Q.
Match List-I with List-II.
List-I Reaction |
List-II Catalyst |
A |
4NH3(g)+5O2(g)→4NO(g)+6H2O(g) |
I |
NO(g) |
B |
N2(g)+3H2(g)→2NH3(g) |
II |
H2SO4(l) |
C |
C12H22O11(aq)H2O(l)→ (Glucose) +C6H12O6C6H12O6 |
III |
Pt(s) |
D |
2SO2(g)+O2(g)→2SO3(g) |
IV |
Fe(s) |
Choose the correct answer from the options
given below :
Solution:
(a)4NH3(g)+5O2(g)Pt(s)4NO(g)+6H2O(g)
Ostwald process 500K
(b) N2+3H2Fe(s)2NH3(g)
Haber's process
(c)
C12H22O11 (aq.) +H2O(ℓ)H+ (glucose) C6H12O6+ (fructose) C6H12O6
Inversion of sugar cane
(d) 2SO2(g)+O2(g)NO(g)2SO3(g)