Physical Properties of Organic Compounds

Boiling Point and Vapor Pressure

The transition between liquid and gas is very important for organic liquids.  It is useful for purification through distillation, and is responsible for evaporation.  Finally, without some vapor pressure, no compound could reach your nose to produce those wonderful or terrible smells we associate with organic chemistry.
The vapor pressure is the actual pressure of vapor over a liquid.  Measurable at all temperatures, it is smaller at lower temperatures, and increases with higher temperatures.  Eventually, the vapor pressure will meet or exceed atmospheric pressure, and bubbles will form in the liquid—a process we call “boiling.”  The temperature at which the vapor pressure equals atmospheric pressure is the official boiling point.
Let’s think about the physical process involved in boiling.  The figure below shows molecules (represented by “X”) in the liquid on the left, and in the gas on the right:
Physical process:


X--X is composite of polar, H-bonding, and Van der Waals forces.
The big change on boiling is that the intermolecular forces (X---X) are broken, and not replaced by anything.  [Note, however, that the molecule X stays intact, and the covalent bonds are not broken.]  This means that the stronger the intermolecular forces (X—X) are, the harder it will be to disrupt them.  This will yield a lower vapor pressure at a given temperature, and higher boiling point.  These intermolecular forces will be a composite of polar, H-bonding, and Van der Waals forces.

Predicting Boiling Points

Even for professional chemists, it is very difficult to predict the boiling point of a single compound.  However, it is not too difficult in simple cases to predict the relative boiling points of two compounds (that is, which compound boils at the higher temperature). 

Boiling points of homologous series of compounds with 8 functional groups.  Note the trend towards higher boiling points as compounds get longer within a series, and the four classes of functional groups by strength of intermolecular interaction.

A quick look at  shows that there are four classes of functional groups:  Acids have the highest boiling points at a given size, followed by alcohols, the the group containing ketones, amines and esters, and finally, much lower, are alkanes, alkenes, and alkynes.  This may be easily related to the intermolecular forces available to the different molecules.  Acids have very strong, double hydrogen bonding in the liquid phase which must be disrupted to produce the gas phase.  Liquid phase alcohols have hydrogen bonding as well, but not a strong as acids.  The polar functional groups (ketones, etc.) have dipolar interactions, but no hydrogen bonding, making their intermolecular forces of intermediate strength.  Finally, the hydrocarbons (alkanes, etc.) have only weak van der Waals interactions to hold them together, leading to low boiling points.
Within a series, there is a steady increase in boiling points as compounds get larger.  This is due to an increase in surface area, and therefore in van der Waals interactions.  The addition of one or two carbons will not cause much of a rise in boiling point, but it will be noticeable and can add up with large changes in size.  Note also that the lines get somewhat closer as the molecules get larger, because as one adds more alkane content to the molecule, the molecules increasingly resemble alkanes.
Another important factor in boiling point prediction relates to the difference between branched and linear carbon skeletons.  Since branching allows the compound to be more compact, it reduces surface area, reducing the van der Waals interactions, and therefore reducing the boiling point. Note the effects of branching on the following three alcohols:

If you look around the organic literature, you see almost nothing about boiling points of ionic compounds.  There is a very good reason:  they are held together so tightly that they are destroyed before you can get them to boil.  For example, while acetic acid has a boiling point of a little over 100°C, sodium acetate has a melting point of 285-6°, and a boiling point much higher than that (can’t be measured due to decomposition, no doubt).  In the same vein, most ionic compounds have essentially no vapor pressure.

