RSS

Monthly Archives: August 2012

Using Chemical Equations in Calculations

Using Chemical Equations in Calculations

 

The following sections are concerned with the amounts of substances which participate in chemical reactions, the quantities of heat given off or absorbed when reactions occur, and the volumes of solutions which react exactly with one another. These seemingly unrelated subjects are discussed together because many of the calculations involving them are almost identical in form. The same is true of the density calculations and of the calculations involving molar mass and the Avogadro constant. In each case one quantity is defined as the ratio of two others.

TABLE 1 Summary of Related Quantities and Conversion Factors.

Related Quantities Conversion Factor Definition Road Map
Volume ↔ mass Density, ρ \rho=\frac{m}{V} V\text{ }\overset{\rho }{\longleftrightarrow}\text{ }m
Amount of substance ↔ mass Molar Mass, M M=\frac{m}{n} n\text{ }\overset{M}{\longleftrightarrow}\text{ }m
Amount of substance ↔ number of particles Avogadro constant, NA N_{\text{A}}=\frac{N}{n} n\text{ }\overset{N_{\text{A}}}{\longleftrightarrow}\text{ }N
Amount of X consumed or produced ↔ amount of Y consumed or produced Stoichiometric ratio, S(Y/X) S\text{(Y/X)}=\frac{n_{\text{Y}}}{n_{\text{X}}} n_{\text{X}}\text{ }\overset{S\text{(Y/X)}}{\longleftrightarrow}\text{ }n_{\text{Y}}
Amount of X consumed or produced ↔ quantity of heat absorbed during reaction ΔHm for thermochemical equation \Delta H_{\text{m}}=\frac{q}{n_{\text{X}}} n_{\text{X}}\text{ }\overset{\Delta H_{m}}{\longleftrightarrow}\text{ }q
Volume of solution ↔ amount of solute Concentration of solute, cX c_{\text{X}}=\frac{n_{\text{X}}}{V} V\text{ }\overset{c_{\text{X}}}{\longleftrightarrow}\text{ }n_{\text{X}}

 

The first quantity serves as a conversion factor relating the other two. A summary of the relationships and conversion factors we have encountered so far is given in Table 1.

An incredible variety of problems can he solved using the conversion factors in Table 1. Sometimes only one factor is needed, but quite often several are applied in sequence, as in Example 3 in Titrations. In solving such problems, it is necessary first to think your way through, perhaps by writing down a road map showing the relationships among the quantities given in the problem. Then you can apply conversion factors, making sure that the units cancel, and calculate the result.

The examples in these sections should give you some indication of the broad applications of the problem-solving techniques we have developed here. Once you have mastered these techniques, you will be able to do a great many useful computations which are related to problems in the chemical laboratory, in everyday life, and in the general environment. You will find that the same type of calculations, or more complicated problems based on them, will be encountered again and again throughout your study of chemistry and other sciences.

\underset{\text{Gasoline}}{\mathop \text{2C}_{\text{8}}\text{H}_{\text{12}}}\,\text{ + }\underset{\text{Air}}{\mathop \text{25O}_{\text{2}}}\,\text{ }\to \text{ }\underset{\begin{smallmatrix} 
 \text{Carbon} \\ 
 \text{dioxide} 
\end{smallmatrix}}{\mathop \text{16CO}_{\text{2}}}\,\text{ + }\underset{\text{Water}}{\mathop \text{18H}_{\text{2}}\text{O}}\,

There are a great many circumstances in which you may need to use a balanced equation. For example, you might want to know how much air pollution would occur when 100 metric tons of coal were burned in an electric power plant, how much heat could be obtained from a kilogram of natural gas, or how much vitamin C is really present in a 300-mg tablet. In each instance someone else would probably have determined what reaction takes place, but you would need to use the balanced equation to get the desired information.

