A calorimeter is a device used to measure the quantity of heat flow in a chemical reaction. Two of the most common types of calorimeters are the coffee cup calorimeter and the bomb calorimeter.
Coffee Cup Calorimeter
A coffee cup calorimeter is essentially a polystyrene (Styrofoam) cup with a lid. The cup is partially filled with a known volume of water and a thermometer is inserted through the lid of the cup so that its bulb is below the water surface. When a chemical reaction occurs in the coffee cup calorimeter, the heat of the reaction if absorbed by the water. The change in the water temperature is used to calculate the amount of heat that has been absorbed (used to make products, so water temperature decreases) or evolved (lost to the water, so its temperature increases) in the reaction.
Heat flow is calculated using the relation:
q = (specific heat) x m x Dt
where q is heat flow, m is mass in grams, and Dt is the change in temperature. The specific heat is the amount of heat required to raise the temperature of 1 gram of a substance 1 degree Celsius. The specific heat of water is 4.18 J/(g·°C).
For example, consider a chemical reaction which occurs in 200 grams of water with an initial temperature of 25.0°C. The reaction is allowed to proceed in the coffee cup calorimeter. As a result of the reaction, the temperature of the water changes to 31.0°C. The heat flow is calculated:
qwater = 4.18 J/(g·°C) x 200 g x (31.0°C - 25.0°C)
qwater = +5.0 x 103 J
In other words, the products of the reaction evolved 5000 J of heat, which was lost to the water. The enthalpy change, DH, for the reaction is equal in magnitude but opposite in sign to the heat flow for the water:
DHreaction = -(qwater)
Recall that for an exothermic reaction, DH <>water is positive. The water absorbs heat from the reaction and an increase in temperature is seen. For an endothermic reaction, DH > 0; qwater is negative. The water supplies heat for the reaction and a decrease in temperature is seen.
Bomb Calorimeter
A coffee cup calorimeter is great for measuring heat flow in a solution, but it can't be used for reactions which involve gases, since they would escape from the cup. The coffee cup calorimeter can't be used for high temperature reactions, either, since these would melt the cup. A bomb calorimeter is used to measure heat flows for gases and high temperature reactions.
A bomb calorimeter works in the same manner as a coffee cup calorimeter, with one big difference. In a coffee cup calorimeter, the reaction takes place in the water. In a bomb calorimeter, the reaction takes place in a sealed metal container, which is placed in the water in an insulated container. Heat flow from the reaction crosses the walls of the sealed container to the water. The temperature difference of the water is measured, just as it was for a coffee cup calorimeter. Analysis of the heat flow is a bit more complex than it was for the coffee cup calorimeter because the heat flow into the metal parts of the calorimeter must be taken into account:
qreaction = - (qwater + qbomb)
where qwater = 4.18 J/(g·°C) x mwater x Dt
The bomb has a fixed mass and specific heat. The mass of the bomb multiplied by its specific heat is sometimes termed the calorimeter constant, denoted by the symbol C with units of joules per degree Celsius. The calorimeter constant is determined experimentally and will vary from one calorimeter to the next. The heat flow of the bomb is:
qbomb = C x Dt
Once the calorimeter constant is known, calculating heat flow is a simple matter. The pressure within a bomb calorimeter often changes during a reaction, so the heat flow may not be equal in magnitude to the enthalpy change.
Understanding Enthalpy and Thermochemical Equations
Thermochemical equations are just like other balanced equations except they also specify the heat flow for the reaction. The heat flow is listed to the right of the equation using the symbol ΔH. The most common units are kilojoules, kJ. Here are two thermochemical equations:
H2 (g) + ½ O2 (g) → H2O (l); ΔH = -285.8 kJ
HgO (s) → Hg (l) + ½ O2 (g); ΔH = +90.7 kJ
When you write thermochemical equations, be sure to keep the following points in mind:
Coefficients refer to the number of moles. Thus, for the first equation, -282.8 kJ is the ΔH when 1 mol of H2O (l) is formed from 1 mol H2 (g) and ½ mol O2.
Enthalpy changes for a phase change, so the enthalpy of a substance depends on whether is it is a solid, liquid, or gas. Be sure to specify the phase of the reactants and products using (s), (l), or (g) and be sure to look up the correct ΔH from heat of formation tables. The symbol (aq) is used for species in water (aqueous) solution.
The enthalpy of a substance depends upon temperature. Ideally, you should specify the temperature at which a reaction is carried out. When you look at a table of heats of formation, notice that the temperature of the ΔH is given. For homework problems, and unless otherwise specified, temperature is assumed to be 25°C. In the real world, temperature may different and thermochemical calculations can be more difficult.
Certain laws or rules apply when using thermochemical equations:
ΔH is directly proportional to the quantity of a substance that reacts or is produced by a reaction.
Enthalpy is directly proportional to mass. Therefore, if you double the coefficients in an equation, then the value of ΔH is multiplied by two. For example:
H2 (g) + ½ O2 (g) → H2O (l); ΔH = -285.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (l); ΔH = -571.6 kJ
ΔH for a reaction is equal in magnitude but opposite in sign to ΔH for the reverse reaction.
For example:
HgO (s) → Hg (l) + ½ O2 (g); ΔH = +90.7 kJ
Hg (l) + ½ O2 (l) → HgO (s); ΔH = -90.7 kJ
This law is commonly applied to phase changes, although it is true when you reverse any thermochemical reaction.
ΔH is independent of the number of steps involved.
This rule is called Hess's Law. It states that ΔH for a reaction is the same whether it occurs in one step or in a series of steps. Another way to look at it is to remember that ΔH is a state property, so it must be independent of the path of a reaction.
