CARA MEMBUAT PROBIOTIK BONGGOL PISANG

Cara membuat probiotik dengan bonggol pisang adalah sebagai berikut :

Pastikan bahwa pohon pisang tsb benar dalam keadaan sehat tidak terkena suatu penyakit apapun
Cari pohon pisang yg hidup berdekatan dengan pohon bambu
Ambil bonggol pisang dari pohon yg sudah berbuah tapi belum mati
Bonggol pisang tidak boleh disimpan lebih dari semalam dari penebangan
Pengambilan bonggol pada pagi atau sore hari
Kupas bonggol pisang (3 kg) sampai bersih, cuci dengan air sumur
Cacah halus kalau bisa di blender tapi jgn terlalu halus, asal hancur saja
Didihkan air sebanyak 5 liter, kemudian masukan gula merah 1 kg dalam air mendidih tersebut. Didihkan selama 30 menit
Dinginkan ... harus benar2 dingin
Sterilkan wadah (jerigen/botol aqua galon .. apa saja) bisa dgn disiram air panas, atau semprot dengan alkohol 70% sebelumnya
Campurkan bonggol pisang yg sudah dicacah/blender dengan larutan gula kemudian masukan dalam wadah yg sdh di sterilkan tadi
Tutup rapat wadah dan simpan minimal selama 15 hari baiknya 1 bulan
Jika ada indikasi timbul gas sebaiknya pakai aerator sederhana dari slang dan botol berisi air

selamat mencoba....

Absolute Zero

Question: What Is Absolute Zero?

Answer: Absolute zero is the point where no more heat can be removed from a system, according to the absolute or thermodynamic temperature scale. This corresponds to 0 K or -273.15°C. In classical kinetic theory, there should be no movement of individual molecules at absolute zero, but experimental evidences shows this isn't the case.

Temperature is used to describe how hot or cold an object it. The temperature of an object depends on how fast its atoms and molecules oscillate. At absolute zero, these oscillations are the slowest they can possibly be. Even at absolute zero, the motion doesn't completely stop.

It's not possible to reach absolute zero, though scientists have approached it. The NIST achieved a record cold temperature of 700 nK (billionths of a Kelvin) in 1994. MIT researchers set a new record of 0.45 nK in 2003.

Single Bond Energies Table

Knowing the values for bond energy helps us to predict whether a reaction will be exothermic or endothermic. For example, if the bonds in the product molecules are stronger than the bonds of the reactant molecules, then the products are more stable and have a lower energy than the reactants, and the reaction is exothermic. If the reverse is true, then energy (heat) must be absorbed in order for the reaction to occur, making the reaction endothermic. In this case, the products have a higher energy than the reactants. Bond energies may be used to calculate change in enthalpy, ΔH, for a reaction by applying Hess's Law. ΔH can be obtained from the bond energies only when all of the reactants and products are gases.

Single Bond Energies (kJ/mol) at 25°C
	H	C	N	O	S	F	Cl	Br	I
H	436	414	389	464	339	565	431	368	297
C		347	293	351	259	485	331	276	238
N			159	222	—	272	201	243	—
O				138	—	184	205	201	201
S					226	285	255	213	—
F						153	255	255	—
Cl							243	218	209
Br								193	180
I									151

Nuclear Reactions - Mass - Energy Relations

The energy change in a nuclear reaction is considerably greater than that of a normal chemical reaction. This change can be calculated using Einstein's equation:

ΔE = Δmc²

where ΔE is the change in energy, Δm is the change in mass (mass of products - mass of reactants), and c is the speed of light (3.00 x 10⁸ m/s).

As written, this relation gives the energy change in joules and the mass change in kilograms. Usually small quantities of a sample decay, and the energy change is very large, so it's more common to get an energy change in kilojoules (kJ) corresponding to a mass change in grams. Using the relations

1 kJ = 10³ J and 1 kg = 10³ g

Einstein's equation may be rewritten

ΔE (in kJ) = 9.00 x 10¹⁰ Δm (in grams)

For example, to calculate the ΔE in kJ for the radioactive decay of radium:

²²⁶₈₈Ra --> ²²²₈₆Rn + ⁴₂He

when one mole of radium decays, we first calculate Äm for the reaction and then obtain ΔE using the equation.