 
Leave a comment

Posted by on August 2, 2012 in Uncategorized

 

CHEMICAL BONDING

Chemical Bonding

We have covered the basic ideas of atomic structure, but it is worth realizing that of all the elements only the noble gases are found naturally in a form such that their atoms occur as individuals, widely separated from all other atoms. Under the conditions that prevail on the surface of the earth, almost all atoms are linked by chemical bonds to other atoms. Oxygen, for example, is the most common element on earth. It is found in combination with metals in rocks, with hydrogen in water, with carbon and hydrogen in living organisms, or as the diatomic molecule O2 in the atmosphere, but individual oxygen atoms are quite rare. Most other elements behave in a similar way. Thus, if we want to understand the chemistry of everyday matter, we need to understand the nature of the chemical bonds which hold atoms together.

Theories of chemical bonding invariably involve electrons. When one atom approaches another, the valence electrons, found in the outermost regions of the atoms, interact long before the nuclei can come close together. Electrons are the least massive components of an atom, and so they can relocate to produce electrostatic forces which hold atoms together. According to Coulomb’s law, such electrostatic or coulombic forces are quite large when charges are separated by distances of a few hundred picometers—the size of an atom. Coulombic forces, then, are quite capable of explaining the strengths of the bonds by which atoms are held together.

An important piece of evidence relating electrons and chemical bonding was noted by G. N. Lewis shortly after the discovery that the atomic number indicated how many electrons were present in each kind of atom. Most chemical formulas correspond to an even number of electrons summed over all constituent atoms. Thus H2O has 2 electrons from the 2 H’s and 8 from O for a total of 10, NCl3 has 7 + (3 × 17) = 58 electrons, and so on. This is a bit surprising when you consider that half the elements have odd atomic numbers so that their atoms have an odd number of electrons. Lewis suggested that when atoms are bonded together, the electrons occur in pairs, thus accounting for the predominance of even numbers of electrons in chemical formulas. These pairs often leave bonded atoms with eight electrons in the outer most shell, known as the octet rule.

There are two important ways in which the valence shells of different atoms can interact to produce electron pairs and chemical bonds. When two atoms have quite different degrees of attraction for their outermost electrons, one or more electrons may transfer their allegiance from one atom to another, pairing with electrons already present on the second atom. The atom to which electrons are transferred will acquire excess negative charge, becoming a negative ion, while the atom which loses electrons will become a positive ion. These oppositely charged ions will be held together by the coulombic forces of attraction between them, forming an ionic bond. Since electrons are shifted so that one neutral atom becomes positively charged, and the other becomes negatively charged, substances formed from ionic bonding are often in pairs and are then referred to as binary ionic compounds. Binary ionic compounds are not too common, but the existence of polyatomic ions greatly extends the number of ionic substances. Because oppositely charged ions are held in crystal lattices by strong coulombic forces, physical properties of ionic compounds include hardness, brittleness, and having high melting and boiling points. The majority of them dissolve in water, and in solution each ion exhibits its own chemical properties. Ionic compounds obey the octet rule, which explains why ions are generally in noble-gas electron configurations.

On the other hand, when two atoms have the same degree of attraction for their valence electrons, it is possible for them to share pairs of electrons in the region between their nuclei. Such shared pairs of electrons attract both nuclei, holding them together with a covalent bond. Sharing one or more pairs of electrons between two atoms attracts the nuclei together and usually results in an octet around each atom. Covalent bonding often produces individual molecules, like CO2 or CH3CH2OH, which have no net electrical charge and little attraction for each other. Thus covalent substances often have low melting and boiling points and are liquids or gases at room temperature. Occasionally, as in the case of SiO2, an extended network of covalent bonds is required to satisfy the octet rule. Such giant molecules result in solid compounds with high melting points.
A number of atomic properties, such as ionization energy, electron affinity, van der Waals radii, covalent atomic radii, and ionic radii, are important in determining whether certain elements will form covalent or ionic compounds and what properties those compounds will have. In the next sections we will consider the formation of covalent and ionic bonds and the properties of some substances containing each type of bond. The figure below previews how each of atomic properties important to understanding bonding varies according to an atom’s position in the periodic table. The figure also includes one atomic property, electronegativity, which will be covered when we explore further aspects of covalent bonding in more depth.

Periodic variations of atomic properties.
 