If Reaction (1) + Reaction (2) = Reaction (3), then ΔH3 = ΔH1 + ΔH2
The molar heat of formation of a compound (ΔHf) is equal to its enthalpy change (ΔH) when one mole of compound is formed at 25°C and 1 atm from elements in their stable form. This is a table of the heats of formation for a variety of common compounds. As you can see, most heats of formation are negative quantities, which implies that the formation of a compound from its elements usually is an exothermic process.
Compound
ΔHf (kJ/mol)
Compound
ΔHf (kJ/mol)
AgBr(s)
-99.5
C2H2(g)
+226.7
AgCl(s)
-127.0
C2H4(g)
+52.3
AgI(s)
-62.4
C2H6(g)
-84.7
Ag2O(s)
-30.6
C3H8(g)
-103.8
Ag2S(s)
-31.8
n-C4H10(g)
-124.7
Al2O3(s)
-1669.8
n-C5H12(l)
-173.1
BaCl2(s)
-860.1
C2H5OH(l)
-277.6
BaCO3(s)
-1218.8
CoO(s)
-239.3
BaO(s)
-558.1
Cr2O3(s)
-1128.4
BaSO4(s)
-1465.2
CuO(s)
-155.2
CaCl2(s)
-795.0
Cu2O(s)
-166.7
CaCO3
-1207.0
CuS(s)
-48.5
CaO(s)
-635.5
CuSO4(s)
-769.9
Ca(OH)2(s)
-986.6
Fe2O3(s)
-822.2
CaSO4(s)
-1432.7
Fe3O4(s)
-1120.9
CCl4(l)
-139.5
HBr(g)
-36.2
CH4(g)
-74.8
HCl(g)
-92.3
CHCl3(l)
-131.8
HF(g)
-268.6
CH3OH(l)
-238.6
HI(g)
+25.9
CO(g)
-110.5
HNO3(l)
-173.2
CO2(g)
-393.5
H2O(g)
-241.8
H2O(l)
-285.8
NH4Cl(s)
-315.4
H2O2(l)
-187.6
NH4NO3(s)
-365.1
H2S(g)
-20.1
NO(g)
+90.4
H2SO4(l)
-811.3
NO2(g)
+33.9
HgO(s)
-90.7
NiO(s)
-244.3
HgS(s)
-58.2
PbBr2(s)
-277.0
KBr(s)
-392.2
PbCl2(s)
-359.2
KCl(s)
-435.9
PbO(s)
-217.9
KClO3(s)
-391.4
PbO2(s)
-276.6
KF(s)
-562.6
Pb3O4(s)
-734.7
MgCl2(s)
-641.8
PCl3(g)
-306.4
MgCO3(s)
-1113
PCl5(g)
-398.9
MgO(s)
-601.8
SiO2(s)
-859.4
Mg(OH)2(s)
-924.7
SnCl2(s)
-349.8
MgSO4(s)
-1278.2
SnCl4(l)
-545.2
MnO(s)
-384.9
SnO(s)
-286.2
MnO2(s)
-519.7
SnO2(s)
-580.7
NaCl(s)
-411.0
SO2(g)
-296.1
NaF(s)
-569.0
So3(g)
-395.2
NaOH(s)
-426.7
ZnO(s)
-348.0
NH3(g)
-46.2
ZnS(s)
-202.9
Reference: Masterton, Slowinski, Stanitski, Chemical Principles, CBS College Publishing, 1983.
These are molar heats of formation for anions and cations in aqueous solution. In all cases, the heats of formation are given in kJ/mol at 25°C for 1 mole of the ion.
Cations
ΔHf (kJ/mol)
Anions
ΔHf (kJ/mol)
Ag+ (aq)
+105.9
Br- (aq)
-120.9
Al3+ (aq)
-524.7
Cl- (aq)
-167.4
Ba2+ (aq)
-538.4
ClO3- (aq)
-98.3
Ca2+ (aq)
-543.0
ClO4- (aq)
-131.4
Cd2+ (aq)
-72.4
CO32- (aq)
-676.3
Cu2+ (aq)
+64.4
CrO42- (aq)
-863.2
Fe2+ (aq)
-87.9
F- (aq)
-329.1
Fe3+ (aq)
-47.7
HCO3- (aq)
-691.1
H+ (aq)
0.0
H2PO4- (aq)
-1302.5
K+ (aq)
-251.2
HPO42- (aq)
-1298.7
Li+ (aq)
-278.5
I- (aq)
-55.9
Mg2+ (aq)
-462.0
MnO4- (aq)
-518.4
Mn2+ (aq)
-218.8
NO3- (aq)
-206.6
Na+ (aq)
-239.7
OH- (aq)
-229.9
NH4+ (aq)
-132.8
PO43- (aq)
-1284.1
Ni2+ (aq)
-64.0
S2- (aq)
+41.8
Pb2+ (aq)
+1.6
SO42- (aq)
-907.5
Sn2+ (aq)
-10.0
Zn2+ (aq)
-152.4
Reference: Masterton, Slowinski, Stanitski, Chemical Principles, CBS College Publishing, 1983.
An endothermic process or reaction absorbs energy in the form of heat (endergonic processes or reactions absorb energy, not necessarily as heat). Examples of endothermic processes include the melting of ice and the depressurization of a pressurized can. In both processes, heat is absorbed from the environment. You could record the temperature change using a thermometer or by feeling the reaction with your hand. The reaction between citric acid and baking soda is a highly safe example of an endothermic reaction, commonly used as a chemistry demonstration. Do you want a colder reaction? Solid barium hydroxide reacted with solid ammonium thiocyanate produces barium thiocyanate, ammonia gas, and liquid water. This reaction gets down to -20°C or -30°C, which is more than cold enough to freeze water. It's also cold enough to give you frostbite, so be careful! The reaction proceeds according to the following equation:
Ba(OH)2.8H2O (s) + 2 NH4SCN (s) --> Ba(SCN)2 (s) + 10 H2O (l) + 2 NH3 (g)
Here's what you need to use this reaction as a demonstration:
32g barium hydroxide octahydrate
17g ammonium thiocyanate (or could use ammonium nitrate or ammonium chloride)
Pour the barium hydroxide and ammonium thiocyanate into the flask.
