A definite volume of H 2 O 2 under going spontaneous decomposition required 25 ml of standard permanganate solution for titration. After 10 and 20 minutes respectively the volumes of permanganate

Question

A definite volume of H 2 ​ O 2 ​ under going spontaneous decomposition required 25 ml of standard permanganate solution for titration. After 10 and 20 minutes respectively the volumes of permanganate required were 15 ml and 9 ml , respectively. Calculate the fraction of H 2 ​ O 2 ​ decomposed after 25 minutes [Given : ( 0.6 ) 2.5 = 0.28 ]

TestHub · Accepted Answer

For a first-order reaction: k = t 2.303 ​ lo g a − x a ​ Permanganate volume ∝ amount of H₂O₂ left. At 10 min: a − x a ​ = 15 25 ​ k = 10 2.303 ​ lo g 15 25 ​ At 20 min: 15 9 ​ = ( 25 15 ​ ) 2 = ( 0.6 ) 2 Therefore, after 25 min: [ H 2 ​ O 2 ​ ] 0 ​ [ H 2 ​ O 2 ​ ] 25 ​ ​ = ( 0.6 ) 2.5 = 0.28 Fraction decomposed: = 1 − 0.28 = 0.72 Fraction of H₂O₂ decomposed = 0.72

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