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First Order Reaction

t 1 / 4 can b...

Question

$t_{1 / 4}$ can be taken as the time taken for the concentration of a reactant to drop to 3/4 of its value. If the rate constant for a first order reaction is k, the $t_{1 / 4}$ can be written as [ln2 = 0.695, ln = 1.1]

A

0.69 / k

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B

0.75 / k

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C

0.10 / k

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D

0.29 / k

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Solution

The correct option is D 0.29 / k
$t_{\frac{1}{4}} = \frac{2.303}{K} l o g \frac{4}{3} = \frac{2.303}{K O} l o g 4 - l o g 3 = \frac{0.29}{K}$

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Similar questions

t_{1 / 4}

can be taken as the time taken for the concentration of a reactant to drop to

\frac{3}{4}

of its value. If the rate constant for a first order reaction is

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t_{1 / 4}

can be written as:

Q. The rate constant for a first order reaction is

k = 7 \times 10^{- 4} s^{- 1}

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Q. First order decomposition of

R ⟶ P

The time taken for its decomposition to

{\frac{1}{2}}^{t h}

{\frac{3}{4}}^{t h}

of its initial concentration are

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t_{\frac{3}{4}}

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\frac{[t_{1 / 2}]}{[t_{3 / 4}]} \times 10 ? (l o g_{10} 2 = 0.3)

Q. First order decomposition of

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The time taken for its decomposition to

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{\frac{3}{4}}^{t h}

of its initial concentration are

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