The rate constant of reaction is 1.5 x 107 sec–1 at 50°C and 4.5 x107 sec–1 at 100°C. Calculate the value of activation energy, Ea for the reaction (R =8.314 JK–1 mol–1).

by using of arrhenius of equation 

log k2k1 = Ea2.303 RT2-T1T2T1

or           

log 4.5 × 1071.5 ×107 = Ea2.303 × 8.314373-323373 × 323

or                    

log 3 = Ea2.303 × 8.31450373 × 323

or              

                    Ea = 0.4771 × 2.303 × 8.314 × 373 × 32350      = 22011.75 J mol-1 = 22.01 kJ mol-1.

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The rate of a particular reaction doubles when temperature changes from 27°C to 37°C. Calculate the energy of activation of such a reaction.


When              T1 = 27°C = 300 kk1 = k (say)

when,             T2 = 37°C=310 Kk2 = 2k

Substituting these values in the equation

                       logk2k1 = Ea2.303 RT2-T1T1T2
We get

                 log2kk=Ea2.303 × 8.314×310-300300×310

or

log 2 = Ea2.303 × 8.314×10300 × 310


or                    Ea = 53.6 kJ mol-1

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From the data given below, show that the decomposition of hydrogen peroxide in aqueous solution is a first order reaction:.

Time (min)           0            10            20
N (mL)               22.8        13.8          8.20

where N is the volume of standard potassium permanganate solution in mL required to decompose definite volume of the peroxide solution.

The volume of standard KMnO4 is proportional to the amount of H2O2. It can therefore be used in place of concentration of H,O2.
So, 
[A]0 = 22.8 mL,
[A]10 min = 13.8 mL
[A]20 min = 8.20 mL

Substituting the values of concentration and time in the first order rate equation

        k=2.303tlogA0At

(i)    
                    k=2.30310log22.8 mL13.8 mL=5.11 × 10-2 min-1


(ii)    
                    k=2.30310log22.8 mL8.30 mL=5.11 × 10-2 min-1

Since k has the same value in both cases, decomposition of hydrogen peroxide is a first order reaction.
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The activation energy of a reaction is 94.14 kJ mol–1 and the value of rate constant at 313 K is 1.8 x 10–5 sec–1. Calculate the frequency factor A.

We have given 

Ea = 94.14 kJ mol-1 = 94140 J mol-1T = 313,  k = 1.8 × 10-5 sec-1

thus by usingv Arrhenius equation, 
 
Now,        log k = -Ea2.303 RT+ log A

We get,     
                 logA = log(1.8 × 10-5) + 941402.303 × 8.314 × 313         = (log 1.8) - 5 + 15.7082        = 0.2553 - 5 + 15.7082 = 10.9635

Therefore,
          A=anti log(10.9635) = 9.194 × 1010 collisions/sec.
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Calculate the half-life period of a first order reaction when the rate constant is 5 year–1.

For half- life first order reaction

 t1/2 = 0.693k

k= 5years-1
or                       t1/2 = 0.6935 years-1      = 0.1368 year = 50 days 14 hrs. 8 min 9.6 s.

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