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Jetzt kostenlos anmeldenImagine 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: constantvolume calorimetry.
In this article, we will discuss how constantvolume calorimetry is used to measure the heat released by chemical reactions.
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 constantvolume 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.
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 pressurevolume work: $$\Delta U=q+P\Delta V$$ However, in constantvolume 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 q_{v} 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, q_{rxn}, will be absorbed by the calorimeter, q_{cal}.
We calculate the heat absorbed by the calorimeter, q_{cal}, 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 C_{substance}.
So the formula for the heat absorbed by the calorimeter, q_{cal}, is: $$ q_{cal}=C_{cal}*\Delta T$$ where, ΔT, is the measured temperature change. We calculate the specific heat of the calorimeter, C_{cal}, 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 C_{cal }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 q_{cal}: $$q_{cal}=C_{bomb}*\Delta T+mC_{water}*\Delta T$$
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, Δn_{gases}, 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 
$$\Delta U=q_{v}$$ 

$$ q_{cal}=q_{rxn}$$ 

$$ q_{cal}=C_{cal}*\Delta T$$ 

$$q_{cal}=C_{bomb}*\Delta T+mC_{water}*\Delta T$$ 

$$\Delta H=q+\Delta n_{gases}RT$$ 

Here is what a bomb calorimeter looks like:
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.
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:
\(PV=nRT\)
\(P=\frac{nRT}{V}\)
\(P=\frac{5\,mol*0.0821 \frac{Latm}{molK}*1,573\,K}{2\,L}\)
\(P=322.9\,atm\)
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.
Now that we know what constantvolume calorimetry is, let's move on to some examples!
For calibration purposes, a 0.76 g sample of benzoic acid (C_{6}H_{5}COOH) 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 (C_{3}H_{8}) 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, m_{benzoic 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 q_{cal}.
\(q_{cal}=C_{cal}*\Delta T\)
\(q_{cal}=11.28\frac{kJ}{K}*1890\,K\)
\(q_{cal}=21,319\,kJ\)
We want our answer to be in kJ/mol, so we have to divide q_{cal} 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 Δn_{gas}, 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 O_{2} 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 Δn_{gas}:
\(\Delta n_{gas}=n_{products}n_{reactants}\)
\(\Delta n_{gas}=3(5+1)\)
\(\Delta n_{gas}=3\)
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:
\(\Delta H=q+\Delta n_{gas}RT\)
\(\Delta H=21,319\frac{kJ}{mol}+(3*8.314\frac{J}{molK}*((298.15K))\)
\(\Delta H=21,319\frac{kJ}{mol}7,432.7\frac{J}{mol}=21,319\frac{kJ}{mol}7.432.7\frac{J}{mol}*\frac{1\,kJ}{1000\,J}\)
\(\Delta H=21,319\frac{kJ}{mol}7.43\frac{kJ}{mol}\)
\(\Delta H=21,326\frac{kJ}{mol}\)
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 (C_{7}H_{5}N_{2}O_{6}) is ignited in a bomb calorimeter, C_{bomb}= 359 J/°C. The calorimeter contains 865 g of water (C_{w }= 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 :
H_{2}O  TNT/bomb 
m = 865 g  m = 0.764 g 
C = 4.184 J/g°C  C = 359 J/°C 
ΔT = T_{f }–30°C  ΔT = T_{f }–30°C 
First, we need to calculate q_{rxn}. 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).
\(n_{TNT}=\frac{m_{TNT}}{molar\,mass_{TNT}}=\frac{0.764\,g}{\frac{213.12\,g}{mol}}=0.00358\,mol\,TNT\)
\(0.00358\,mol*3374\frac{kJ}{mol}=12.08\,kJ\)
\(q_{rxn}=12.08\,kJ*\frac{1000\,J}{1\,kJ}=12,080\,J\)
Now we can set up our expression for q_{rxn} and solve for T_{f}.
\(q_{rxn}=C_{bomb}\Delta T+mC_{water}\Delta T\)
\(12,080\,J=(359\frac{J}{^\circ C}*\Delta T)+(865\,g*4.184\frac{J}{g^\circ C}*\Delta T)\)
\(12,080\,J=359\frac{J}{^\circ C}*\Delta T+3,619\frac{J}{^\circ C}*\Delta T\)
\(12,080\,J=3,978\frac{J}{^\circ C}*\Delta T\)
\(\Delta T=\frac{12,080\,J}{3,978\frac{J}{^\circ C}}\)
\(\Delta T=3.04^\circ C\)
\(\Delta T=T_f30^\circ C\)
\(3.04^\circ C=T_f30^\circ C\)
\(T_f=33.04^\circ C\)
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 "q_{cal}" instead of "q_{bomb}", so for example, you may see: $$q_{cal}=q_{cal}+q_{water}$$ Here the first q_{cal} 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.
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 coffeecup calorimeter since it is a styrofoam cup similar to a coffee cup.
In constantvolume calorimetry, the volume is fixed during the experiment. This is done to measure the amount of heat transferred to the calorimeter from the reaction. 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).
We perform constantvolume calorimetry using a bomb calorimeter. The bomb calorimeter is insulated so that no heat can escape. Inside the calorimeter, a chamber that contains the reaction is submerged in water. When the reaction occurs, a thermometer measures the change in temperature of the water, as a stirrer keeps the temperature uniform. This change in temperature is used to calculate the amount of heat produced from the reaction.
During constant pressure calorimetry, the pressure within the calorimeter is kept stable. The type of calorimeter used is often called a coffeecup calorimeter since it is a styrofoam cup similar to a coffee cup.
Calorimeters are designed to keep either the pressure or volume constant. Bomb calorimeters keep a constant volume, while coffee cup calorimeters keep a constant pressure.
When a calorimeter is kept at constant pressure, the change in heat (q) is equal to the change in enthalpy (Delta H). For constantvolume calorimetry, the change in heat is equal to the change in internal energy (Delta U).
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