Kinetics

Kinetics is the area of chemistry concerned with reaction rates. The rate can be expressed as:
rate = change in substance/time for change to occur (usually in M/s) There are several factors that determine the rate of a specific reaction and those are expressed in the "collision theory" that states that for molecules to react, they must:

1. collide
2. have the right energy
3. have the right geometry

To increase the rate, you must make the above more likely to occur. This is possible by changing other factors such as:

• increasing the surface area (of solids)-this allows for more collisions and gives more molecules the right geometry
• increasing the temperature-this gives more molecules the right energy (also called the activation energy, E_a)
  
  
• increasing the concentration (of gases and solutions)-this allows for more collisions and more correct geometry
• using a catalyst-helps molecules achieve the correct geometry by providing a different way to react
  
  

The reaction rate can also be expressed by using a "rate law" and is written as follows:
For the general reaction: aA + bB + ... ----> gG + hH + ..... the reaction rate can be calculated by:
Reaction rate = k[A]^m[B]ⁿ .... Where:
[A], [B], etc. are the concentrations of the reactants
k is the rate constant or rate coefficient, a value dependent on temperature.
m,n, etc. are exponents that correspond to a, b, etc. The concentration is raised to the power of its coefficient in the balanced equation.
Reaction order is a topic that comes with reaction rates. If you have a reaction in that A, B, and C are possible reactants, then we can describe the order of the reaction following this chart:
The order of the reaction is defined as the sum of the exponents of the coefficients. In general, first order reactions are most commonly seen, but reactions of other orders are also important. Zero-order reactions -- those for that the change in the reaction is independent of the concentration of the reaction -- are also possible.
It is possible to determine the order of a reactant, and eventually the reaction rate, using initial rate information that includes the concentration of the reactants and the rate at that the product is formed. If you double the concentration of reactant X and the rate increases by 2^a, then the order of reactant X is "a". If you triple the concentration of reactant Y and the rate increases by 3^b, then the order of Y is "b". For example, if you have a reaction with one reactant, A, and you double [A] and the rate doubles, then the rate=k[A]¹. If, instead, you double [A] and the rate quadruples, the rate=k[A]². If you double [A] and the rate stays the same, then the rate=k[A]⁰.
To find the rate constant, k, using initial rate information, just plug in from the experiment one of the concentrations and rate into the rate law and solve. The units of k are trickier:
units of k=units of rate/(units of concentration)^{reaction order}
Ex: for 2nd order reaction, k=(M/s)/M²=M^-1s^-1 Example Problem: Find the rate law and rate constant of A + B --> C using the following data

When dealing with reaction rates, it is sometimes important to know how to graph a straight line with the data you have. When graphing concentration versus time, there are two ways to graph a line. If you have a first order reaction, then the graph of ln[A] vs. time is a line. If you have a second order reaction, then the graph of 1/[A] vs. time produces a line.
A quantitative way to examine reaction rates is through Arrhenius Equation that states:
k=Ae^-E_a/RT Where:
A is a constant related to the geometry needed
e is a constant, approximately 2.7281
E_a is the activation energy
R is the gas law constant, 8.314 J/mol-K
T is the temperature in kelvins
If it is a simple geometry to attain, A will be large. If a large E_a is needed then the exponent becomes more negative and therefore decreases k. If the temperature increases then the exponent becomes less negative and therefore increases k. A pop-up calculator is available to help practice using Arrhernius' Equation to make calculations.
The following are two (2) energy profile graphs that help demonstrate energy changes during a reaction.
Not all reactions happen exactly as they are written. Most, in fact, go through an intermediate step. Reaction mechanism studies look at how a reaction actually occurs. Defined, a reaction mechanism is a series of elementary reactions that are proposed to account for the rate law (kinetics) of a particular reaction. The diagram below shows the two steps involved in a particular mechanism, and it shows how we get the reaction from the mechanism.
It is helpful to remember certain terms and facts when dealing with mechanisms. You cannot derive a mechanism from the equation and when you combine the steps of a mechanism, you end up with the reaction. The molecularity of a step tells how many molecules are involved (most involve two (2) molecules so they are bimolecular). An intermediate product is a molecule formed in one step and then used in another. One of the most important concepts to keep in mind is that the steps are not equally important. To speed up the reaction, you must speed up the slowest step (also called the rate-determining step).
rate law of slow step=rate law of reaction When determining the rate of a step, simply make the exponent of the reactant's concentration in the rate law the same as the coefficient of the reactant in the step.
Example problem: Find the slow step of the following reaction mechanism
A catalyst is a substance added to a reaction that comes out of the reaction unchanged. As mentioned earlier, catalysts help lower the activation energy as shown in the following graph. They do this by changing the reaction mechanism.

