In a first order reaction- the concentration of the reactant decreases from 800 mol dm -3 to 50 mol dm -3 in 2 - 104 s- The rate constant of the reaction in s -1 jIIT-JEE- 2003A- 2 - 104B- 3-45 - 10-5- 1-386 - 10-4D- 2 - 10-4

The correct option is C 1.386 × 10^ 4k= 2.303/tlog [A]_0/[A]_1 = log 800/50 = 1.386 × 10^4 s^ 1 ...

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Standard XII

Chemistry

Half life

In a first or...

Question

In a first order reaction, the concentration of the reactant decreases from 800 mol $d m^{- 3}$ to 50 mol $d m^{- 3}$ in $2 \times 10^{4} s .$ The rate constant of the reaction in $(s^{- 1})$ is
(IIT-JEE, 2003)

2 \times 10^{4}

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3.45 \times 10^{- 5}

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1.386 \times 10^{- 4}

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2 \times 10^{- 4}

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Solution

The correct option is C $1.386 \times 10^{- 4}$
$k = \frac{2.303}{t} l o g \frac{[A]_{0}}{[A]_{1}} = l o g \frac{800}{50} = 1.386 \times 10^{4} s^{- 1}$

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In a first order reaction, the concentration of the reactant decreases from 800 mol $d m^{- 3}$ to 50 mol $d m^{- 3}$ in $2 \times 10^{4} s .$ The rate constant of the reaction in $(s^{- 1})$ is
(IIT-JEE, 2003)

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Q. In a first order reaction the concentration of reactant decreases from 800

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In a first order reaction, the concentration of the reactant decreases from 800 mol dm−3 to 50 mol dm−3 in 2×104s. The rate constant of the reaction in (s−1) is (IIT-JEE, 2003)

The correct option is C 1.386×10−4k=2.303tlog[A]0[A]1=log80050=1.386×104 s−1

In a first order reaction, the concentration of the reactant decreases from 800 mol $d m^{- 3}$ to 50 mol $d m^{- 3}$ in $2 \times 10^{4} s .$ The rate constant of the reaction in $(s^{- 1})$ is
(IIT-JEE, 2003)

The correct option is C $1.386 \times 10^{- 4}$
$k = \frac{2.303}{t} l o g \frac{[A]_{0}}{[A]_{1}} = l o g \frac{800}{50} = 1.386 \times 10^{4} s^{- 1}$