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)
The order of a reaction is the sum of the powers of the concentration terms in the experimental rate law. It need not follow the stoichiometric coefficients, and it can be a fraction. A pseudo first order reaction is one in which one reactant is present in such large excess that its concentration hardly changes, so the rate depends on the other reactant alone.
(A) Third order — This matches (II) \(2NO(g) + O_2 (g) → 2NO_2 (g)\). The rate law is second order in NO and first order in \( \ce{O2} \):
\( \displaystyle \text{Rate} = k[\ce{NO}]^{2}[\ce{O2}], \quad 2 + 1 = 3 \)
(B) First order — This matches (I) \(C_2H_4 (g) + H_2 (g) → C_2H_6 (g)\). Hydrogenation of ethene depends on ethene alone:
\( \displaystyle \text{Rate} = k[\ce{C2H4}] \)
(C) One and half order — This matches (IV) \(CHCl_3 + Cl_2 → CCl_4 + HCl\). The rate law has a half power of chlorine:
\( \displaystyle \text{Rate} = k[\ce{CHCl3}][\ce{Cl2}]^{1/2}, \quad 1 + \frac{1}{2} = \frac{3}{2} \)
(D) Pseudo first order — This matches (III) \(CH_3COOC_2H_5 + H_2O → CH_3COOH + C_2H_5OH\). In acid hydrolysis water is in large excess, so its concentration stays constant and is taken into k:
\( \displaystyle \text{Rate} = k'[\ce{CH3COOC2H5}][\ce{H2O}] = k[\ce{CH3COOC2H5}] \)
Shortcut: (C)-(IV) appears in Options (2) and (3); (A)-(II) then leaves only Option (2).