Order of reaction numerical part 3

Order of reaction

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Order of reaction-

Q 1)

(i) The reaction .   A + 3B ———> 2C ,  obeys the rate equation ,

Rate = K [A]^1/2 [B]^3/2. Find order of reaction ?

(ii) The reaction  A + B —–> C  has zero order . Write rate equation ?

Solution )

(i) Rate = dx / dt = K [A] ^1/2 [B]^3/2

order of reaction = 1/2 + 3/2 =2

 Reaction is second order.

(ii) If reaction is zero order then,

Rate = K [A]⁰[B]⁰   Ans.

Q 2) Find order of reaction and unit of rate constant-

(i) 2 H₂ + 2 NO ——–> 2 H₂O + N₂

Rate = K [H₂][NO]²

(ii) H₂O₂ + 3 I^– + 2 H⁺ ——–> 2 H₂O + I₃^– (aq.)

Rate = K [H₂O₂][I^–]

(iii) CH₃CHO ———> CH₄ + CO

Rate = K [CH₃CHO]^3/2

Solution)

(i) Rate = K [H₂][NO]²

Order of reaction = 1 + 2 = 3

Reaction is third order.

Unit of  K –

Rate = K [H₂][NO]²

K = Rate /  [H₂][NO]²

unit of K = (mole litre^-1 sec^-1) / (mole litre^-1 )(mole litre^-1 )²

Unit of K = litre² mole ^-2sec^-1 Ans.

(ii)

Rate = K [H₂O₂][I^–]

Order of reaction = 1 + 1 = 2

Reaction is second order.

Unit of  K-

Rate = K [H₂O₂][I^–]

K = Rate /  [H₂O₂][I^–]

unit of K = (mole litre^-1 sec^-1) / (mole litre^-1 )(mole litre^-1 )

Unit of K = litre mole^-1 sec^-1 Ans.

(iii)

Rate = K [CH₃CHO] ^3/2

Order of reaction = 3/2

Unit of  K-

Rate = K [CH₃CHO]^3/2
K = Rate / [CH₃CHO]^3/2

unit of K = (mole litre^-1 sec^-1) / (mole litre-1 )^3/2

Unit of K = litre^1/2 mole ^-1/2 sec^-1 Ans.

Q 3) -The rate of reaction,

2A + B ——–> A₂B has the rate law,

Rate = K[A]² with rate constant equal to 0.50 mole ^-1 litre sec^-1. Calculate the rate of reaction when,

(i) [A] = 0.60 mole / litre , [B] = 0.05 mole/litre

(ii) When conc. of A and B have been reduced to 1/4.

Solution)

(i)  K = 0.50 mole ^-1 litre sec^-1 , [A] = 0.60 mole /litre

Rate of reaction = K[A]² = 0.50 x 0.60 x 0.60

 Rate of reaction = 0.18 Ans.

(ii)

K = 0.50 mole ^-1 litre sec^-1 , [A] = 0.60 x 1/4  mole /litre

Rate of reaction = K[A]²

= 0.50 x 0.60 x 0.60/ 4 x 4

 Rate of reaction = 0.011  Ans.

Q 4) The reaction  ,  2NO + O₂ ——> 2 NO₂ follows the rate law,

Rate = K [NO]²[O₂]. If K = 2 x 10 ^-6 litre² mole^-2sec ^-1. Calculate rate of reaction when [NO] = 0.04 mole /litre and

[O₂] = 0.2 mole / litre.

Solution )

[NO] = 0.04 mole /litre ,  [O₂] = 0.2 mole / litre , K = 2 x 10 ^-6 litre² mole^-2sec ^-1.

Rate = K [NO]²[O₂] = 2 x 10 ^-6  [0.04]² [0.2]

Rate =6.4 x 10 ^-10 mole litre^-1 sec^-1 Ans.