## Arrhenius equation-

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## Arrhenius equation-

### 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

### 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

### 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

### Solution )

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

#### 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.

### Solution )

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

#### 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

### 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 = 2.314

K2 = 7.8 x 2.314

### 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