Stir the mixture.
The odor of ammonia should become evident within about 30 seconds. If you hold a piece of dampened litmus paper over the reaction you can watch a color change showing that the gas produced by the reaction is basic.
Liquid will be produced, which will freeze into a slush as the reaction proceeds.
If you set the flask on a damp block of wood or piece of cardboard while performing the reaction you can freeze the bottom of the flask to the wood or paper. You can touch the outside of the flask, but don't hold it in your hand while performing the reaction.
After the demonstration is completed, the contents of the flask can be washed down the drain with water. Do not drink the contents of the flask. Avoid skin contact. If you get any solution on your skin, rinse it off with water.
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Sodium acetate or hot ice is an amazing chemical you can prepare yourself from baking soda and vinegar. You can cool a solution of sodium acetate below its melting point and then cause the liquid to crystallize. The crystallization is an exothermic process, so the resulting ice is hot. Solidification occurs so quickly you can form sculptures as you pour the hot ice.
In a saucepan or large beaker, add baking soda to the vinegar, a little at a time and stirring between additions. The baking soda and vinegar react to form sodium acetate and carbon dioxide gas. If you don't add the baking soda slowly, you'll essentially get a baking soda and vinegar volcano, which would overflow your container. You've made the sodium acetate, but it is too dilute to be very useful, so you need to remove most of the water.
Here is the reaction between the baking soda and vinegar to produce the sodium acetate:
Na+[HCO3]– + CH3–COOH → CH3–COO– Na+ + H2O + CO2
Boil the solution to concentrate the sodium acetate. You could just remove the solution from heat once you have 100-150 ml of solution remaining, but the easiest way to get good results is to simply boil the solution until a crystal skin or film starts to form on the surface. This took me about an hour on the stove over medium heat. If you use lower heat you are less likely to get yellow or brown liguid, but it will take longer. If discoloration occurs, it's okay.
Once you remove the sodium acetate solution from heat, immediately cover it to prevent any further evaporation. I poured my solution into a separate container and covered it with plastic wrap. You should not have any crystals in your solution. If you do have crystals, stir a very small amount of water or vinegar into the solution, just sufficient to dissolve the crystals.
Place the covered container of sodium acetate solution in the refrigerator to chill.
Activities Involving Hot Ice
The sodium acetate in the solution in the refrigerator is an example of a supercooled liquid. That is, the sodium acetate exists in liquid form below its usual melting point. You can initiate crystallization by adding a small crystal of sodium acetate or possibly even by touching the surface of the sodium acetate solution with a spoon or finger. The crystallization is an example of an exothermic process. Heat is released as the 'ice' forms. To demonstrate supercooling, crystallization, and heat release you could:
Drop a crystal into the container of cooled sodium acetate solution. The sodium acetate will crystallize within seconds, working outward from where you added the crystal. The crystal acts as a nucleation site or seed for rapid crystal growth. Although the solution just came out of the refrigerator, if you touch the container you will find it is now warm or hot.
Pour the solution onto a shallow dish. If the hot ice does not spontaneously begin crystallization, you can touch it with a crystal of sodium acetate (you can usually scrape a small amount of sodium acetate from the side of the container you used earlier). The crystallization will progress from the dish up toward where you are pouring the liquid. You can construct towers of hot ice. The towers will be warm to the touch.
You can re-melt sodium acetate and re-use it for demonstrations.
Hot Ice Safety
As you would expect, sodium acetate is a safe chemical for use in demonstrations. It is used as a food additive to enhance flavor and is the active chemical in many hot packs. The heat generated by the crystallization of a refrigerated sodium acetate solution should not present a burn hazard.
Exothermic chemical reactions produce heat. In this reaction vinegar is used to remove the protective coating from steel wool, allowing it to rust. When the iron combines with oxygen, heat is released.
Difficulty: Average
Time Required: 15 minutes
Here's How:
Place the thermometer in the jar and close the lid. Allow about 5 minutes for the thermometer to record the temperature, then open the lid and read the thermometer.
Remove the thermometer from the jar (if you didn't already in Step 1).
Soak a piece of steel wool in vinegar for 1 minute.
Squeeze the excess vinegar out of the steel wool.
Wrap the wool aroung the thermometer and place the wool/thermometer in the jar, sealing the lid.
Allow 5 minutes, then read the temperature and compare it with the first reading.
Chemistry is Fun!
Tips:
Not only does the vinegar remove the protective coating on the steel wool, but once the coating is off its acidity aids in oxidation (rust) of the iron in the steel.
The thermal energy given off during this chemical reaction causes the mercury in the thermometer to expand and rise up the column of the thermometer tube.
In the rusting of iron, four atoms of solid iron react with three molecules of oxygen gas to form two molecules of solid rust (iron oxide).
Many chemical reactions release energy in the form of heat, light, or sound. These are exothermic reactions. Exothermic reactions may occur spontaneously and result in higher randomness or entropy (ΔS > 0) of the system. They are denoted by a negative heat flow (heat is lost to the surroundings) and decrease in enthalpy (ΔH <>There are other chemical reactions that must absorb energy in order to proceed. These are endothermic reactions. Endothermic reactions cannot occur spontaneously. Work must be done in order to get these reactions to occur. When endothermic reactions absorb energy, a temperature drop is measured during the reaction. Endothermic reactions are characterized by positive heat flow (into the reaction) and an increase in enthalpy (+ΔH).
