Question

Match List-I with List-II.

List-I Reaction		List-II Catalyst
A	$ 4 NH _3( g )+5 O _2( g ) \rightarrow 4 NO ( g )+6 H _2 O ( g )$	I	NO(g)
B	$N _2( g )+3 H _2( g ) \rightarrow 2 NH _3( g )$	II	$H_2SO_4(l)$
C	$ C _{12} H _{22} O _{11}( aq ) H _2 O (l) \rightarrow \text { (Glucose) }+ C _6 H _{12} O _6 C _6 H _{12} O _6$	III	Pt(s)
D	$2 SO _2( g )+ O _2( g ) \rightarrow 2 SO _3( g )$	IV	Fe(s)

Choose the correct answer from the options given below :

• A. (A) - (II), (B) - (III), (C) - (I), (D) - (IV)
• B. (A) - (III), (B) - (II), (C) - (I), (D) - (IV)
• C. (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
• D. (A) - (III), (B) - (II), (C) - (IV), (D) - (I)

Answer 1

Answer: (A)

(a)$4 NH _3( g )+5 O _2( g ) \xrightarrow{ Pt ( s )} 4 NO ( g )+6 H _2 O ( g )$
Ostwald process $500 K$
(b) $N _2+3 H _2 \xrightarrow{ Fe ( s )} 2 NH _3( g )$
Haber's process
(c)
$C _{12} H _{22} O _{11} \text { (aq.) }+ H _2 O (\ell) \xrightarrow{ H ^{+}} \underset{\text { (glucose) }}{ C _6 H _{12} O _6}+\underset{\text { (fructose) }}{ C _6 H _{12} O _6}$
Inversion of sugar cane
(d) $2 SO _2( g )+ O _2( g ) \xrightarrow{ NO ( g )}2 SO _3( g )$