Chemistry Chemical Kinetics

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New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

lnK = -E_a/RT + I
-E_a/R = slope is negative
⇒ -E_a/R = (10-0)/ (5-0)
E_a = 2R

New answer posted

a month ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

Without catalyst: K = A e - E a / R T

  Presence catalyst:   10 6 K = A e - E c / R T E q ( B ) - E q ( A ) 10 6 = e - ( E - E c ) / R T E - E c R T = 2.303 * 6 Δ E = E c - E = ( - 2.303 ) * 6 R T

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Al equilibrium

K f H 2 [ N O ] 2 = K b N 2 H 2 O + H 2

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

(a) Rate vs Time for zero order is a horizontal line (Rate = const.)
(b) t?/? vs Concentration for zero order is a straight line through the origin (t?/? ∝ [C?])
(c) Concentration vs Time for first order is an exponential decay (C? = C?e?)
(d) Concentration vs Time for a reaction is a decreasing curve, not a horizontal line.
(e) Rate vs Concentration for first order is a straight line through the origin (Rate = k[Conc.]¹)

New answer posted

a month ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

A → B
t=0: a, 0
t=30min: (a-x), x
given x = 0.2 mole / lit
Average rate = +Δ [B]/Δt = 0.2 mole/lit / (30/60)h = 0.4 mole lit? ¹h? ¹
= 4 * 10? ¹ mole lit? ¹h? ¹

New answer posted

a month ago

0 Follower 2 Views

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

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

X → Y

EafEab=2030Eab=20

Eab=50kJ/mole

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

K = 3.3 * 10-4 s-1

Time for 40% completion ; t

Using K = 2 . 3 0 3 t l o g 1 0 [ R ] 0 [ R ]  

3.3 * 10-4 =   2 . 3 0 3 t l o g 1 0 [ R ] 0 0 . 6 [ R ] 0

t = 2 . 3 0 3 3 . 3 * 1 0 4 * 0 . 2 2 t = 2 5 . 5 8 m i n s  

so; the nearest integer is 26.

New answer posted

2 months ago

0 Follower 2 Views

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

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

log (k2k1)=Ea2.303*8.314 [13001325]

log (5) = Ea2.303*8.314 [13001325]

Ea=0.7*2.303*8.314*300*32525=52271.7

Ea in kJ/mole = 52271.71000=52.2kJ/mol

the nearest integer is 52.

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