Match List I with List II
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LIST I |
LIST II |
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A. \(CH_3CHO\overset{773\,K}{\longrightarrow}CH_4+CO\) |
I. Zero order |
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B. \(2NH_3\overset{Pt}{\longrightarrow}N_2+3H_2\) |
II. Order of 1/2 with respect to one reactant |
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C. \(H_2+Br_2\longrightarrow 2HBr\) |
III. 1.5 |
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D. \(SO_2Cl_2\longrightarrow SO_2+Cl_2\) |
IV. First order |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-I, C-II, D-IV
The order of a reaction is found from its experimental rate law, not from the balanced equation, and it can be zero or a fraction.
(A) \( \ce{CH3CHO} \xrightarrow{\text{773 K}} \ce{CH4 + CO} \) — This matches III. 1.5. The rate law is rate = k[CH3CHO]3/2, so the order is 3/2 = 1.5.
(B) \( \ce{2NH3} \xrightarrow{\ce{Pt}} \ce{N2 + 3H2} \) — This matches I. Zero order. At high pressure the platinum surface is fully covered with \( \ce{NH3} \) molecules, so the rate does not depend on its concentration: rate = k[NH3]0 = k.
(C) \( \ce{H2 + Br2 -> 2HBr} \) — This matches II. Order of 1/2 with respect to one reactant. The rate law is rate = k[H2][Br2]1/2: order 1 in \( \ce{H2} \) and 1/2 in \( \ce{Br2} \).
(D) \( \ce{SO2Cl2 -> SO2 + Cl2} \) — This matches IV. First order. Its thermal decomposition follows rate = k[SO2Cl2].