The following is an example of how this can be done.

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Practice Kinetic Quantitative Problem:

Kinetic quantitative solution. Practice Kinetic Qualitative Problems:

Kinetic qualitative solution. [Advanced Index] [Gas Laws] [Thermodynamics] [Kinetics] [Equilibria]
[Redox Reactions] [Nuclear Chemistry]

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Chemical Thermodynamics

The scientific discipline that intersects the areas of chemistry and physic is commonly known as physical chemistry, and it is in that area that a thorough study of thermodynamics takes place. Physics concerns itself heavily with the mechanics of events in nature. Certainly changes in energy -- however measured, whether it be heat, light, work, etc. -- are clearly physical events that also have a chemical nature to them. Thermodynamics is the study of energy changes accompanying physical and chemical changes. The term itself clearly suggests what is happening -- "thermo", from temperature, meaning energy, and "dynamics", which means the change over time. Thermodynamics can be roughly encapsulated with these topics:

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• Heat and Work
• Energy
• Enthalpy
• Entropy
• Gibbs Free Energy

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Heat and Work

Heat and work are both forms of energy. They are also related forms, in that one can be transformed into the other. Heat energy (such as steam engines) can be used to do work (such as pushing a train down the track). Work can be transformed into heat, such as might be experienced by rubbing your hands together to warm them up. Work and heat can both be described using the same unit of measure. Sometimes the calorie is the unit of measure, and refers to the amount of heat required to raise one (1) gram of water one (1) degree Celsius. Heat energy is measured in kilocalories, or 1000 calories. Typically, we use the SI units of Joules (J) and kilojoules (kJ). One calorie of heat is equivalent to 4.187 J. You will also encounter the term specific heat, the heat required to raise one (1) gram of a material one (1) degree Celsius. Specific heat, given by the symbol "C", is generally defined as:

C =

q MΔT

Where:
C = specific heat in cal/g-°C
q = heat added in calories,
m = mass in grams
ΔT = rise in temperature of the material in °C.
The value of C for water is 1.00 cal/g-°C.
The values for specific heat that are reported in the literature are usually listed at a specific pressure and/or volume, and you need to pay attention to these settings when using values from textbooks in problems or computer models.
Example Problem: If a 2.34 g substance at 22°C with a specific heat of 3.88 cal/g-°C is heated with 124 cal of energy, what is the new temperature of the substance?

Answer:

ΔT = q MC

ΔT = (124) (2.34)(3.88) = 13.7°C

new T = 22 + 13.7 = 35.7°C

Two other common heat variables are the heat of fusion and the heat of vaporization. Heat of fusion is the heat required to melt a substance at its melting temperature, while the heat of vaporization is the heat required to evaporate the substance at its boiling point.
Chemical work is primarily related to that of expansion. In physics, work is defined as:
w = d × f Where:
w = work, in joules (N×m) (or calories, but we are using primarily SI units)
d = distance in meters
f = opposing force in Newtons (kg*m/s²)
In chemical reactions, work is generally defined as :
w = distance × (area × pressure) The value of distance times area is actually the volume. If we imagine a reaction taking place in a container of some volume, we measure work by pressure times the change in volume.
w = ΔV × P Where:
ΔV is the change in volume, in liters
If ΔV=0, then no work is done.
Example Problem: Calculate the work that must be done at standard temperature and pressure (STP is 0°C and 1 atm) to make room for the products of the octane combustion:
2 C₈H₁₈ + 25 O₂ --> 16 CO₂ + 18 H₂O

