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BITSAT 2014 Paper

Section: Chemistry

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Question : 63 of 150

Marks: +1, -0

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

twice the half life of R
twice the half life of S
the half life of S
the half life of R

Solution:

Substance R

2k rate constant

t_{1/2}

Half life period

Substance S

k rate constant

2 t_{1/2}

Half life period

T = n ×

t_{1/2}

where n = number of half life period

Amount of R left =

\frac{0.5}{(2)^{\frac{T}{t_{1/2}}}}

;

Amount of S left =

\frac{0.25}{(2)^{\frac{T}{2t_{1/2}}}}

Equating both

\frac{0.5}{0.25}

=

\frac{(2)^{\frac{T}{t_{1/2}}}}{(2)^{\frac{T}{2t_{1/2}}}}

or 2 =

(2)^{\frac{T}{t_{1/2}}}

∴ T =

2 t_{1/2}

.

2t_{1/2}

s half life of S and twice the half -life of R

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