Examples of Endothermic and Exothermic Processes
Photosynthesis is an example of an endothermic chemical reaction. In this process, plants use the energy from the sun to convert carbon dioxide and water into glucose and oxygen. This reaction requires 15MJ of energy (sunlight) for every kilogram of glucose that is produced:
An example of an exothermic reaction is the mixture of sodium and chlorine to yield table salt. This reaction produces 411 kJ of energy for each mole of salt that is produced:
Na(s) + 0.5Cl2(s) = NaCl(s)
Demonstrations You Can Perform
Many exothermic and endothermic reactions involve toxic chemicals, extreme heat or cold, or messy disposal methods. These demonstrations are safe and easy:
Most endothermic reactions contain toxic chemicals, but this reaction is safe and easy. Use it as a demonstration or vary the amounts of citric acid and sodium bicarbonate to make an experiment.
Difficulty: Average
Time Required: Minutes
Here's How:
Pour the citric acid solution in a styrofoam coffee cup. Use a thermometer or other temperature probe to record the initial temperature.
Stir in the baking soda (sodium bicarbonate). Track the change in temperature as a function of time.
The reaction is: H3C6H5O7(aq) + 3 NaHCO3(s) --> 3 CO2(g) + 3 H2O(l) + NaC6H5O7(aq)
When you have completed your demonstration or experiment, simply wash the cup out in a sink. No toxic chemicals to mess with!
Tips:
Feel free to vary the concentration of the citric acid solution or the quantity of sodium bicarbonate.
An endothermic is a reaction that requires energy to proceed. The intake of energy may be observed as a decrease in temperature as the reaction proceeds. Once the reaction is complete, the temperature of the mixture will return to room temperature.
ΔH for a reaction is equal to the sum of the heats of formation of the product compounds minus the sum of the heats of formation of the reactant compounds:
ΔH = Σ ΔHf products - Σ ΔHf reactants
Omitting terms for the elements, the equation becomes:
ΔH = 4 ΔHf Al2O3(s) - 3 ΔHf Fe3O4(s)
The values for ΔHf may be found in the Heats of Formation of Compounds table. Plugging in these numbers:
ΔH = 4(-1669.8 kJ) - 3(-1120.9 kJ)
ΔH = -3316.5 kJ
Answer
ΔH = -3316.5 kJ
2. Heat of Formation Problem
Calculate ΔH for the ionization of hydrogen bromide:
HBr(g) --> H+(aq) + Br-(aq)
Heat of Formation Solution
ΔH for a reaction is equal to the sum of the heats of formation of the product compounds minus the sum of the heats of formation of the reactant compounds:
ΔH = Σ ΔHf products - Σ ΔHf reactants
Remember, the heat of formation of H+ is zero. The equation becomes:
ΔH = ΔHf Br-(aq) - ΔHf HBr(g)
The values for ΔHf may be found in the Heats of Formation of Compounds of Ions table. Plugging in these numbers:
Hydrogen peroxide decomposes according to the following thermochemical reaction:
H2O2(l) → H2O(l) + 1/2 O2(g); ΔH = -98.2 kJ
Calculate the change in enthalpy, ΔH, when 1.00 g of hydrogen peroxide decomposes.
Solution
The thermochemical equation tells us that ΔH for the decomposition of 1 mole of H2O2 is -98.2 kJ, so this relationship can be used as a conversion factor. Using the Periodic Table, the molecular mass of H2O2 is 34.0, which means that 1 mol H2O2 = 34.0 g H2O2.
Using these values:
ΔH = 1.00 g H2O2 x 1 mol H2O2 / 34.0 g H2O2 x -98.2 kJ / 1 mol H2O2
ΔH = -2.89 kJ
Answer
The change in enthalpy, ΔH, when 1.00 g of hydrogen peroxide decomposes = -2.89 kJ
Estimate the change in enthalpy, ΔH, for the following reaction:
H2 (g) + Cl2 (g) → 2 HCl (g)
Solution
To work this problem, think of the reaction in terms of simple steps:
Step 1 The reactant molecules, H2 and Cl2, break down into their atoms
H2(g) → 2 H(g) Cl2(g) → 2 Cl(g)
Step 2 These atoms combine to form HCl molecules
2 H (g) + 2 Cl (g) → 2 HCl (g)
In the first step, the H-H and Cl-Cl bonds are broken. In both cases, one mole of bonds is broken. When we look up the single bond energies for the H-H and Cl-Cl bonds, we find them to be +436 kJ/mol and + 243 kJ/mol, therefore for the first step of the reaction:
ΔH1 = +(436 kJ + 243 kJ) = +679 kJ
Bond breaking requires energy, so we expect the value for ΔH to be positive for this step. In the second step of the reaction, two moles of H-Cl bonds are formed. Bond breaking liberates energy, so we expect the ΔH for this portion of the reaction to have a negative value. Using the table, the single bond energy for one mole of H-Cl bonds is found to be 431 kJ:
Values of thermodynamic quantities are commonly expressed for standard state conditions, so it is a good idea to understand what the standard state conditions are.
A superscript circle is used to denote a thermodynamic quantity that is under standard state conditions:
ΔH = ΔH° ΔS = ΔS° ΔS = ΔS°
Standard State Conditions
Certain assumptions apply to standard state conditions. Standard temperature and pressure commonly is abbreviated as STP.
The standard state temperature is 25°C (298 K). It is possible to calculate standard state values for other temperatures.
All liquids are pure.
The concentration of all solutions is 1 M (1 molar).
All gases are pure.
All gases are at 1 atm pressure.
The energy of formation of an element in its normal state is defined as zero.