Answer:

Knowing that 25 moles of gas are replaced by 34 moles of gas in this reaction, we can   calculate a net increase of 9 moles of gas. Knowing the molar volume of an ideal gas at   STP (22.4 L/mol), the change in volume and the work of expansion can be calculated   dV = 9 moles ∗ 22.4 L/mol = 202 L   The external pressure is 1.0 atm (standard pressure), so the work required is:   w = dV ∗ P = 202 L ∗ 1.00 atm = 202 l-atm   Using the conversion factor of 1 L-atm = 101 J, the amount of work in joules is:   w = 202 L-atm ∗ 101 j/L-atm = 2000 J, or 2kJ of energy

Energy

You might remember the first law of thermodynamics: energy cannot be created or destroyed. Energy can only change form. Chemically, that usually means energy is converted to work, energy in the form of heat moves from one place to another, or energy is stored up in the constituent chemicals. You have seen how to calculate work. Heat is defined as that energy that is transferred as a result of a temperature difference between a system and its surroundings. Mathematically, we can look at the change in energy of a system as being a function of both heat and work: ΔE = q - w Where:
ΔE is the change in internal energy of a system
q is the heat flowing into the system
w is the work being done by the system
If q is positive, we say that the reaction is endothermic, that is, heat flows into the reaction from the outside surroundings. If q is negative, then the reaction is exothermic, that is, heat is given off to the external surroundings.
You might also remember the terms kinetic energy and potential energy. Kinetic energy is the energy of motion -- the amount of energy in an object that is moving. Potential energy is stationary, stored energy. If you think of a ball sitting on the edge of a table, it has potential energy in the energy possible if it falls off the table. Potential energy can be transformed into kinetic energy if and when the ball actually rolls off the table and is in motion. The total energy of the system is defined as the sum of kinetic and potential energies.
In descriptions of the energy of a system, you will also see the phrase "state properties". A state property is a quantity whose value is independent of the past history of the substance. Typical state properties are altitude, pressure, volume, temperature, and internal energy.

Enthalpy

Enthalpy is an interesting concept: it is defined by its change rather than a single entity. A state property, the word enthalpy comes from the Greek "heat inside". If you have a chemical system that undergoes some kind of change but has a fixed volume, the heat output is equal to the change in internal energy (q = ΔE). We will define the enthalpy change, ΔH, of a system as being equal to its heat output at constant pressure:
dH = q at constant pressure Where:
ΔH = change in enthalpy
We define enthalpy itself as:
H = E + PV Where:
H = enthalpy
E = energy of the system
PV = pressure in atm times volume in liters
You will not need to be able to calculate the enthalpy directly; in chemistry, we are only interested in the change in enthalpy, or ΔH.
ΔH = H_final - H_initial or ΔH = H(products) - H(reactants) Tables of enthalpies are generally given as ΔH values. Example Problem: Calculate the ΔH value of the reaction:

HCl + NH₃ → NH₄Cl (ΔH values for HCl is -92.30; NH₃ is -80.29; NH₄Cl is -314.4)

Answer:

ΔH = ΔHproducts - ΔHreactants

ΔHproducts = -314.4

ΔHreactants = -92.30 + (-80.29) = -172.59

ΔH = -314.4 - 172.59 = 141.8

We can also represent enthalpy change with the equation:
ΔH = ΔE + P ΔV Where:
ΔV is the change in volume, in liters
P is the constant pressure
If you recall, work is defined as P ΔV, so enthalpy changes are simply a reflection of the amount of energy change (energy going in or out, endothermic or exothermic), and the amount of work being done by the reaction. For example, if ΔE = -100 kJ in a certain combustion reaction, but 10 kJ of work needs to be done to make room for the products, the change in enthalpy is:
ΔH = -100 kJ + 10 kJ = -90 kJ This is an exothermic reaction (which is expected with combustion), and 90 kJ of energy is released to the environment. Basically, you get warmer. Notice the convention used here -- a negative value represents energy coming out of the system.
You can also determine ΔH for a reaction based on bond dissociation energies. Breaking bonds requires energy while forming bonds releases energy. In a given equation, you must determine what kinds of bonds are broken and what kind of bonds are formed. Use this information to calculate the amount of energy used to break bonds and the amount used to form bonds. If you subtract the amount to break bonds from the amount to form bonds, you will have the ΔH for the reaction.
Example Problem: Calculate ΔH for the reaction:

