Match List-I with List-II
List – I
List – II
(A)
4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)
(I)
NO(g)
(B)
N2(g) + 3H2(g) -> 2NH3(g)
(II)
H2SO4(I)
(C)
C12H22O11(aq) + H2O(I) -> C6H12O6 + C6H12O6
Glucose Fructose
(III)
Pt(s)
(D)
2SO2(g) + O2(g) -> 2SO3(g)
(IV)
Fe(s)
Choose the correct answer from the options given below
Match List-I with List-II
List – I |
List – II |
||
(A) |
4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g) |
(I) |
NO(g) |
(B) |
N2(g) + 3H2(g) -> 2NH3(g) |
(II) |
H2SO4(I) |
(C) |
C12H22O11(aq) + H2O(I) -> C6H12O6 + C6H12O6 Glucose Fructose |
(III) |
Pt(s) |
(D) |
2SO2(g) + O2(g) -> 2SO3(g) |
(IV) |
Fe(s) |
Choose the correct answer from the options given below
Option 1 -
(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Option 2 -
(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
Option 3 -
(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Option 4 -
(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
-
1 Answer
-
Correct Option - 3
Detailed Solution:(a) Nitric oxide formation -> Pt is used as catalyst
(b) Haber’s process -> Fe is used as catalyst
(c) Hydrolysis of ester -> Acid (H2SO4) is used as catalyst
(d) SO3 formation -> NO is used as catalyst
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