Thermochemical equations are just like other balanced equations except they also specify the heat flow for the reaction. The heat flow is listed to the right of the equation using the symbol ΔH. The most common units are kilojoules, kJ. Here are two thermochemical equations:
H2 (g) + ½ O2 (g) → H2O (l); ΔH = -285.8 kJ
HgO (s) → Hg (l) + ½ O2 (g); ΔH = +90.7 kJ
When you write thermochemical equations, be sure to keep the following points in mind:
Coefficients refer to the number of moles. Thus, for the first equation, -282.8 kJ is the ΔH when 1 mol of H2O (l) is formed from 1 mol H2 (g) and ½ mol O2.
Enthalpy changes for a phase change, so the enthalpy of a substance depends on whether is it is a solid, liquid, or gas. Be sure to specify the phase of the reactants and products using (s), (l), or (g) and be sure to look up the correct ΔH from heat of formation tables. The symbol (aq) is used for species in water (aqueous) solution.
The enthalpy of a substance depends upon temperature. Ideally, you should specify the temperature at which a reaction is carried out. When you look at a table of heats of formation, notice that the temperature of the ΔH is given. For homework problems, and unless otherwise specified, temperature is assumed to be 25°C. In the real world, temperature may different and thermochemical calculations can be more difficult.
Certain laws or rules apply when using thermochemical equations:
ΔH is directly proportional to the quantity of a substance that reacts or is produced by a reaction.
Enthalpy is directly proportional to mass. Therefore, if you double the coefficients in an equation, then the value of ΔH is multiplied by two. For example:
H2 (g) + ½ O2 (g) → H2O (l); ΔH = -285.8 kJ
2 H2 (g) + O2 (g) → 2 H2O (l); ΔH = -571.6 kJ
ΔH for a reaction is equal in magnitude but opposite in sign to ΔH for the reverse reaction.
For example:
HgO (s) → Hg (l) + ½ O2 (g); ΔH = +90.7 kJ
Hg (l) + ½ O2 (l) → HgO (s); ΔH = -90.7 kJ
This law is commonly applied to phase changes, although it is true when you reverse any thermochemical reaction.
ΔH is independent of the number of steps involved.
This rule is called Hess's Law. It states that ΔH for a reaction is the same whether it occurs in one step or in a series of steps. Another way to look at it is to remember that ΔH is a state property, so it must be independent of the path of a reaction.
If Reaction (1) + Reaction (2) = Reaction (3), then ΔH3 = ΔH1 + ΔH2
Polynucleotides, which are more often called nucleic acids, are polymeric glycosylamines in which the polymer backbone consists of sugars that are linked together by phosphodiester bonds. There are two types of nucleic acids, those in which the polymer backbone is composed of D-ribose units and those in which it is composed of D-2-deoxyribose units. Nucleic acids based on these two sugars are called ribonucleic acids (RNA) and deoxyribonucleic acids (DNA), respectively. Figure 1 identifies the essential structural features of the repeat unit of RNA.
Figure 1
The Repeat Unit of RNA
The D-ribose portion of the structure is shown in red. The blue B represents any of the four heterocyclic bases shown in Figure 2. The phospho diester group that links C3 of one D-ribose unit with C5 of the next is shown in green. The term nucleic acid reflects the fact that phosphoric acid, H3PO4, is a tribasic acid, i.e. it has three acidic OH groups. Two of them are involved in the formation of the phosphodiester bonds that link one sugar to the next. The third is available to act as an acid.
Figure 2
The Heterocyclic Bases in RNA
One of these four heterocyclic bases is attached to the anomeric carbon of each D-ribose unit in the polymer chain. Adenine and guanine are members of a class of heterocyclic bases called purines. Cytosine and uracil belong to the family called pyrimidines.
Exercise 1 Draw the structure of the two dinucleotidesthat contain adenine and cytosine. In DNA, the sugar is D-2-deoxyribose. Instead of having uracil as one of the four heterocyclic bases attached to each anomeric carbon of the sugar, DNA has thymine. Figure 3 presents the structure of D-2-deoxyribose and thymine for comparison to Figures 1 and 2.
Figure 3
The Structures of D-2-Deoxyribose and Thymine
Keto-Enol Tautomerization
Each of the 5 heterocyclic bases mentioned above contains an aromatic system of pi electrons, although it is not obvious in guanine, cytosine, uracil, and thymine because these bases exist as keto tautomers. Figure 4 shows the two tautomeric forms of thymine.
Figure 4
Tautomerization in Heterocyclic Bases
The aromatic six pi electron system is more obvious in the enol tautomer of thymine.
Exercise 2 Using water as both an acid and a base, write equations to show how the protons are transferred from the nitrogens to the oxygen atoms in thymine.
Exercise 3Ultraviolet radiation is known to damage DNA. One UV-induced reaction that has been implicated in such damage involves a 2+2 cycloaddition of a thymine molecule on one sugar with a second thymine on an adjacent sugar. The formation of this "cyclobutylthymine dimer" is thought to cause a local distortion of the shape of the DNA, thereby altering its activity. Draw the structure of the dimer that is formed by the 2+2 cycloaddition of the keto forms of two thymine molecules.
The Primary Structure of Nucleic Acids
The primary structure of nucleic acids describes the sequence of heterocyclic bases attached to the anomeric carbon of each sugar in the polymer backbone. A compact description of the primary sequence is afforded by using the 1-letter abbreviation of each of the heterocyclic bases. For example, AAGCUC describes an oligoribonucleotide in which the heterocylcic bases are adenine, adenine, guanine, cytosine, uracil, and cytosine. Figure 5 shows the structure of the compound.
Figure 5
The Structure of AAGCUC
The 6 letter string is read from the 5' end towards the 3' end of the chain. Note that since uracil occurs in RNA and thymine in DNA, the sequence AAUCGC identifies this nucleic acid chain as a ribonucleic acid.
