Chemical Kinetics

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a month ago

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V
Vishal Baghel

Contributor-Level 10

Rate = K [A]?
K = Rate / [A]?
= (mole L? ¹s? ¹) / (mole L? ¹)?
= mole¹? L? ¹ s? ¹
Unit of K: mole¹? L? ¹ s? ¹

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Fraction of molecules having enough energy to form product = e E a / R T

Fraction of molecules having enough energy to form product = e 8 0 . 9 * 1 0 3 8 . 3 1 4 * 7 0 0

= e 1 3 . 8 e 1 4

So, x = 14

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

C 1 2 H 2 2 O 1 1 + H 2 O C 6 H 1 2 O 6 G l u c o s e + C 6 H 1 2 O 6 F r u c t o s e

K = 2 . 3 0 3 t l o g a a x

k t 2 . 3 0 3 = l o g a a x

l n 2 * 9 1 0 3 * 2 . 3 0 3 = l o g ( 1 f )

l o g ( 1 f ) = 8 1 . 2 4 * 1 0 2 = 8 1 * 1 0 2

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

l o g K = l o g A E a 2 . 3 0 3 R T

S l o p e = E a 2 . 3 0 3 R = 1 0 , 0 0 0 K  

E a 2 . 3 0 3 R = 1 0 4            

l o g ( 1 0 5 ) = l o g A 1 0 4 * 1 5 0 0   (at 500 K temperature)

T = 1 0 4 1 9 K = 1 0 0 1 9 * 1 0 2 K = 5 . 2 6 3 1 * 1 0 2 K = 5 2 6 . 3 1 K 5 2 6 K

New question posted

2 months ago

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New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

log k = 20.35 - ( 2 . 4 7 * 1 0 3 ) T  

Comparing with,

l o g k = l o g A E a 2 . 3 0 3 R T

E a 2 . 3 0 3 R = 2 . 4 7 * 1 0 3

= 47.29 kJ/mole

Ans. = 47

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

t1/2 = 340 sec      P0 = 55.5 kPa

t1/2 = 170 sec      P0 = 27.8 kPa

? t 1 / 2 ? ( P 0 ) 1 ? n

? 1 ? n = 1 n = 0

? zero order reaction

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

K = 0 . 6 9 3 t 1 / 2 = 0 . 6 9 3 7 0 * 6 0 = 6 9 3 0 7 * 6 * 1 0 6

= 165 * 10-6 s-1

New answer posted

2 months ago

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Vishal Baghel

Contributor-Level 10

( t 1 / 2 ) l ( t 1 / 2 ) I I = ( P I P I I ) 1 n n = 1

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

t1/2 = 0.301 min

t = 2 min

K = 2 . 3 0 3 t l o g ( C o C t )

0 . 6 9 3 0 . 3 0 1 = 2 . 3 0 3 2 l o g ( C o C t )

2 . 3 0 3 * 0 . 3 0 1 0 . 3 0 1 = 2 . 3 0 3 2 l o g ( C o C t )

2 = l o g ( C o C t )

C o C t = 1 0 2 = 1 0 0

Ans. 100

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