Match List-I with List-II
|
List-I Order of the reaction |
List-II Example |
|
(A) Third order |
(I) $C_2H_4 (g) + H_2 (g) → C_2H_6 (g)$ |
|
(B) First order |
(II) $2NO(g) + O_2 (g) → 2NO_2 (g)$ |
|
(C) One and half order |
(III) $CH_3COOC_2H_5 + H_2O → CH_3COOH + C_2H_5OH$ |
|
(D) Pseudo first order |
(IV) $CHCl_3 + Cl_2 → CCl_4 + HCl$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
|
List-I Order of the reaction |
List-II Example |
|
(A) Third order |
(II) $2NO(g) + O_2 (g) → 2NO_2 (g)$ |
|
(B) First order |
(I) $C_2H_4 (g) + H_2 (g) → C_2H_6 (g)$ |
|
(C) One and half order |
(IV) $CHCl_3 + Cl_2 → CCl_4 + HCl$ |
|
(D) Pseudo first order |
(III) $CH_3COOC_2H_5 + H_2O → CH_3COOH + C_2H_5OH$ |
(A) Third order - (II)
Reaction:
2NO(g) + O₂(g) → 2NO₂(g)
Rate law:
Rate = k[NO]²[O₂]
Total order = 2 + 1 = 3 (third order).
(B) First Order - (I)
Reaction:
C₂H₄(g) + H₂(g) → C₂H₆(g)
Hydrogenation reactions on catalyst surfaces often depend on only one reactant concentration, so the rate behaves as first order.
(C) One and half order - (IV)
Reaction:
CHCl₃ + Cl₂ → CCl₄ + HCl
Rate law:
Rate = k[CHCl₃][Cl₂]¹ᐟ²
Total order = 1 + 1/2 = 3/2 (one and half order).
(D) Pseudo first order - (III)
Reaction:
CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
Hydrolysis occurs in large excess of water, so [H₂O] remains almost constant. Thus the reaction behaves as pseudo first order.