The Secondary Structure of Nucleic Acids
It is hard to imagine anyone who hasn't heard of the a-helix of DNA. This is the dominant secondary structure of this polynucleotide. But the a-helix of DNA is, in fact, a double helix, i.e. one strand of DNA is wrapped around another. In proteins the association of two polypeptide chains is considered an aspect of quaternary structure, but in nucleic acids, the intertwining of two a-helices is regarded as part of the polymer's secondary structure.
Hydrogen Bonding: Inter-Chain Association
The elucidation of the structure of DNA required the efforts of many people and the interpretation of huge amounts of data. A key part of that data was the observation that the ratio of certain pairs of heterocyclic bases was nearly 1/1 regardless of the source of the DNA. For example, the A/T ratio in DNA obtained from wheat germ was 1.01/1, while that from human liver tissue was 1.00/1. The G/C ratio in DNA from these two species was 1.00/1 and 0.98/1, respectively. These ratios suggested a 1/1 correspondence between adenine bases on one chain of the double helix and thymine bases on the other. Similarly, it seemed reasonable to assume that each cytosine on one chain was associated with a guanine on the other. The "association" assumed was hydrogen bonding. Figure 6 shows idealized representations of the inter-chain H-bonding between A and T and G and C bases.
Figure 6
H-Bonding Between Heterocyclic Bases in DNA
Note that in both cases a purine base is associated with a pyrimidine base. Note, too, that the A/T base pair involves two H-bonding interactions, while the G/C pairing entails three.
If you would like to explore the structures of some DNA molecules, go to Molecules to Explore at USM's Biochemistry web site. There you will find four interactive structures of DNA molecules. For the brave of heart there is also a RasMol tutorial if you are interested in learning how to view all of the structural features of a protein or DNA molecule.
One of the most common reactions of carboxylic acids and related compounds is nucleophilic acyl substitution. Figure 1 depicts this transformation in general terms.
Figure 1
Nucleophilic Acyl Substitution
When Y represents the nitrogen atom of an amine, the transformation converts a carboxylic acid or a derivative of a carboxylic acid into an amide. Equation 1 provides a simple example.
As we have seen, extension of reaction 1 to a bifunctional acid chloride and a bifunctional amine may be used to prepare polyamides such as nylon:
An alternative method of making polyamides is available when an acyl and an amino group are part of the same molecule:
The latter approach is the one involved in the formation of polyamides known as polypeptides or proteins. This topic looks briefly at the chemistry of these biopolymers. Before we consider that chemistry, you should familiarize yourself with the chemistry of the bifunctional molecules from which these polymers are made, amino acids.
Peptides
Peptides are amides formed by the reaction of the a-amino group of one amino acid with the carboxylate group of another amino acid. Peptides which contain two amino acids are called dipeptides. Figure 2 shows the structure of a dipeptide formed from glycine and alanine.
Figure 2
Glycylalanine
Peptides made from three amino acids are called tripeptides, etc. Peptides containing from 2-10 amino acids are arbitrarily called oligopeptides. Compounds containing more amino acids are called polypeptides or proteins.
In order to simplify the representation of peptides, chemists assigned a 3-letter code to each amino acid. By this convention the dipeptide in Figure 6 would be Gly-Ala; the N-terminal amino acid is written first. Bradykinin, a nonapeptide which is involved in regulating blood pressure, has the structure Arg-Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg.
Peptide Structure
Peptides are structurally complex. However, two aspects of peptide structure make it easier to understand this complexity.
The H-N-C=O fragment of each peptide unit is planar.
The stereochemistry at each chiral a-carbon is the same, i.e. the R group always projects in the same direction.
Exercise 1 According to VSEPR theory, the geometry around the nitrogen atom of an amide should be approximately
trigonal planar pyramidal which means that the hybridization of the nitrogen atom should be approximately
sp2 sp3
Exercise 2 Resonance theory rationalizes the planarity of the amide group by suggesting that the lone pair of electrons on the nitrogen atom interacts with the pi system of the carbonyl group. Use curved arrows to depict this interaction. Draw the structure of the resonance contributor that is produced when the lone pair is fully delocalized onto the oxygen atom of the carbonyl group. Indicate all formal charges. What is it about this resonance contributor that is consistent with a planar fragment?
The structural features of proteins have been broken down into 4 categories, primary, secondary, tertiary, and quaternary structure.
Primary Structure
This is simple. The primary structure of a peptide is the sequence in which the amino acids are connected, starting with the N-terminal amino acid. In the case of bradykinin mentioned above, the primary structure is Arg-Pro-Pro-Gly-Phe-Ser-Pro-Phe-Arg. A Lewis structure of bradykinin is shown in Figure 3. Bradykinin is a nonapeptide. The 8 peptide bonds that connect the 9 amino acids are shown in red in Figure 3.
Figure 3
Bradykinin: A Nonapeptide
The presence of the substituents attached to the a-carbon can obscure the primary structure of even a simple peptide like bradykinin. Figure 4 shows the "backbone" of a nonapeptide stripped of its Ca substituents.
Figure 4
A Peptide Backbone
There are several features about this representation that are noteworthy. First, it represents an idealized conformation in which all the "backbone" atoms lie in the same plane. (The substituents attached to each a-carbon project in front of and behind that plane.) Second, if you "read" the structure from left-to-right, the carbonyl group of the each amide unit points in the opposite direction of the N-H bond of that unit. This is called the "all-trans" conformation. It's a fantasy. Doesn't happen. Third, since the bond between the carbonyl carbon and the nitrogen atom of each amide unit has significant double bond character, rotation about that bond is restricted. However, "free" rotation about the N-Ca bond as well as the C-Ca bond of the O=C-Ca group is possible.
