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Chemical Kinetics

Question
CBSEENCH12010617

Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:

  • 6.93×10−4 mol min−1

  • 2.66 L min−1 at STP

  • 1.34×10−2 mol min−1

  • 6.93×10−2 mol min−1

Solution

A.

6.93×10−4 mol min−1

straight k space equals space open parentheses fraction numerator 0.693 minus 1 over denominator 25 end fraction close parentheses space min to the power of negative 1 end exponent

or space fraction numerator straight d left square bracket straight O subscript 2 right square bracket over denominator dt end fraction space equals space minus 1 half fraction numerator straight d left square bracket straight H subscript 2 straight O subscript 2 right square bracket over denominator dt end fraction
space equals space fraction numerator straight k left square bracket straight H subscript 2 straight O subscript 2 right square bracket over denominator 2 end fraction space equals 6.93 space straight x space 10 to the power of negative 4 end exponent space mol space min to the power of negative 1 end exponentFor first order reaction
straight k space equals fraction numerator 2.303 over denominator straight t end fraction log space fraction numerator straight a over denominator straight a minus straight x end fraction
Given colon
straight t equals space 50 space min comma
straight a equals space 0.5 straight M
straight a minus straight x space equals space 0.125 space straight M

therefore space straight k space equals space fraction numerator 2.303 over denominator 50 end fraction space log fraction numerator 0.5 over denominator 0.125 end fraction space equals space 0.0277 space min to the power of negative 1 end exponent
Now comma space as space per space reaction

2 straight H subscript 2 straight O subscript 2 space rightwards arrow with space on top space 2 straight H subscript 2 straight O space plus space straight O subscript 2
fraction numerator negative 1 over denominator 2 end fraction fraction numerator straight d left square bracket straight H subscript 2 straight O subscript 2 right square bracket over denominator dt end fraction space equals space 1 half fraction numerator straight d left square bracket straight H subscript 2 straight O right square bracket over denominator dt end fraction space equals space fraction numerator straight d left square bracket straight O subscript 2 right square bracket over denominator dt end fraction
Rate space of space reaction comma space
fraction numerator negative straight d left square bracket straight H subscript 2 straight O subscript 2 right square bracket over denominator dt end fraction space equals space straight k left square bracket straight H subscript 2 straight O subscript 2 right square bracket
therefore space fraction numerator straight d left square bracket straight O subscript 2 right square bracket over denominator dt end fraction space equals space minus space 1 half fraction numerator straight d left square bracket straight H subscript 2 straight O subscript 2 right square bracket over denominator dt end fraction space equals space 1 half straight k left square bracket straight H subscript 2 straight O subscript 2 right square bracket space... space left parenthesis straight i right parenthesis
When space the space concentration space of space straight H subscript 2 straight O subscript 2 space reaches space 0.05 space straight M comma
fraction numerator straight d left square bracket straight O subscript 2 right square bracket over denominator dt end fraction space equals space 1 half space straight x space 0.0277 space straight x space 0.05 space space space left square bracket from space equ space left parenthesis straight i right parenthesis right square bracket
or space
fraction numerator straight d left square bracket straight O subscript 2 right square bracket over denominator dt end fraction space equals space 6.93 space straight x space 10 to the power of negative 4 end exponent space mol space min to the power of negative 1 end exponent
Alternative Method:
If fifty minutes, the concentration of H2O2 decreases from 0.5 to 0.125 M or in one half-life, concentration of H2O2 decreases from 0.5 to 0.25 M. In two half-lives, concentration of H2O2 decreases from 0.5 to 0.125 M or 2t1/2 = 50 min
t1/2 = 25 min