--%>

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 : Freezing point of equimolal aqueous

    The freezing point of equi-molal aqueous solution will be maximum for:            (a) C6H5NH3+Cl-(aniline hydrochloride)  (b) Ca(NO3

  • 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 : Molecular energies and speeds The

    The average translational kinetic energies and speeds of the molecules of a gas can be calculated.The result that the kinetic energy of 1 mol of the molecules of a gas is equal to 3/2 RT can be used to obtain numerical values for the

  • Q : Question related to molarity Help me to

    Help me to go through this problem. Molarity of a solution containing 1g NaOH in 250ml of solution: (a) 0.1M (b) 1M (c) 0.01M (d) 0.001M

  • Q : Determining mole fraction of water in

    A mixture has 18 g water and 414 g ethanol. What is the mole fraction of water in mixture (suppose ideal behaviour of mixture): (i) 0.1  (ii) 0.4  (iii) 0.7  (iv) 0.9 Choose the right answer from abo

  • Q : Problem on preparing of a solution Give

    Give me answer of this question. How many grams of CH3OH should be added to water to prepare 150 solution of@M CH3 OH: (a) 9.6 (b) 2.4 (c) 9.6x 103 (d) 2.4 x103

  • Q : Mole fraction of solute The mole

    The mole fraction of the solute in 1 molal aqueous solution is: (a) 0.027 (b) 0.036 (c) 0.018 (d) 0.009What is the correct answer.

  • Q : Mass percent Help me to go through this

    Help me to go through this problem. 10 grams of a solute is dissolved in 90 grams of a solvent. Its mass percent in solution is : (a) 0.01 (b) 11.1 (c)10 (d) 9

  • Q : Describe Enzyme Catalyzed reactions

    Many enzyme catalyzed reactions obeys a complex rate equation that can be written as the total quantity of enzyme and the whole amount of substrate in the reaction system. Many rate equations that are more complex than first and se

  • Q : Question based on normality Provide

    Provide solution of this question. A 5 molar solution of H2SO4 is diluted from 1 litre to 10 litres. What is the normality of the solution : (a) 0.25 N (b) 1 N (c) 2 N (d) 7 N