Rotation around single bonds in alkanes is described in terms of a dihedral angle. In the case of peptides, this parameter is called the torsional angle. There are two torsional angles of interest in peptides. The first, designated F, defines the angle of rotation about the N-Ca bond, while the second, which is labeled Y, is the angle of rotation about the C-Ca bond. Figure 5 defines these angles for an "all-trans" segment of polypeptide.
Figure 5
Torsional Angles
Note that F and Y are the rotational angles around the two main-chain bonds to Ca. Rotation around these bonds gives rise to the secondary structure of proteins.
Secondary Structure
The primary structure of a peptide doesn't convey any information about the 3-dimensional shape of the peptide. The secondary structure does. In talking about secondary structure, we are actually referring to the topology of regions within the peptide. Such localized structure is a reflection of the conformations about a sequence of peptide bonds. These conformations are determined in large part by the nature of the R groups attached to the a carbon. Biochemists have identified three common topologies, helices, pleated sheets, and turns. Figure 7 presents an interactive model of Gly-Gly-Gly-Gly-Gly-Gly-Gly-Gly-Gly which clearly indicates the helical nature of this synthetic nonapeptide.
Tertiary Structure
The tertiary structure of a polypeptide describes the way in which the chain loops and twists and bends. While the primary structure usually depicts the backbone of a polypeptide as an extended chain in which the N-terminal amino acid is far away from the C-terminus, the fact is that the chain is not extended, and it is possible for the 5th amino acid to be spatially quite close to the 45th or the 450th, for example. Figure 6 shows the structure of one polypeptide chain from human insulin. This chain contains 51 amino acids.
Figure 6
Tertiary Strucutre in a Simple Protein
Quaternary Structure
One of the roles that proteins play is that of reaction catalyst. In this role proteins are more commonly referred to as enzymes. Most enzymes are comprised of two or more polypeptide chains that are held together by non-covalent forces. The individual polypeptide chains are called sub-units. Hemoglobin, for example, contains four sub-units, each of which is organized around a central iron atom. Human insulin is a hexamer, i.e. it contains 6 sub-units. (Note-When you click on the Human insulin link, you will open a file. Before viewing the file enter 1 into the text field that asks you how many models you want to display. Click OK. Click OK on the next window that appears. You should now see a wireframe model of human insulin.) The spatial relationship of one sub-unit to another is called the quaternary structure of a protein.
Protein Function
Proteins serve two main functions, structural and catalytic. Structural proteins are generally fibrous in nature. Hair and smooth muscle are examples of structural proteins. Catalytic proteins generally have a globular quaternary structure that is more or less spherical. The catalytic site is buried in the interior of the structure.
By far the most important polysaccharides are polymers of D-glucose. These materials are generally divided into two categories depending upon 1. the nature of the glycosidic bond that connects one monomeric unit to the next 2. whether the "backbone" of the polymer is branched or unbranched. In this topic we will focus on the most common polysaccharides, cellulose and starch. We will also look briefly at an interesting class of polysaccharides known as cyclodextrins.
Cellulose
Cellulose is a "straight chain" polymer of D-glucose in which the monomeric units are connected together by b-1,4 linkages. This means that the oxygen at C-4 of one D-glucose is connected to C-1 of another D-glucose, and that the C1-OC4 bond, the glycosidic linkage, occupies an equatorial position in the D-glucopyranose ring. This bond is highlighted in red in Figure 1.
Figure 1
A b 1,4 Linkage
Cellulose is not a single compound, but rather a mixture of polymers of D-glucose in which the chain length varies from one molecule to the next. A typical polymer chain may contain 10,000-15,000 D-glucose units. Because each monomer unit contains several OH groups, inter-chain hydrogen bonding is extensive. Figure 2 illustrates the secondary bonding interactions between polymer chains.
Figure 2
Between the Sheets
The figure shows fragments of four polymer chains, one pair of which lie in the roughly same plane, the second pair lying in a parallel plane. The dashed red lines indicate inter-chain H bonds for those molecules that lie in the same plane. The dashed blue lines suggest inter-chain H bonds between chains in different planes.
Starch
Starch is a mixed polysaccharide consisting of two main components, a-amylose and amylopectin. The primary difference between cellulose and a-amylose is the nature of the glycosidic linkage. In cellulose it is a b-1,4 linkage, while in a-amylose the D-glucose units are joined together by a-1,4 linkages. While this is a seemingly small difference, it has a major impact on the shapes and functions of these two polymers. Unlike cellulose, the polymer strands of a-amylose do not assemble in planar sheets. Rather, as Figure 3 suggests, they adopt a helical structure similar to that found in nucleic acids. Typical chain lengths for a-amylose are approximately 1000 monomer units.
Figure 3
Around the Bend
The second component of starch is amylopectin. Like a-amylose, the D-glucose units in amylopectin are connected by a-1,4 linkages. The major difference between a-amylose and amylopectin is that amylopectin is a branched polymer; at irregular intervals there are branch points where a secondary polysaccharide chain is connected to the main chain by a-1,6 linkages. In amylopectin the branches occur, on average, every 24-30 D-glucose units along the main chain. Starch is the form in which plants store excess D-glucose. In animals it is stored as glycogen, which is similar in structure to amylopectin, except that it is more highly branched. Typically there is an a-1,6 linkage to a side chain every 8-12 D-glucose units along the main chain. Figure 4 offers a generic structure for amylopectin and glycogen. Two side chains are shown in color.
Figure 4
Energy Storage in Plants and Animals
The highest concentrations of glycogen are found in muscle and liver cells.