N₂ + 3H₂ → 2NH₃ (The bond dissociation energy for N-N is 163 kJ/mol; H-H is 436 kJ/mol; N-H is 391 kJ/mol)

Answer:

ΔH = ΔHproducts - ΔHreactants

To use the bond dissociation energies, we must determine how many bonds   are in the products and the reactants. In NH3 there are 3 N-H bonds so in 2 NH3   there are 6 N-H bonds. In N2 there is 1 N-N bond and in 3H2 there are 3 H-H bonds.

ΔHproducts = 6(391) = 2346

ΔHreactants = 163 + 3(436) = 1471

ΔH = 2346 - 1471 = 875

Entropy

Gas Laws

Gases behave differently from the other two commonly studied states of matter, solids and liquids, so we have different methods for treating and understanding how gases behave under certain conditions. Gases, unlike solids and liquids, have neither fixed volume nor shape. They are molded entirely by the container in which they are held. We have three variables by which we measure gases: pressure, volume, and temperature. Pressure is measured as force per area. The standard SI unit for pressure is the pascal (Pa). However, atmospheres (atm) and several other units are commonly used. The table below shows the conversions between these units.

Units of Pressure
1 pascal (Pa)	1 N*m-2 = 1 kg*m-1*s-2
1 atmosphere (atm)	1.01325*105 Pa
1 atmosphere (atm)	760 torr
1 bar	105 Pa

Volume is related between all gases by Avogadro's hypothesis, which states: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. From this, we derive the molar volume of a gas (volume/moles of gas). This value, at 1 atm, and 0° C is shown below.

Vm =

V n

= 22.4 L at 0°C and 1 atm

Where:
V_m = molar volume, in liters, the volume that one mole of gas occupies under those conditions
V=volume in liters
n=moles of gas
An equation that chemists call the Ideal Gas Law, shown below, relates the volume, temperature, and pressure of a gas, considering the amount of gas present.
PV = nRT Where:
P=pressure in atm
T=temperature in Kelvins
R is the molar gas constant, where R=0.082058 L atm mol^-1 K^-1.
The Ideal Gas Law assumes several factors about the molecules of gas. The volume of the molecules is considered negligible compared to the volume of the container in which they are held. We also assume that gas molecules move randomly, and collide in completely elastic collisions. Attractive and repulsive forces between the molecules are therefore considered negligible.
Example Problem: A gas exerts a pressure of 0.892 atm in a 5.00 L container at 15°C. The density of the gas is 1.22 g/L. What is the molecular mass of the gas?

Answer:

PV = nRT

T = 273 + 15 = 228

(0.892)(5.00) = n(.0821)(288)

n = 0.189 mol

.189 mol 5.00L x x grams 1 mol = 1.22 g/L

x = Molecular Weight = 32.3 g/mol

We can also use the Ideal Gas Law to quantitatively determine how changing the pressure, temperature, volume, and number of moles of substance affects the system. Because the gas constant, R, is the same for all gases in any situation, if you solve for R in the Ideal Gas Law and then set two Gas Laws equal to one another, you have the Combined Gas Law:

P1V1 n1T1

=

P2V2 n2T2

Where: values with a subscript of "1" refer to initial conditions
values with a subscript of "2" refer to final conditions
If you know the initial conditions of a system and want to determine the new pressure after you increase the volume while keeping the numbers of moles and the temperature the same, plug in all of the values you know and then simply solve for the unknown value.
Example Problem: A 25.0 mL sample of gas is enclosed in a flask at 22°C. If the flask was placed in an ice bath at 0°C, what would the new gas volume be if the pressure is held constant?

