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Arrhenius equation numerical part 1

Arrhenius equation-

source: Slideshare.net

Arrhenius equation-

Q 1) The slope of line in the graph of ‘log K ‘ Vs  ‘1/T’  for a reaction is -5400 K. Calculate  the energy of activation ?

(R = 8.314 JK-1Mole-1)

Solution )

Slope = -5400 K ; Ea = ?  ;  R = 8.314 JK-1Mole-1

According to Arrhenius equation,

slope = -Ea / 2.303 R

-5400 = – Ea / 2.303 x 8.314

Ea = 103394.57 J mole-1

Ea = 103.39 KJ mole-1 Ans.

Q 2) The rate constant of a first order reaction becomes 6 times when the temperature is raised from 350 K to 410 K .

Calculate  the energy of activation ? (R = 8.314 JK-1Mole-1)

Solution )

T1 = 350 K ; T2 = 410 K ; K2 = 6K1

According to Arrhenius equation,

log [K2 / k1] = [Ea/ 2.303 R][(1 / T1) –  (1 / T2)]     

log [6K1 / k1] = [Ea/ 2.303 x 8.314 ][(1 / 350) –  (1 / 410)]    

log 6 = [Ea/ 2.303 x 8.314 ][(410 -350) /  (350 x 410)]    

0.7782 =Ea x 60 / 2.303 x 8.314 x 410 x 350

Ea = 35636.56 J mole-1

Ea = 35.64  KJ mole-1 Ans.

Q 3 ) In the Arrhenius equation for a reaction , the value of A and Ea are 4 x 10 13 sec-1 and 98.6 K J mole-1 respectively. If the reaction is of first order , at what temperature will it half life period be 10 minutes ?

Solution )

t 1/2= 10 minute = 10 x 60 sec = 600 sec  ; A = 4 x 10 13 sec-1 ; Ea = 98.6 K J mole-1 = 98.6 x 103  J mole-1

K = 0.693 / t1/2 =  0.693 / 600

K = 1.155 x 10-3 sec-1

According to Arrhenius equation,

log K = log A – Ea / 2.303 RT

log (1.155 x 10 -3) = log (4 x 10 13) – [98.6 x 10-3 / 2.303 x 8.314 x T]

log 1.155 +l og 10 -3 = log 4 + log 10 13 – [98.6 x 10 3 / 2.303 x 8.314 x T]

log 1.155 – 3l og 10  = log 4 + 13 log 10  – [98.6 x 10 3 / 2.303 x 8.314 x T]

0.06258 – 3 x 1 = 0.6020 + 13 x 1 –  [ 98.6 x 10 3 / 2.303 x 8.314 x T]

– 2.9374 = 13.6020 – 5149.59 / T

– 2.9374 – 13.6020 =  -5149.59 / T

16.5394 = 5149.59 / T

T = 311.35 K Ans.

Q 4 ) The rate constant of a reaction is 1.2 x 10 -3 sec-1 at 303 K and 2.1 x 10 -3 sec-1 at 313 K .Calculate the energy of activation?

( R = 8.314 J K -1 mole-1)

Solution )

T1 = 303 K ; T2 = 313  K ; K1 =  1.2 x 10 -3  ; K2 = 2.1 x 10 -3

According to Arrhenius equation,

log [K2 / k1] = [Ea/ 2.303 R][(1 / T1) –  (1 / T2)]     

log [ 2.1 x 10-3 / 1.2 x 10-3] = [Ea/ 2.303 x 8.314 ][(1 / 303) –  (1 / 313)]    

(log 2.1 log 1.2)  = [Ea/ 2.303 x 8.314 ][(313 -303) /  (303 x 313)]    

0.3222 – 0.0792  =Ea x 10 / (19.15 x 303 x 313)

Ea = 0.243 x 19.15 x 303 x 313 /10

Ea = 44132.8 J mole-1

Ea = 44.13  KJ mole-1 Ans.

Q 5) The rate of a reaction triples when temperature changes from 200 C to 500 C . Calculate  the energy of activation ? (R = 8.314 JK-1Mole-1)

Solution )

T1 = 20 + 273 = 293 K ; T2 = 50 + 273 = 323  K ; K2 = 3 K1

According to Arrhenius equation,

log [K2 / k1] = [Ea/ 2.303 R][(1 / T1) –  (1 / T2)]     

log [3 K1 / k1] = [Ea/ 2.303 x 8.314 ][(1 / 293) –  (1 / 323)]    

log 3 = [Ea/ 2.303 x 8.314 ][(323 -293) /  (323 x 293)]    

0.4771 =Ea x 30 / 2.303 x 8.314 x 293 x 323

Ea =  28817.89 J mole-1

Ea = 28.82  KJ mole-1 Ans.

Q 6 ) The reaction  2NO2 ——->  2 NO + O2    has an Ea of 110 K J mole-1. At 4000C the rate constant is

7.8 mole -1 litre sec -1. Calculate rate constant at 4300 C ?

Solution )

T1 = 400 + 273 = 673 K ; T2 = 430 + 273 = 703  K ; K2 =  ? ; K1 = 7.8 ; Ea = 110 K J mole-1 = 110 x 10 3  J mole-1

log [K2 / k1] = [Ea/ 2.303 R][(1 / T1) –  (1 / T2)]     

log [ K2 / 7.8 ] = [110 x 10 3/ 2.303 x 8.314 ][(1 / 673) –  (1 / 703)]    

log (K2 / 7.8) = [110 x 10 3 / 2.303 x 8.314 ][(703 -673) /  (703 x 673)]    

log (K2 / 7.8)  = 0.3643

Taking antilog of RHS,

 K2 / 7.8  =  antilog 0.3643

K2 / 7.8 = 2.314

K2 = 7.8 x 2.314

K2 = 18.05   mole-1 litre sec-1 Ans.

Q 7 ) A first order reaction is 50 % complete in 30 minutes at 27 0C and in 10 minutes at 470C .Calculate Ea of reaction ?

Solution )

At 270C ,

t 1/2 = 30 min

K1 = 0.693 / t 1/2 = 0.693 / 30

K1 = 0.0231 min-1

At 47 0 C ,

t 1/2 = 10 min

K2 = 0.693 / t 1/2 = 0.693 / 10

K2 = 0.0693 min-1

T1 = 27 + 273 = 300 K ; T2 = 47 + 273 = 320  K ;

log [K2 / k1] = [Ea/ 2.303 R][(1 / T1) –  (1 / T2)]     

log [0.0693 / 0.0231] = [Ea/ 2.303 x 8.314 ][(1 / 300) –  (1 / 320)]    

log 3 = [Ea/ 2.303 x 8.314 ][(320 -300) /  (320 x 300)]    

0.4771 =Ea x 20 / 2.303 x 8.314 x 320 x 300

Ea =  43848.49 J mole-1

Ea = 43.85  KJ mole-1 Ans.

 

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