State the role of activated complex in a reaction and state the relation with energy of activation.

Activated complex is the intermediate compound formed by reactants, which is highly unstable and readily changes into product. Those reactants which possess activation energy and collide in proper orientation can form activated complex which can easily form products.
Lower the activation energy, more easily activated complex will be formed and faster will be the reaction.
Activation energy = energy of activated complex - energy of reactants.


Energy of activation. The minimum energy over average energy which must be gained by the molecules before they could react to form products is called the energy of activation. It is denoted by Ea.
According to the Arrhenius theory, activation energy is independent of temperature. However, precise measurements indicate that the activation energy tends to decrease slightly with a rise in temperature.

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The gas phase decomposition of dimethyl ether follows first order kinetics:
CH3 – O CH3 (g) → CH4 (g) + H2(g) + CO(g)
The reaction is carried out in a constant volume container at 500°C and has a half life of 14.5 minutes. Initially only dimethyl ether is present at a present of 0.40 atmosphere. What is the total pressure of the system after 12 minutes?
(Assume ideal gas behaviour).

The gas phase decomposition of dimethyl ether follows first order kinetics:
CH3 – O CH3 (g) → CH4 (g) + H2(g) + CO(g)

we have given 
Volume is constant 
temperature is 5000 C
half life is =14.5 min

PV = nRT

or           nV=PRT

   a=PRT=0.40.082×773   = 6.31 × 10-3 mol L-1K = 0.693t1/2 = 0.69314.5    = 4.78 × 10-2 min-1K = 2.30312logaa-x



4.78×10-2 = 2.30312 log aa-xaa-x = 1.77446      a-x = 6.31 × 10-31.77446 moles L-1                = 3.556 × 10-3 moles L-1  x = (6.310 -3.556) × 10-3 moles/L         =2.754 × 10-3 moles/L  After 12 minutes.

Total number of moles L-1

                     = a+2x= 6.31×10-3+2×2.754×10-3= 11.818×10-3

P = nVRT  = 11.818 × 10-3 ×0.082 × 773  = 0.749 atm.
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What are Pseudo chemical reaction? Give examples.


These are the reactions which follow kinetics of lower order whereas by appearance these appear to be reactions of higher order. In other words, these are false or pseudo in nature. In these reactions one of the reactants (genrally water) is reprsent in such a large excess that its concentration remain practially unchanged during the reaction. Therefore, the concentration of such a reactant does not appear in the rate law for the reaction.
Examples: (i)

C12H22O11+H2OH+ C6H12O6 + C6H12O6                                        Glucose      Fructose      reaction rate (r)  = k[Sucrose] [H2O] [H3O]+                                   = k1(Sucrose)

(ii) 
              

CH3COOC2H5+H2OH+ Ethylacetate                                      CH3COOH+C2H5OH                                           Acetic            Ethyl                                            acid               alcohol

reaction rate (r) = kCH3COOC2H5                                                         [H2O] {H3O+]                            = k1(CH3COOC2H5)

          


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How will you verify whether a reaction is first order or not?

Whether a reaction is of first order or not can be verified by using the analytical methods based on the rate equations. The reagents are allowed to react, the change of concentration is measured at different time intervals. For a first order reaction:

(i) Plot of log [A0]/[A] versus t gives a straight line passing through the origin.

(ii) Plot of log[A] versus t gives a straight line with an intercept log [A]0 and a slope = – k /2.303.

(iii) Plot of [A] versus t is exponential.

(iv) The value of k calculated by using kt = 2.303 log [A]/ [A] for different time intervals come out to
be same.

(v) The half life period is independent of the initial concentration of the reactant.

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For the first order reaction R → P, deduce the integrated form of the rate law. What will be the nature of the curve when concentration is plotted against time for such a reaction?

For the reaction

R → P

Rate =-d[R]dt =K[R]Or 1[R]d[R] =-kdtOn integrating both sides, we get1d[R]R =-kdtIn [R] =-kt +I .....(1)Where I =integration constantAt t=0, [R] =[R]0 intial conc of the reactantpurtting this in eq (1) givesIn[R]0 =-k xo+Iand I =In[R]0substituting the value of I in eq(1) gives In[R] =-Kt +In[R]0Changing into log 10 we havelog[R] = -k2.303t+log[R]0When log[R] is plotted against tit gives a straight line with slope =-k2.303 and y intercept =log[R]0k=2.303tlog [R][R]0



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