Answer:

Because the pressure and the number of moles are held constant, we do not   need to represent them in the equation because their values will cancel. So the   combined gas law equation becomes:

V1 T1 = V2 T2

25.0 mL 295 K = V2 273 K

V2 = 23.1 mL

We can apply the Ideal Gas Law to solve several problems. Thus far, we have considered only gases of one substance, pure gases. We also understand what happens when several substances are mixed in one container. According to Dalton's law of partial pressures, we know that the total pressure exerted on a container by several different gases, is equal to the sum of the pressures exerted on the container by each gas.
P_t = P₁ + P₂ + P₃ + ... Where:
P_t=total pressure
P₁=partial pressure of gas "1"
P₂=partial pressure of gas "2"
and so on
Using the Ideal Gas Law, and comparing the pressure of one gas to the total pressure, we solve for the mole fraction.

P1 Pt

=

n2 RT/V nt RT/V

=

n1 nt

= X1

Where:
X₁ = mole fraction of gas "1"
And discover that the partial pressure of each the gas in the mixture is equal to the total pressure multiplied by the mole fraction.

P1 =

n1 nt

Pt = X1Pt

Example Problem: A 10.73 g sample of PCl₅ is placed in a 4.00 L flask at 200°C.
a) What is the initial pressure of the flask before any reaction takes place?
b) PCl₅ dissociates according to the equation: PCl₅(g) --> PCl₃(g) + Cl₂(g). If half of the total number of moles of PCl₅(g) dissociates and the observed pressure is 1.25 atm, what is the partial pressure of Cl₂(g)?

Answer:
a) 10.73 g PCl5 x 1 mol 208.5 g = 0.05146 mol PCl5
PV = nRT
T = 273 + 200 = 473
P(4.00) = (.05146)(.0821)(473)
P = 0.4996 atm
b)	PCl5	→	PCl3	+	Cl2
Start:	.05146 mol		0 mol		0 mol
Change:	-.02573 mol		+.02573 mol		+.02573 mol
Final:	.02573 mol		.02573 mol		.02573 mol
XCl2 = nCl2 ntotal = PCl2 Ptotal
PCl2 1.25 atm = .02573 mol .07719 mol
PCl2 = .4167 atm

As we stated earlier, the shape of a gas is determined entirely by the container in which the gas is held. Sometimes, however, the container may have small holes, or leaks. Molecules will flow out of these leaks, in a process called effusion. Because massive molecules travel slower than lighter molecules, the rate of effusion is specific to each particular gas. We use Graham's law to represent the relationship between rates of effusion for two different molecules. This relationship is equal to the square-root of the inverse of the molecular masses of the two substances.

r1 r2

=

μ1 μ1

Where:
r₁=rate of effusion in molecules per unit time of gas "1"
r₂=rate of effusion in molecules per unit time of gas "2"
u₁=molecular mass of gas "1"
u₂=molecular mass of gas "2"
Previously, we considered only ideal gases, those that fit the assumptions of the ideal gas law. Gases, however, are never perfectly in the ideal state. All atoms of every gas have mass and volume. When pressure is low and temperature is low, gases behave similarly to gases in the ideal state. When pressure and temperature increase, gases deviate farther from the ideal state. We have to assume new standards, and consider new variables to account for these changes. A common equation used to better represent a gas that is not near ideal conditions is the van der Waals equation, seen below.

P +

n2a V2

V n

- b

= RT

Where the van der Waals constants are:
a accounts for molecular attraction
b accounts for volume of molecules
The table below shows values for a and b of several different compounds and elements.