Δm = mass of 1 mol ⁴₂He + mass of 1 mol ²²²₈₆Rn - mass of 1 mol ²²⁶₈₈Ra

Δm = 4.0015 g + 221.9703 g - 225.9771 g

Δm = -0.0053 g

Note that Δm may be an extremely small quantity, so it is important to know the masses of products and reactants with a high degree of accuracy in order to know the mass difference to two significant figures.

ΔE (in kJ) = 9.00 x 10¹⁰ x (-0.0053)

ΔE = -4.8 x 10⁸kJ

ΔE in kJ when one gram of radium (one mole weighs 226 g) decays would be:

ΔE = 1 g Ra x (-4.8 x 10⁸ kJ)/226 g Ra

ΔE = -2.1 x 10⁶ kJ

Nernst Equation

The Nernst equation is used to calculate the voltage of an electrochemical cell or to find the concentration of one of the components of the cell. Here is a look at the Nernst equation and an example of how to apply it to solve a problem.

The Nernst Equation

E_cell = E⁰_cell - (RT/nF)lnQ

E_cell = cell potential under nonstandard conditions (V)
E⁰_cell = cell potential under standard conditions
R = gas constant, which is 8.31 (volt-coulomb)/(mol-K)
T = temperature (K)
n = number of moles of electrons exchanged in the electrochemical reaction (mol)
F = Faraday's constant, 96500 coulombs/mol
Q = reaction quotient, which is the equilibrium expression with initial concentrations rather than equilibrium concentrations

Sometimes it is helpful to express the Nernst equation differently:

E_cell = E⁰_cell - (2.303*RT/nF)logQ

at 298K, E_cell = E⁰_cell - (0.0591 V/n)log Q

Nernst Equation Example

A zinc electrode is submerged in an acidic 0.80 M Zn²⁺ solution which is connected by a salt bridge to a 1.30 M Ag⁺ solution containing a silver electrode. Determine the initial voltage of the cell at 298K.

Unless you've done some serious memorizing, you'll need to consult the standard reduction potential table, which will give you the following information:

E⁰_red: Zn²⁺_aq + 2e^- → Zn_s = -0.76 V

E⁰_red: Ag⁺_aq + e^- → Ag_s = +0.80 V

E_cell = E⁰_cell - (0.0591 V/n)log Q

Q = [Zn²⁺]/[Ag⁺]²

The reaction proceeds spontaneously so E⁰ is positive. The only way for that to occur is if Zn is oxidized (+0.76 V) and silver is reduced (+0.80 V). Once you realize that, you can write the balanced chemical equation for the cell reaction and can calculate E⁰:

Zn_s → Zn²⁺_aq + 2e^- and E⁰_ox = +0.76 V

2Ag⁺_aq + 2e^- → 2Ag_s and E⁰_red = +0.80 V

which are added together to yield:

Zn_s + 2Ag⁺_aq → Zn²⁺_a + 2Ag_s with E⁰ = 1.56 V

Now, applying the Nernst equation:

Q = (0.80)/(1.30)²

Q = (0.80)/(1.69)

Q = 0.47

E = 1.56 V - (0.0591 / 2)log(0.47)

E = 1.57 V

Measurement of Heat Flow & Enthalpy Change

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·&degC).

For example, consider a chemical reaction which occurs in 200 grams of water with an initial temperature of 25.0&degC. 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&degC. The heat flow is calculated:

q_water = 4.18 J/(g·&degC) x 200 g x (31.0&degC - 25.0&degC)

q_water = +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:

DH_reaction = -(q_water)

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; q_water 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:

q_reaction = - (q_water + q_bomb)

where q_water = 4.18 J/(g·&degC) x m_water 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:

q_bomb = 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.

