For a reaction 2 H 2 O 2 2 H 2 O 2 → Alkaline medium I - 2 H 2 O + O 2 The proposed mechanism is as given below: 1. H 2 O 2 + I - → H 2 O + IO - (slow) 2. H 2 O 2 + IO - → H 2 O + I - + O 2 (fast) Write the overall order of a reaction.

Step number 1 is the slowest among two steps, hence this is the rate-determining step. i.e., H 2 O 2 + I - → H 2 O + IO - (slow) Rate law of the reaction is given by: Rate = k [ H 2 O 2 ] [ I – ] As the rate law shows that the reaction is first order with H 2 O 2 and I – both, hence the overall order of the reaction becomes 1 + 1 = 2 .

Find g o f and f o g when f : R → R and g : R → R is defined by f ( x ) = x 2 + 2 x + 3 and g ( x ) = 3 x − 4 .

Given f ( x ) = x 2 + 2 x + 3 and g ( x ) = 3 x - 4 Therefore, f o g x = f g x = f 3 x - 4 = 3 x - 4 2 + 2 3 x - 4 + 3 = 9 x 2 - 24 x + 16 + 6 x - 8 + 3 = 9 x 2 - 18 x + 11 And g ∘ f x = g f x = g x 2 + 2 x + 3 = 3 x 2 + 2 x + 3 - 4 = 3 x 2 + 6 x + 9 - 4 = 3 x 2 + 6 x + 5

If g : R → R be two functions defined as f x = | x | + x and g x = | x | - x , ∀ x ∈ R . Then find f o g and g o f . Hence find f o g ( – 3 ) , f o g ( 5 ) and g o f ( – 2 ) .

Given : f ( x ) = | x | + x and g ( x ) = | x | - x , ∀ x ∈ R To find: f o g , and g o f We know that f o g = f ( g ( x ) ) = | g ( x ) | + g ( x ) = | | x | − x | + ( | x | − x ) . . . . i [Since g x = x - x ] Now, we know that x = x if x ≥ 0 - x if x < 0 Thus, for x ≥ 0 f o g x = x - x + x - x = 0 [from i ] For x < 0 , f o g x = - x - x + - x - x = - 4 x [From i ] Therefore, f o g = f g x = 0 x ≥ 0 - 4 x x < 0 . . . . . . i i Similarly, g o f = g ( f ( x ) ) = | f ( x ) | − f ( x ) = | | x | + x | − ( | x | + x ) [Since f x = x + x ] g f x = x + x - x + x = 0 for x ≥ 0 - x + x - - x + x = 0 for x < 0 Therefore, g ( f ( x ) ) = g o f = 0 ∀ x Now, f o g ( − 3 ) = ( 4 ) ( − 3 ) = − 12 (Since from i i , f o g = 4 x f o r x < 0 ) f o g ( 5 ) = 0 (since, f o g = 0 f o r x ≥ 0 ) g o f ( − 2 ) = 0 (since, g o f = 0 ∀ x )

Let f : R → R and g : R → R be defined by f ( x ) = x + 1 and g ( x ) = x − 1 . Show that f o g = g o f = I R .

Given, f : R → R and g : R → R ⇒ f o g : R → R and g o f : R → R (Also, we know that I R : R → R ) So, the domains of all f o g , g o f and I R are the same. ( f o g ) ( x ) = f ( g ( x ) ) = f ( x − 1 ) = x − 1 + 1 = x = I R ( x ) . . . ( 1 ) ( g o f ) ( x ) = g ( f ( x ) ) = g ( x + 1 ) = x + 1 − 1 = x = I R ( x ) . . . ( 2 ) From ( 1 ) and ( 2 ) , ( f o g ) ( x ) = ( g o f ) ( x ) = I R ( x ) , ∀ x ∈ R Hence , f o g = g o f = I R

Let f : R → R and g : R → R be defined by f ( x ) = x 2 and g ( x ) = x + 1 . Show that f o g ≠ g o f .

Given, f : R → R and g : R → R . So, the domains of f and g are the same. ( f o g ) ( x ) = f ( g ( x ) ) = f ( x + 1 ) = ( x + 1 ) 2 = x 2 + 1 + 2 x ( g o f ) ( x ) = g ( f ( x ) ) = g ( x 2 ) = x 2 + 1 So, f o g ≠ g o f

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Important Questions on Chemical Kinetics

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Write a possible RNA strand that can be transcribed from the above DNA molecule formed.

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Which of the following statements is correct?

a. rRNA is the most abundant RNA.
b. Ribosomes are composed of approximately 80% rRNA and 20% ribosomal proteins.
c. rRNA is a type of coding RNA.

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Which one of the following is wrongly matched?

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Explain the mechanism of translation that occurs in the ribosomes in a prokaryote.

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Describe an inverted pyramid of biomass with the help of an example.

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What is transcription?