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QJEEM22Chc5ce: JEE Main 2022 Chemistry - Class 12 JEEM22Chc5ce Solution: Match List - I with List - II. .tg…

Match List - I with List - II.

List - I
Reaction
List - II
Catalyst
(A) 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4N{H_3}(g) + 5{O_2}(g) \to 4NO(g) + 6{H_2}O(g) (I) NO(g)NO(g)
(B) N2(g)+3H2(g)→2NH3(g){N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g) (II) H2SO4(l){H_2}S{O_4}(l)
(C) C12H22O11(aq)+H2O(l)→C6H12O6Glucos⁡e+C6H12O6Fructose{C_{12}}{H_{22}}{O_{11}}(aq) + {H_2}O(l) \to \mathop {{C_6}{H_{12}}{O_6}}\limits_{Glu\cos e} + \mathop {{C_6}{H_{12}}{O_6}}\limits_{Fructose} (III) Pt(s)Pt(s)
(D) 2SO2(g)+O2(g)→2SO3(g)2S{O_2}(g) + {O_2}(g) \to 2S{O_3}(g) (IV) Fe(s)Fe(s)

Choose the correct answer from the options given below :

Multiple Choice Options

  • A (A)−(II),(B)−(III),(C)−(I),(D)−(IV)(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{III}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{IV})
  • B (A)−(III),(B)−(II),(C)−(I),(D)−(IV)(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{IV})
  • C (A)−(III),(B)−(IV),(C)−(II),(D)−(I)(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{I})
  • D (A) - (III), (B) - (II), (C) - (IV), (D) - (I)

Step-by-Step Verified Solution

(A) 4NH3( g)+5O2( g)⟶Pt(s)4NO(g)+6H2O(g)4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \stackrel{\mathrm{Pt}(\mathrm{s})}{\longrightarrow} 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})

(B) N2( g)+3H2( g)⟶Fe(s)2NH3( g)\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \stackrel{\mathrm{Fe}(\mathrm{s})}{\longrightarrow} 2 \mathrm{NH}_{3}(\mathrm{~g})

(C) C12H22O11(aq)+H2O(l)⟶H2SO4(l)\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \stackrel{\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{l})}{\longrightarrow} C6H12O6 Glucose +C6H12O6 Fructose \underset{\text { Glucose }}{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}+\underset{\text { Fructose }}{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}}

(D) 2SO2( g)+O2( g)⟶NO(g)2SO32 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \stackrel{\mathrm{NO}(\mathrm{g})}{\longrightarrow} 2 \mathrm{SO}_{3}

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