Exercise 1 Classify the glycosidic linkage highlighted in red an blue in the following polysaccharide fragment:
The bond highlighted in red isa-1,2 b-1,2 a-1,3 b-1,3 a-1,4 b-1,4 a-1,6 b-1,6
The bond highlighted in blue isa-1,2 b-1,2 a-1,3 b-1,3 a-1,4 b-1,4 a-1,6 b-1,6
Cyclodextrins
Linking D-gluocse units together with a-1,4 linkages means that the growth of the polysaccharide follows a helical path. Occassionally, this coiling brings the D-glucose at the end of the growing polymer chain close enough to the one at the beginning that a glycosidic bond can form between them, thereby creating a cyclic polysaccharide. These structures are known as cyclodextrins. Figure 5 presents the structure of one such compound which contains a ring comprised of eight D-glucose units. This compound is known as g-cyclodextrin.
Figure 5
A Cyclodextrin Molecule
Cyclodextrins are natural products formed by the action of enzymes called cycloglucosyltransferases, CGTases, on starch. These enzymes are found in a microorganism called Bacillus macerans. Cyclodextrins participate in host-guest interactions, serving as hosts for a variety of small molecules. The number of monomer units in the macrocyclic ring determines the size of the cavity the host makes available to the guest. The ability of cyclodextrins to "encapsulate" small molecules has led to the development of a number of interesting applications. They have been used to
enhance the chromatographic separation of chiral drugs
protect the active ingredients in perfumes
increase the solubility of antineoplastic drugs for use in chemotherapy
separate cholesterol from dairy products
In Figure 5 you are looking down on the cavity from above. Figure 6 presents a perspective drawing of the 3-dimensional structure of g-cyclodextrin. Note that the polar OH groups project to the exterior of the structure while the hydrogens attached to the glucose units point into the cavity. Thus the interior is comparatively non-polar. These structural features make the polymer water soluble while still able to transport non-polar materials such as cholesterol.
One of the characteristic reactions of carboxylic acids is esterification, a process that involves the condensationof a carboxylic acid with an alcohol. This transformation is depicted in general terms in Equation 1. The word condensation refers to the fact that the formation of the ester is accompanied by the formation of a molecule of water. In order to drive equilibrium 1 to the right the water is distilled and the vapors are condensed in a receiving flask. A more general definition of a condensation reaction is one in which the formation of the desired product is accompanied by the formation of another small molecule such as H2O, HCl, or NH3.
Reaction 1proceeds by a mechanism known as nucleophilic acyl substitution. As Equation 2 indicates, the process may be extended to the formation of polymers by the simple expedient of reacting a dicarboxylic acid with a diol.
Polymers that are formed by this pathway are commonly referred to as condensation polymers. In this topic we will examine the formation of two main classes of condensation polymers, polyesters, and polyamides.
Condensation Polymers
Polyesters
One of the most ubiquitous polymers in our society today is poly(ethylene terephthalate), PET. It is used to make fibers for clothing as well as containers for carbonated beverages. When used to make clothing it's known as dacron. Figure 1 shows the repeat unit of this material.
Figure 1
Anyone Care for a Soda?
Exercise 1 Draw the structures of the dicarboxylic acid and the diol from which PET is made.
Carboxylic acids are easily converted into acyl chlorides as shown in Equation 2.
Since acyl chlorides undergo facile nucleophilic acyl substitution reactions, many polyesters are prepared by treatment of a diacyl chloride with a diol. Figure 2 shows the repeat unit of the polycarbonate ester Lexan™ which is formed by the reaction of the diacid chloride of carbonic acid (H2CO3) and a diphenol.
Figure 2
Let's Repeat
Exercise 2Draw a valid Lewis structure for carbonic acid.
Exercise 3 Draw the structures of the diacid chloride and the diphenol from which Lexan™ is made.
Exercise 4 The diphenol referred to above is prepared by the acid-catalysed reaction of phenol with acetone. What type of reaction is this? Write an equation for it.
Polyurethanes
Isocyanates, R-N=C=O, react with alcohols to form compounds known as urethanes. Equation 3 describes the reaction in general terms.
While this process is not technically a nucleophilic acyl substitution reaction, it is treated as such because isocyanates are derivatives of carbonic acid.
Exercise 5 Using curved arrows to depict the movement of electrons, show how reaction 3 occurs.
If water is used instead of an alcohol, the product, a carbamic acid, spontaneously decomposes into an amine and carbon dioxide as shown in Equation 4.
Extension of the chemistry described in Equation 3 to a diisocyanate and a diol provides a route to polyurethanes as shown in Equation 5.
If a small amount of water is added to the diisocyanate/diol mix, some of the diisocyanate undergoes a reaction analogous to that shown in Equation 4 and the CO2 that is liberated becomes entrapped in the polymer. This is the basis of polyurethane foams that are use as insulating materials.
Spandex™, a stretchable fabric made by DuPont, is an interesting example of a "mixed" polyurethane. As outlined in Figure 3, it is prepared in a 2-stage process, the first stage involving the condensation of two moles of methylene diisocyanate with one mole of poly(butylene oxide). This forms an intermediate known as a prepolymer. Treatment of this material with ethylenediamine yields the final product.
Figure 3
STRETCH
Exercise 6 Draw the structure of the product of the following reaction: Polyamides
In the same way that they react with diols to produce polyesters, diacyl chlorides condense with diamines form polyamides. Commercially these compounds are more commonly called nylon. Equation 6 describes the process in general terms, while Figure 3 shows the repeat units of two common forms of nylon.
Figure 3
Commercially Important Nylons
Nylon 6,6 is used to make fibers that are used to make carpets and fabrics. Kevlar is also spun into fibers which are woven into fabrics that are used in bullet-proof vests, as well as the cloth that is used to make high-end canoes, kayaks, and sailboats. It is also found in fishing poles and tennis racquets.
Natural fibers such as hair, silk, and wool are polyamides, too. We will discuss these materials more in the topic Biopolymers.