Species	a (dm6 bar mol-2)	b (dm3 mol-1)
Helium	0.034598	0.023733
Hydrogen	0.24646	0.026665
Nitrogen	1.3661	0.038577
Oxygen	1.3820	0.031860
Benzene	18.876	0.11974

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Practice Ideal Gas Law Problem:
2.00 g of hydrogen gas and 19.2 g of oxygen gas are placed in a 100.0 L container. These gases react to form H₂O(g). The temperature is 38°C at the end of the reaction.
a) What is the pressure at the conclusion of the reaction?
b) If the temperature was raised to 77° C, what would the new pressure be in the same container? Ideal gas law solution. Practice Pressure Problem:
1 mole of oxygen gas and 2 moles of ammonia are placed in a container and allowed to react at 850°C according to the equation:

4 NH₃(g) + 5 O₂(g) --> 4 NO(g) + 6 H₂O(g)
a) If the total pressure in the container is 5.00 atm, what are the partial pressures for the three gases remaining?
b) Using Graham's Law, what is the ratio of the effusion rates of NH₃(g) to O₂(g)?
Pressure solution.

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Compressibility and Ideal Gas Approxim

Accelerate Rusting

Purpose

To demonstrate how a chemical reaction from vinegar, water, and bleach can accelerate the rusting process of steel.

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Additional information

Rust is a series of red oxides that form from the reaction of iron with the presence of oxygen and water. The rusting process is electrochemical, meaning it forms from a chemical reaction which takes place in a solution where there is an electron transfer between the electrode and the electrolyte or species in solution. In the case of rusting, the transfer of electrons from iron to oxygen begins the electrochemical process. Electrolytes accelerate the corrosion process when combined with water, similar to the rusting that occurs on vehicles due to road salt in winter conditions. 
 Read more at :

Properties of Water

Water has a number of unusual properites which come about because the atoms that comprise a water molecule form apolar covalent bonds. The main properties of water that are important for living things are the following:
Ice floats. Usually when a liquid freezes the solid ismore dense than the liquid. In water thesolidphase is less dense than the liquid phase under normal conditions. If ice sank, ponds and lakes would freeze solid to the bottom making things very difficult for fish and other creatures to obtain food and oxygen in the winter.
Water is a great solvent especially for polar molecules of ionically bonded substances like salt. That's because the charged ends of the water molecules are attracted to the charges on other molecules. Water molecules will surround soluable molecules or ions in a hydration shell. This helps keep the molecules in solution.
Many gases dissolve in water because the the combination of attractive and repulsive forces among water molecules leaves space for the gas molecules that in most other liquids would not be there. The two gasses we are most concerned about dissolving are carbon dioxide and oxygen.
Water has a high latent heat. Water can store more heat per volume than just about any common substance and releases the heat slowly. This helps moderate the climate.
Has a high heat of vaporization. It takes a lot of heat to break the hydrogen bonds in water and when they braek, the heat energy is carried off as water vapor. This makes water useful as a coolent for large critters like mammals.
Water is cohesive. Water molecules tend to stick together, up to a point. Many insects take advantage of the surface tension brought about at the surface of lakes and pond by cohesion to walk on the surface of the water.
Water is adhesive. Sicks to surfaces with polar groups such as cellulose.
Water disassociates to form hydrogen ions(H+) and hydroxyl (OH-) in small quanitities. This is important for living things because the number of ions in the solutions inside organisms affects the shape and functioning of many of the molecules. The concentration of hydrogen ions in a solution is so important that scientists use a special scale, the pH scale to measure this concentration

The pH Scale

The concentration of hydrogen ions in a solution is very important for living things. This is because, since the hydrogen ions are positively charged they alter the charge environment of other molecules in solution. By putting different forces on the molecules, the molecules change shape from their normal shape. This is particularly important for proteins in solution because the shape of a protein is related to its function.
The concentration of hydrogen ions is commonly expressed in terms of the pH scale. Low pH corresponds to high hydrogen ion concentration and vice versa. A substance that when added to water increases the concentration of hydrogen ions(lowers the pH) is called an acid. A substance that reduces the concentration of hydrogen ions(raises the pH) is called a base. Finally some substances enable solutions to resist pH changes when an acid or base is added. Such substances are called buffers. Buffers are very important in helping organisms maintain a relatively constant pH.
Study the pH chart given below carefully. Note that each decrease in pH by one pH unit means a tenfold increase in the concentration of hydrogen ions.

