--%>
Assignment Help - homework help

Describe First Order Rate Equation

The integrated forms of the first order rate equations are conveniently used to compare concentration time results with this rate equation.

Rate equations show the dependence of the rate of the reaction on concentration can be integrated to give expressions for the dependence of the concentrations on time. We generally use the integrated rate equation that is obtained to deduce the order of a reaction.

A first order reaction is one for which, at a given temperature, the rate of the reaction depends only on the first power of the concentration of a single reacting species. If the concentrations of this species is represented by c (for solutions, the units of moles per litre are ordinarily used), and if the volume of the system remains essentially constant during the course of the reaction, the first order rate equation can be written

-dc/dt = kc

The rate of constant k is then a positive quantity and has the units of the reciprocal of time.

Integrated rate equation: the experimental results obtained in a study of the rate of a reaction are usually values of c or some related to c at various times. Such data can best be compared with the integrated form of the first order rate equation. If the concentration at time t = 0 is c0, and if at some later time t the concentration has fallen to c, the integration gives

481_First order reactions.png 

With -In (c/c0) = In (c0/c), the integration can be written as

In c0/c = kt

Sometimes a more convenient form is

In c = -kt + In c0

A reaction can therefore be said to be first order if a plot of In (c0/c) or In c versus t gives a straight line. If a straight line is obtained, the slope of the line can be used to give the value of the rate constant k. an alternative to this graphical procedure is the calculation of a value of kfrom the individual measurements of c at the various times t, for example. The reaction is classified as first order if all the data lead to essentially the same values for k, that is, if it is satisfies with k as a constant.

Example: the rate of conversion of tert-butyl bromide to tert-butyl alcohol, (CH3)3CBr + H2O 1598_First order reactions1.png (CH3)3COH + HBr, has been studied and some concentration-time results are found in table given below. Verify that the reaction is first order, and deduce the values of the rate constant at the two temperatures.

Solution: from the data at each temperature we calculate In (c0/c) values. Then the graphical display shows a plot of In (c0/c) versus t is constructed. The straight lines, each going through the origin show that at both temperatures the data conform to the integrated first order relation. The slopes give the values of the rate constants 

K = 0.00082 min-1 = 0.137 × 10-4 s-1 [25°C]    

K = 0.0142 min-1 = 2.37 × 10-4 s-1 [50°C]    

Concentration of tert butyl bromide as a function of time for the reaction (CH3)3CBr + H2O 1598_First order reactions1.png (CH3)3COH + HBr in a 10% water, 90% acetone solvent

At 25°C  At 50°C
Time, h (CH3)3CBr, Mol L-1 Time, min (CH3)3CBr, Mol L-1
0 0.1039 0 0.1056
3.15 0.0896 9 0.0961
6.20 0.0776 18 0.0856
10.0 0.0639 27 0.0767
13.5 0.0529 40 0.0645
18.3 0.0353 54 0.0536
26.0 0.0270 72 0.0432
30.8 0.0207 105 0.0270
37.3 0.0142 135 0.0174
43.8 0.0101 180 0.0089

   Related Questions in Chemistry

  • Q : Molarity of the final mixture Can

    Can someone please help me in getting through this problem. Two solutions of a substance (that is, non electrolyte) are mixed in the given manner 480 ml of 1.5M first solution + 520 ml of 1.2M second solution. Determine the molarity of the final mixture

    		•
  • Q : Meaning of Molar solution Molar

    Molar solution signifies 1 mole of solute present/existed in: (i) 1000g of solvent (ii) 1 litre of solvent (iii) 1 litre of solution (iv) 1000g of solution

    		•
  • Q : Calculation of molecular weight Provide

    Provide solution of this question. In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol. mass = 58) at 298K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of ace

    		•
  • Q : How can enzymes act as catalyst?

    Enzymes are complex proteinous substances, produced by living bodies, such as act as catalysis in the physiological reactions. The enzymes are, also called biochemical catalysts and the phenomenon is known as bio-chemical catalysis because numerous reactions that occur the bodies of animals and p

    		•
  • Q : Mole fraction in vapours Choose the

    Choose the right answer from following. If two substances A and B have P0A P0B= 1:2 and have mole fraction in solution 1 : 2 then mole fraction of A in vapours: (a) 0.33 (b) 0.25 (c) 0.52 (d) 0.2

    		•
  • Q : Explain Photoelectron Spectroscopy. The

    The energies of both the outer and inner orbitals of atoms and molecules can be determined by photoelectron spectroscopy.Energy changes of the outermost or highest energy electron of molecules were dealt with here in a different passion. The energies of ot

    		•
  • Q : Dipole attractions for london dispersion

    Illustrate how are dipole attractions London dispersion forces and hydrogen bonding similar?

    		•
  • Q : Crystals of covalent compounds Crystals

    Crystals of the covalent compounds always contain:(i) Atoms as their structural units  (ii) Molecules as structural units  (iii) Ions held altogether by electrostatic forces (iv) High melting pointsAnswer: (i)

    		•
  • Q : Lab question Explain how dissolving the

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid, establishes a buffer with a pH of approximately 5.

    		•
  • Q : What is Flash Photolysis Reactions.

    An example illustrates the type of mechanism that can be written to explain the development of flash photolysis reactions. Often, as the reactions in the ozone layer of the earth's atmosphere, we are interested in the kinetic behavior of species that are not a

    		•