1890 K. Using this information, what is the enthalpy of combustion for propane in kJ/mol? The combustion reaction is \(C_3H_{8\,(g)}+5O_{2\,(g)}\rightarrow 3CO_{2\,(g)}+4H_2O_{(l)}\)
Using the calibration data, we can calculate the specific heat of the calorimeter. Our first step is to convert from grams to moles for the benzoic acid sample mass, mbenzoic acid:
\(n_{benzoic\, acid}=\frac{m_{benzoic\, acid}}{molar\,mass_{bezoic\, acid}}=\frac{0.76\,g}{\frac{122.12\,g}{mol}}=0.00622\,mol\,benzoic\, acid\)
Next, we multiply the molar amount by ΔU to get the amount of heat released into the calorimeter:
\(q_{cal}=0.00622\,mol*3,226\frac{kJ}{mol}=20.07\,kJ\)
Now we can calculate the specific heat of the calorimeter:
\(q_{cal}=C_{cal}*\Delta T\)
\(C_{cal}=\frac{q_{cal}}{\Delta T}\)
\(C_{cal}=\frac{20.07\,kJ}{1.78\,K}\)
\(C_{cal}=11.28\frac{kJ}{K}\)
Now onto the second half of the problem. We are told that this reaction was done under the same conditions, meaning the specific heat of the calorimeter is the same. Our first step is to calculate qcal.
\(q_{cal}=C_{cal} \cdot \Delta T\)
\(q_{cal}=11.28\frac{kJ}{K} \cdot 1890\,K\)
\(q_{cal}=21,319\,kJ\)
We want our answer to be in kJ/mol, so we have to divide qcal by our amount of propane.
\(\frac{21,319\,kJ}{1.0\,mol}=21,319\frac{kJ}{mol}\)
\(q_{cal}=-21,319\frac{kJ}{mol}\)
We change the sign to negative since this is an exothermic reaction. Exothermic reactions are a net release of heat, so since the system is losing heat, the sign is negative.
For our next step, we need to know Δngas, so we use the chemical equation to calculate the moles of gas. The equation tells us the ratio of reactants to products, so we can use the amount of propane to calculate how many moles of gas were produced and how many moles of O2 are needed for the reaction. Since we have 1 mol of propane, the moles of the other gases are equal to the coefficient (number in front)
\(C_3H_{8\,(g)}+5O_{2\,(g)} \rightarrow 3CO_{2\,(g)}+4H_2O_{(l)}\)
\(5\,mol\,O_2\,\,\,3\,mol\,CO_2\)
Now we calculate Δngas:
\(\Delta n_{gas}=n_{products}-n_{reactants}\)
\(\Delta n_{gas}=3-(5+1)\)
\(\Delta n_{gas}=-3\)
Here the negative sign means that the volume of the system would decrease, due to the reaction, if it were not kept at a constant volume.
Now we can finally calculate ΔH. Converting the initial temperature of the calorimeter, 25°C, to Kelvins, we get: 25°C + 273.15 K = 298.15 K, then:
\(\Delta H=q+\Delta n_{gas}RT\)
\(\Delta H=-21,319\frac{kJ}{mol}+(-3*8.314\frac{J}{molK}*((298.15K))\)
\(\Delta H=-21,319\frac{kJ}{mol}-7,432.7\frac{J}{mol}=-21,319\frac{kJ}{mol}-7.432.7\frac{J}{mol} \cdot \frac{1\,kJ}{1000\,J}\)
\(\Delta H=-21,319\frac{kJ}{mol}-7.43\frac{kJ}{mol}\)
\(\Delta H=-21,326\frac{kJ}{mol}\)
Thus, the enthalpy of combustion of one mole of propane is, ΔH = -21,326 kJ/mol.
That was a bit of a long one, so let's work on a shorter problem.
A 0.764 g sample of TNT (C7H5N2O6) is ignited in a bomb calorimeter, Cbomb= 359 J/°C. The calorimeter contains 865 g of water (Cw = 4.184 J/g °C), initially at 30 °C. TNT has a heat of combustion of 3374 kJ/mol and a molar mass of 213.12 g/mol. Using this data, calculate the final temperature of the calorimeter/water.
Let's start by writing out our known variables. Here the TNT is the sample, so it is considered the "bomb". Also :
H2O | TNT/bomb |
m = 865 g | m = 0.764 g |
C = 4.184 J/g°C | C = 359 J/°C |
ΔT = Tf –30°C | ΔT = Tf –30°C |
First, we need to calculate qrxn. The heat of reaction is the heat of combustion of TNT multiplied by the molar amount (this is because the explosion of TNT is the reaction).
\(n_{TNT}=\frac{m_{TNT}}{molar\,mass_{TNT}}=\frac{0.764\,g}{\frac{213.12\,g}{mol}}=0.00358\,mol\,TNT\)
\(0.00358\,mol \cdot 3374\frac{kJ}{mol}=12.08\,kJ\)
\(q_{rxn}=12.08\,kJ \cdot \frac{1000\,J}{1\,kJ}=12,080\,J\)
Now we can set up our expression for qrxn and solve for Tf.
\(q_{rxn}=C_{bomb}\Delta T+mC_{water}\Delta T\)
\(12,080\,J=(359\frac{J}{^\circ C}\cdot \Delta T)+(865\,g \cdot 4.184\frac{J}{g^\circ C} \cdot \Delta T)\)
\(12,080\,J=359\frac{J}{^\circ C} \cdot \Delta T+3,619\frac{J}{^\circ C} \cdot \Delta T\)
\(12,080\,J=3,978\frac{J}{^\circ C} \cdot \Delta T\)
\(\Delta T=\frac{12,080\,J}{3,978\frac{J}{^\circ C}}\)
\(\Delta T=3.04^\circ C\)
\(\Delta T=T_f-30^\circ C\)
\(3.04^\circ C=T_f-30^\circ C\)
\(T_f=33.04^\circ C\)
Calorimetry problems can be tricky, so always keep track of your units and variables. Writing a "knowns" chart can be helpful to organize the data given to you in the problem.
Also, some texts may use "qcal" instead of "qbomb", so for example, you may see: $$q_{cal}=q_{cal}+q_{water}$$ Here the first qcal is referring to the whole calorimeter, while the second is referring to the "bomb"/container the reaction is in, which is why I made that distinction to avoid confusion.
Constant Pressure and Constant-volume Calorimetry
There is another type of calorimetry called constant pressure calorimetry.
During constant pressure calorimetry, the pressure within the calorimeter is kept stable. The type of calorimeter used is often called a coffee-cup calorimeter since it is a styrofoam cup similar to a coffee cup.
Constant-volume Calorimetry - Key takeaways
- Calorimetry is the process of measuring the amount of heat transferred to or from a substance. It measures this by exchanging that heat with a calorimeter.
- In constant-volume calorimetry, the volume is fixed during the experiment. The apparatus is typically called a bomb calorimeter since it is typically used to measure reactions that are "explosive" in nature (i.e. combustion reactions).
- Calorimeters are insulated, so neat heat escapes, therefore: \(q_{rxn}=-q_{cal}\)
- The formula for qcal is \(q_{cal}=C_{cal} \cdot \Delta T\) where \(C_{cal}=C_{bomb}+mC_{water}\)
- Calorimetry can also be used to calculate enthalpy (ΔH) where \(\Delta H=q+\Delta n_{gases}RT\)
- In constant-pressure calorimetry, the heat exchange (q) is equal to ΔH, while it is equal to ΔU for constant-volume calorimetry.
Explanations 
Textbooks 
Exams 
Magazine 