Hydrophobic Interactions

When you add some drops of oil to water, the drops combine to form a larger drop. This comes about because water molecules are attracted to each other and are cohesive because they are polar molecules. Oil molecules are non polar and thus have no charged regions on them. This means that they are neither repelled or attracted to each other. The attractiveness of the water molecules for each other then has the effect of squeezing the oil drops together to form a larger drop.
Since it looks like the oil molecules are avoiding the water, this type of interaction is called a hydrophobic interaction. Previous Page
Hydropholbic interactions along with hydrophilic interactions help to determine the three dimensional shape of biologically important molecules and structures such as proteins and cell membranes.
 

Hydrophilic Interactions

Interactions between water and other molecules such that the other molecules are attracted to water are called hydrophilic interactions. Previous Page
Molecules that have charged parts to them are attracted to the charges within the water molecule. This is an important reason why water is such a good solvent. So for instance the picture below shows a glucose molecule in solution. Water molecules surrounding the glucose molecule are shown in blue for clarity. 
Glucose molecules have polar hydroxyl(OH) groups in them and these attract the water to them. When sugar is in a crystal the molecules are attracted to the water and go into solution. Once in solution the molecules stay in solution at least in part because they become surrounded by water molecules. This layer of water molecules surrounding another molecule is called a hydration shell. 

Chemical Bonds.

A chemical bond is an attraction between atoms brought about by a sharing of electrons between to atoms or a complete transfer of electrons. There are three types of chemical bonds: Ionic, Covalent and Polar covalent.  In addition chemists often recognise another type of bond called a hydrogen bond.

Ionic Bonds.  Ionic bonds arise from elements with low electronegativity(almost empty outer shells) reacting with elements with high electronegativity (mostly full outer shells). In this case there is a complete transfer of electrons.  Previous Page A well known example is table salt, sodium chloride. Sodium gives up its one outer shell electron completely to chlorine which needs only one electron to fill its shell. Thus, the attraction between these atoms is much like static electricity since opposite charges attract. Ionic, Covalent, Polar covalent.

 

 Covalent bond.

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Covalent bonds involve a complete sharing of electrons and occurrs most commonly between atoms that have partially filled outer shells or energy levels. Thus if the atoms are similar in negativity then the electrons will be shared. Carbon forms covalent bonds. The electrons are in hybrid orbitals formed by the atoms involved as in this example: ethane. Diamond is strong because it involves a vast network of covalent bonds between the carbon atoms in the diamond. Previous Page

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Hydrogen Bond

The fact that the oxygen end of a water molecule is negatively charged and the hydrogen end positively charged means that the hydrogens of one water molecule attract the oxygen of its neighbor and vice versa. This is because unlike charges attract. This largely electrostatic attraction is called a hydrogen bond and is important in determining many important properties of water that make it such an important liquid for living things. Water can also form this type of bond with other polar molecules or ions such as hydrogen or sodium ions. Further, hydrogen bonds can occurr within and between other molecules. For instance, the two strands of a DNA molecule are held together by hydrogen bonds. Hygrogen bonding between water molecules and the amino acids of proteins are involved in maintaining the protein's proper shape. Previous Page
 

This picture represents a small group of water molecules. Hydrogen bonds between unlike charges are shown as lines without arrows on the ends. The double arrowed lines represent the fact that like charges repell each other. Both hydrogen bonds and the repelling forces balance each other and are both are important in determining the properties of water. Page Top