Question

Match the reaction of Column - I with the catalysts in Column - II:

Column - I (Reaction)	Column - II (Catalyst)
(a) $2KCl{O}_3\left(s\right)\to 2KCl\left(s\right)+3O_2\left(g\right)$	(p) $A{l}_2{O}_3$
(b) $2S{O}_2\left(g\right)+O_2\left(g\right)\to 2S{O}_3\left(g\right)$	(q) $Pt$
(c) $2H_2{O}_2\left(l\right)\to 2H_{2}O\left(l\right)+O_2\left(g\right)$	(r) $V_2{O}_5$
(d) $N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$	(s) $Mn{O}_2$

• A. $a\to s,b\to r,c\to s,d\to p\&q$
• B. $a\to p,b\to r,c\to s,d\to p\&s$
• C. $a\to q,b\to r,c\to s,d\to q$
• D. $a\to s,b\to r,c\to q,d\to p\&q$

Answer 1

The correct option is A $a\to s,b\to r,c\to s,d\to p\&q$

a) For $KCl{O}_3$ decomposition Manganeese(IV) oxide is used as catalyst at an elevated temparature of 150 degree given by the formula $Mn{O}_2$.

b) The catalyst that promotes the $S{O}_2$ oxidation at temparature of 500 degree where $S{O}_2$ is thermodynamically efficient is vanadium pentoxide given by $V_2{O}_5$.

c) The catalyst that provides alternative way with low activation energy for decomposition of $H_2{O}_2$ is $Mn{O}_2$.

d) Synthesis of Ammonia by Haber process uses $Pt$ and $A{l}_2{O}_3$ as a compromise catalysts in place of $Fe$.
Option A is correctt.