--%>

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 : Sugar solution The solution of sugar in

    The solution of sugar in water comprises: (i) Free atoms (ii) Free ions (iii) Free molecules (iv) Free atom and molecules. Choose the right answer from the above.

  • Q : Explain Vapour Pressure Composition A

    A pressure composition diagram for a liquid vapor system can be used to show the composition of the liquid and equilibrium vapor.Vapor equilibrium data are useful in the study of distillations. It is of value to have diagrams showing not only the vapor pre

  • Q : Diffusion Molecular View When the

    When the diffusion process is treated as the movement of particles through a solvent the diffusion coefficient can be related to the effective size of diffusing particles and the viscosity of the medium.To see how the experimental coefficients can be treat

  • Q : How haloalkanes are prepared from

    This is the common method for preparing haloalkanes in laboratory. Alcohols can be converted to haloalkanes by substitution of - OH group with a halogen atom. Different reagents can be used to get haloa

  • Q : Distribution law Help me to go through

    Help me to go through this problem. The distribution law is applied for the distribution of basic acid between : (a) Water and ethyl alcohol (b) Water and amyl alcohol (c) Water and sulphuric acid (d) Water and liquor ammonia

  • Q : Problem on mole fraction of glucose

    Provide solution of this question. While 1.80gm glucose dissolve in 90 of H2O , the mole fraction of glucose is: (a) 0.00399 (b) 0.00199 (c) 0.0199 (d) 0.998

  • Q : Calculating weight of acid Give me

    Give me answer of this question. The formula weight of H2SO4 is 98. The weight of the acid in 400mi of solution is: (a)2.45g (b) 3.92g (c) 4.90g (d) 9.8g

  • Q : Modern periodic table and Mendeleevs

    Differentiate between the modern periodic table and Mendeleevs table?

  • Q : Problem on vapour pressure Choose the

    Choose the right answer from following. If P and P are the vapour pressure of a solvent and its solution respectively N1 and N2 and are the mole fractions of the solvent and solute respectively, then correct relation is: (a) P= PoN1 (b) P= Po N2 (c)P0= N2 (d)

  • 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)