--%>
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 : What are condensation polymers? Give

    These types of polymers are formed as a result of condensation reaction between monomer units. Some common examples are being discussed here: 1. Polyesters 2047_condensat</span></p> </div> <!-- /comment-box --> </li> </td> </tr><tr> <td> <li> <div class=

    Q : Problem based on molarity Select the

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

    		•
  • Q : What is Henry law constant and its

    1. The units of Henry Law constant are same as those of pressure, i.e. torr or h bar. 2. Different gases have dissimilar values of Henry law constant. The values of KH for some gases in water are given in tabl

    		•
  • Q : Effect of addition of mercuric iodide

    Give me answer of this question. When mercuric iodide is added to the aqueous solution of potassium iodide, the:(a) Freezing point is raised (b) Freezing point is lowered (c) Freezing point does not change (d) Boiling point does not change

    		•
  • Q : Explain the process of adsorption of

    The extent of adsorption of a gas on a solid adsorbent is affected by the following factors: 1. Nature of the gas Since physical adsorption is non-specific in nature, every gas will get adsorbed on the

    		•
  • Q : Calculate PH value for a acetic acid 1.

    1. A solution of 0.100 M acetic acid is prepared. a) What is its pH value? b) If 20% of the initial acetic acid is converted to the acetate form by titration with NaOH, what is the resultant pH?

    		•
  • Q : Mcq Give me answer of this question.

    Give me answer of this question. The normality of 10% (weight/volume) acetic acid is: (a)1 N (b)10 N (c)1.7 N (d) 0.83 N

    		•
  • Q : What is synthetic rubber and how it

    To meet human needs, scientists have started preparing synthetic rubbers. Besides having similar properties as natural rubbers they are tougher, more flexible and more durable than natural rubber. They are capable of getting stretched to twice its length. Though, it reverts to its original shape

    		•
  • Q : Molar concentration Choose the right

    Choose the right answer from following. Molar concentration (M) of any solution : a) No. of moles of solute/Volume of solution in litre (b) No. of gram equivalent of solute / volume of solution in litre (c) No. of moles os solute/ Mass of solvent in kg  (

    		•
  • Q : Determining Mole fraction of water Can

    Can someone please help me in getting through this problem. The mole fraction of water in 20% aqueous solution of H2O2 is: (a) 77/68 (b) 68/77 (c) 20/80  (d) 80/20

    		•