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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Chemistry
Chapter
Chemical Kinetics

BITSAT 2014 · Chemistry · Chemical Kinetics

Two substances R and S decompose in solution independently, both...

Two substances R\mathrm{R} and S\mathrm{S} decompose in solution independently, both following first order kinetics. The rate constant of R\mathrm{R} is twice that of S\mathrm{S}. In an experiment, the solution initially contained 0.5 millimoles of R\mathrm{R} and 0.25 of S\mathrm{S}. The molarities of R\mathrm{R} and S\mathrm{S} will be equal just at the end of time equal to

Options

  1. Atwice the half life of R\mathrm{R}
  2. Btwice the half life of SS
  3. Cthe half life of S\mathrm{S}
  4. Dthe half life of R\mathrm{R}

Correct answer

A. twice the half life of R\mathrm{R}

Step-by-step solution

Substance R\mathrm{R} Substance S2kk\mathrm{S}2 \mathrm{k} \quad \mathrm{k} rate constant t1/2=n×t1/2\mathrm{t}_{1/2}=\mathrm{n} \times \mathrm{t}_{1/2} where n=\mathrm{n}= number of half life period Amount of R\mathrm{R} left =0.5(2)T/t/2=\frac{0.5}{(2)^{\mathrm{T} / \mathrm{t}_{/ 2}}} Amount of S\mathrm{S} left =0.25(2)T/2t/2=\frac{0.25}{(2)^{\mathrm{T} / 2 \mathrm{t} / 2}} Equating both 0.50.25=(2)T/t1/2(2)T/2t1/2\frac{0.5}{0.25}=\frac{(2)^{\mathrm{T} / \mathrm{t}_{1/2}}}{(2)^{\mathrm{T} / 2 \mathrm{t}_{1/2}}} or 2=(2)T/t1/2∴T=2t1/2⋅2t1/22=(2)^{\mathrm{T} / \mathrm{t}_{1/2}}\therefore \mathrm{T}=2 \mathrm{t}_{1/2} \cdot 2 \mathrm{t}_{1/2} is half life of S\mathrm{S} and twice the half -life of R\mathrm{R}
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