Leave a comment

Posted by on August 2, 2012 in physical chemistry

 

Tags:

Integrated Rate Law

Integrated Rate Law

For a reaction where the chemical equation has the form

A   ->     products

and the rate law is assumed to be of the form

Rate = – d[A]/dt = k [A]m

where m is the order of the reaction with respect to substance A. Three important cases can be treated: m = 0, m = 1, and m =2. These are called zeroth order, first order, and second order, respectively. The method for finding the rate law involves assuming that the reaction is first order, graphing the concentration-time data in a way that should give a straight line, and rejecting the hypothesis that the reaction is first order if the graph is not linear. Then the process is repeated for second order and zeroth order.

First-Order Reaction, m = 1

In this case the rate law is

– d[A]/dt = k [A].

This equation can be rearranged to

d[A]/[A] = – k t

and then integrated using the methods of the calculus (if you know how to integrate, you can verify the result below). The integration gives

ln[A]t = – k t + ln[A]0

y       =  m x   +    b

where ln represents taking a natural logarithm, t represents a particular time after the reaction began, and [A]t and [A]0 represent the concentration of A at time t and time zero. (It is assumed that the reaction is timed from whenever the reactants were mixed, so time zero, t = 0, is the start of the reaction.) This equation is in the form y = m x + b that represents a straight line. Thus, if the reaction is first order, a graph of the natural logarithm of the concentration of A (on the y axis) versus the time from the start of the reaction (on the x axis) should be a straight line. (If this graph is not straight, the reaction is not first order; perhaps it is second order or zeroth order.) If the graph is linear, then the slope of the graph is – k, the negative of the rate constant, and the intercept is [A]0, the initial concentration of substance A.

Second-Order Reaction, m = 2

For the same reaction, A  ->  products, but where the rate depends on the square of the concentration of A, the rate law becomes

Rate = – d[A]/dt =  k [A]2

As was done with the first-order rate equation, this one can be rearranged and integrated. The result is

1/[A]t = k t + 1/[A]0

This equation is also in the form of a straight line, but this time the slope is k, the rate constant, and the intercept is the reciprocal of the initial concentration of substance A. If a plot of the reciprocal of the concentration of A on the vertical axis versus time on the horizontal axis gives a straight line, then it is reasonable to conclude that the reaction is second order. The rate constant can be calculated from the slope.

Zeroth-Order Reaction, m = 0

For a few reactions,  A   ->  products, the rate does not depend on the concentration of the reactant, A. In such a case the rate is always equal to the rate constant, as shown in the next equation

Rate = – d[A]/dt = k [A]0k

This equation says that the slope of a graph of [A]t on the y-axis and t on the x-axis will be constant (equal to the rate constant, k). A constant slope means that a graph of [A]t versus t will be a straight line. You could also integrate this equation to give

[A]t  =  – k t  +  [A]0

This makes it quite clear that the slope of such a graph is the negative of the rate constant.

To review, the process for finding the order of the reaction (and hence the rate law) and the rate constant involves making up to three graphs. As soon as one of the graphs is found to be a straight line, the order of the reaction is known and the rate constant can be calculated from the slope of the straight line. The units of the rate constant are obtained from the units of the quantities plotted on the graph. For example, for the first-order plot, the y-axis is ln[A]t and the x-axis is t. A logarithm is a number and has no units; t has units of time, usually seconds. The slope is calculated as Δyx so there is a pure number in the numerator and time units in the denominator. This gives the rate constant units of s-1, reciprocal seconds.

Calculating Concentrations from Rate Laws

Once a rate law has been determined for a reaction, it is possible to use the integrated rate equation to calculate the concentration of a reactant or product given the initial concentration of reactant and the time elapsed. For example, in an earlier section the rate law and rate constant were determined for decomposition of a dye at 80 °C. The rate law found was first order

Rate = k [dye]                        k = 0.073 s-1 (at 80 °C)

so the integrated rate law has this form

ln[dye]t  =  – k t  +  ln[A]0            or             ln([dye]t /[dye]0)  =  – k t

Suppose that you started with 0.100 M dye and allowed it to decompose as described earlier. What concentration of dye would be left after 25 s? You can calculate the concentration of dye after 25 s by substituting into the integrated rate equation:

[dye]t /(0.100 M) = exp(- 0.073 s-1 × 25 s)  and [dye]t  = 0.016 M

You can also use the integrated rate law to determine how long it would take for the concentration of the dye to drop to half its initial value, as

ln([dye]t /[dye]0)  =  – k t            and            ln{(0.050 M)/(0.100 M) = – 0.073 s-1 × t

This gives t = 9.5 s, which is in agreement with the figure showing that the concentration had dropped from 1.0 M to 0.49 M (a little less than half the starting concentration)in 10 s.