Laws of Thermochemistry

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:

1. 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.

2. 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.

3. 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:

1. 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

2. 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.

3. 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

Thermochemistry Table for Common Compounds

The molar heat of formation of a compound (ΔH_f) 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.

Cations and Anions in Aqueous Solution

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.

Endothermic Reaction Demonstration

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)
• 125-ml flask
• stirring rod

Perform the Demonstration

• 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.

Hot Ice or Sodium Acetate

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.

Sodium Acetate or Hot Ice Materials

• 1 liter clear vinegar (weak acetic acid)
• 4 tablespoons baking soda (sodium bicarbonate)

Prepare the Sodium Acetate or Hot Ice

1. 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

2. 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.

3. 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.

4. 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.

Create an Exothermic Chemical Reaction

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:

1. 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.
2. Remove the thermometer from the jar (if you didn't already in Step 1).
3. Soak a piece of steel wool in vinegar for 1 minute.
4. Squeeze the excess vinegar out of the steel wool.
5. Wrap the wool aroung the thermometer and place the wool/thermometer in the jar, sealing the lid.
6. Allow 5 minutes, then read the temperature and compare it with the first reading.
7. Chemistry is Fun!

Tips:

1. 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.
2. 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.
3. 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).

What You Need:

• Thermometer
• Jar with Lid
• Steel Wool
• Vinegar

Endothermic and Exothermic Reactions

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:

sunlight + 6CO₂(g) + H₂O(l) = C₆H₁₂O₆(aq) + 6O₂(g)

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:

• Safe Endothermic Reaction
• Safe Exothermic Reaction
• Hot Ice - Exothermic Reaction

Endothermic Chemical Reaction

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:

1. Pour the citric acid solution in a styrofoam coffee cup. Use a thermometer or other temperature probe to record the initial temperature.
2. Stir in the baking soda (sodium bicarbonate). Track the change in temperature as a function of time.
3. The reaction is: H₃C₆H₅O₇(aq) + 3 NaHCO₃(s) --> 3 CO₂(g) + 3 H₂O(l) + NaC₆H₅O₇(aq)
4. When you have completed your demonstration or experiment, simply wash the cup out in a sink. No toxic chemicals to mess with!

Tips:

1. Feel free to vary the concentration of the citric acid solution or the quantity of sodium bicarbonate.
2. 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.

What You Need:

• 25 ml citric acid soln
• 15 g baking soda
• styrofoam cup
• thermometer
• stirring rod

Heat of Formation Worked Example Problem

1. Heat of Formation Problem

Calculate ΔH for the following reaction:

8 Al(s) + 3 Fe₃O₄(s) --> 4 Al₂O₃(s) + 9 Fe(s)

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 ΔH_f products - Σ ΔH_f reactants

Omitting terms for the elements, the equation becomes:

ΔH = 4 ΔH_f Al₂O₃(s) - 3 ΔH_f Fe₃O₄(s)

The values for ΔH_f 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 ΔH_f products - Σ ΔH_f reactants

Remember, the heat of formation of H⁺ is zero. The equation becomes:

ΔH = ΔH_f Br^-(aq) - ΔH_f HBr(g)

The values for ΔH_f may be found in the Heats of Formation of Compounds of Ions table. Plugging in these numbers:

ΔH = -120.9 kJ - (-36.2 kJ)

ΔH = -84.7 kJ

Answer

ΔH = -84.7 kJ

Enthalpy Change Example Problem

• Problem
  
  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

Bond Energies and Enthalpy Change Example Chemistry Problem

Problem

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:

ΔH₂ = -2(431 kJ) = -862 kJ

By applying Hess's Law, ΔH = ΔH₁ + ΔH₂

ΔH = +679 kJ - 862 kJ
ΔH = -183 kJ

Answer

The enthalpy change for the reaction will be ΔH = -183 kJ.

Standard State Conditions - Standard Temperature and Pressure

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.

Laws of Thermochemistry

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:

1. 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.

2. 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.

3. 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:

1. 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

2. 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.

3. 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