Exams › JEE Advanced › Chemistry
Match the products in List-I with the appropriate reactions in List-II. List-I (Products) (P) H2O2 (Q) H2O2 + O2 (R) H2 (S) H2O List-II (Reactions) (1) KO2 + H2O -> (2) Na2O2 + H2O -> (3) NaCl(aq) electrolysis -> (4) NaCl + H2SO4 -> (5) NaHCO3 + heat ->
- P -> 1; Q -> 2; R -> 3; S -> 4
- P -> 2; Q -> 1; R -> 3; S -> 5
- P -> 2; Q -> 1; R -> 3; S -> 4
- P -> 1; Q -> 2; R -> 4; S -> 5
Correct answer: P -> 2; Q -> 1; R -> 3; S -> 4
Solution
KO2 is a superoxide (O2⁻). 4KO2 + 2H2O -> 4KOH + 3O2 gives both H2O2 and O2 as products (actually the reaction gives O2 primarily, with H2O2 as intermediate). More precisely: 2KO2 + 2H2O -> 2KOH + H2O2 + O2. So KO2+H2O -> H2O2 + O2 (Q->1). Na2O2 + H2O -> 2NaOH + H2O2 (P->2). Electrolysis of NaCl(aq): 2H2O + 2e- -> H2 + 2OH- at cathode (R->3). NaCl + H2SO4 (conc) -> NaHSO4 + HCl; water is a product (S->4). NaHCO3 + heat -> Na2CO3 + H2O + CO2 (not in the products list as a standalone answer). Answer: P->2, Q->1, R->3, S->4.
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