The time for the concentration to drop to half its initial value is called the half-life and is represented by t1/2 . For a first-order reaction the half-life is the same regardless of the initial concentration, but this is not true for zeroth-order or second-order reactions. The half-life of a first-order reaction is related to the first-order rate constant and either can be determined if the other is known. The relationship is

ln[A]t1/2  –  ln[A]0  =  – k t1/2

ln{[A]0/2} –  ln[A]0  =  ln[A]0  –  ln2  –  ln[A]0  =  – ln2  =  – k t1/2

t1/2  = ( – ln2)/(- k)  =  0.693/k

 
Leave a comment

Posted by on August 2, 2012 in Uncategorized

 

Reactions in Aqueous Solutions

Reactions in Aqueous Solutions

we emphasized the importance of liquid solutions as a medium for chemical reactions. Water is by far the most important liquid solvent, partly because it is plentiful and partly because of its unique properties. In your body, in other living systems, and in the outside environment a tremendous number of reactions take place in aqueous solutions. Consequently this section, as well as significant portions of many subsequent sections, will be devoted to developing an understanding of reactions which occur in water. Since ionic compounds and polar covalent compounds constitute the main classes which are appreciably soluble in water, reactions in aqueous solutions usually involve these types of substances.

Reactions in aqueous solutions usually involve ionic or polar covalent compounds. The solubilities of such compounds are enhanced because of their interactions with water molecules, especially the hydration of ions. Measurement of the electrical current conducted by a solution of known concentration enables us to determine whether a solute is an electrolyte, and, if so, of what type. Water itself is an extremely weak electrolyte, producing hydronium ions and hydroxide ions, each at a concentration of 1.00 × 10–7 mol dm–3 at 25°C.

There are three important classes of reactions which occur in aqueous solution: precipitation reactions, acid-base reactions, and redox reactions. Whether or not a precipitate will form when solutions of about 0.1 M concentration are mixed can be predicted using the solubility rules. Precipitation reactions are useful for detecting the presence of various ions and for determining the concentrations of solutions.

Acid base reactions and redox reactions are similar in that something is being transferred from one species to another. Acid-base reactions involve proton transfers, while redox reactions involve electron transfers. Redox reactions are somewhat more complicated, though, because proton transfers and other bond-making and bond-breaking processes occur at the same time as electron transfer. Both proton transfer and electron transfer can be broken into half-equations: one to describe donation of a proton (or electrons), and one to describe acceptance of a proton (or electrons). For acid-base reactions, each half-equation involves a conjugate acid-base pair. For oxidation reduction reactions, each half-equation involves a redox couple.

Both conjugate acid-base pairs and redox couples may be tabulated in order of acidic strength, basic strength, oxidizing power and reducing power. Such behavior is also quantified in terms of acid constants, base constants and reduction potentials. In each case such tabulations may be used to determine whether a given reaction will go to completion, occur only to a limited extent, or not occur at all.

 
Leave a comment

Posted by on August 2, 2012 in Uncategorized

 

Tags:

CHEMBOND JUL 2012 RJ COLLEGE

 
Leave a comment

Posted by on August 1, 2012 in Uncategorized

 
 
DRUG REGULATORY AFFAIRS INTERNATIONAL

Drug Regulatory affairs by DR ANTHONY MELVIN CRASTO, Worlddrugtracker

Seven Spheres

Aqua Terra Ignis et Aer

Chemtips

Because Organic Chemistry is Hard Enough

Fathisaffar

A Blog by Mohamed Fathi SAFFAR Pharm.D;Docteur es-sciences Pharmaceutiques;GMP-Inspector; Director at NATIONAL LABORATORY FOR DRUG CONTROL

%d bloggers like this: