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Match List - I with List - II.

List - I

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List - II

Catalyst
\mathrm{(A) \: 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})} \mathrm{(I) \: \mathrm{NO}(\mathrm{g})}
\mathrm{(B) \: \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})} \mathrm{(II) \: \mathrm{H}_{2} \mathrm{SO}_{4}(l)}

\mathrm{(C)\; C_{12}H_{22}O_{11}(aq)+H_{2}O(\ell)\rightarrow C_{6}H_{12}O_{6}+C_{6}H_{12}O_{6}}

                                                                \text{Glucose}            \text{Fructose}

 \mathrm{(III) \: \mathrm{Pt}(\mathrm{s})}
\mathrm{(D) \: 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{3}(\mathrm{~g})} \mathrm{(IV) \: \mathrm{Fe}(\mathrm{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) - (II) - (C) - (IV) - (D) - (I)


Answers (1)

best_answer

\mathrm{(a) \: 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \overset{Pt(s)}{\rightarrow} 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})}
        Catalyst used is Pt(s) which is known as Ostwald process.

\mathrm{(b) \: \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \overset{Fe(s)}{\rightarrow}2 \mathrm{NH}_{3}(\mathrm{g})}
       In presence of Fe(s) , as catalyst, this process is known as Born Haber process.

\mathrm{(c)\; C_{12}H_{22}O_{11}(aq)+H_{2}O(\ell)\overset{H^{+}}{\rightarrow}C_{6}H_{12}O_{6}(Glucose)+C_{6}H_{12}O_{6}(Fructose)}
        \mathrm{H_{2}SO_{4}} is used in the Inversion of Sugar process.

\mathrm{(d) \: 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\overset{NO(g)}{\rightarrow}2 \mathrm{SO}_{3}(\mathrm{g})}
        \mathrm{NO(g)} is used as a Catalyst.

Hence option (C) is correct.
 

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Riya

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