In a reaction involving one single reactant- the fraction of the reactant consumed may be defined as f-1-CC0 where C0 and C are the concentrations of the reactant at the after time- t- For a first order reaction-

The correct option is A dfdt=k(1 f) Integrated rate equation for 1st order reaction K=2.303tlog[A]_1[A]_2 Where [A]_1 is the initial concentrati ...

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Standard XII

Chemistry

First Order Reaction

In a reaction...

Question

In a reaction involving one single reactant, the fraction of the reactant consumed may be defined as $f = (1 - \frac{C}{C_{0}})$ where $C_{0}$ and $C$ are the concentrations of the reactant at the after time, t. For a first order reaction:

\frac{d f}{d t} = k (1 - f)

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- \frac{d f}{d t} = k f

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- \frac{d f}{d t} = k (1 - f)

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\frac{d f}{d t} = k f

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Solution

The correct option is A $\frac{d f}{d t} = k (1 - f)$
Integrated rate equation for 1st order reaction
$K = \frac{2.303}{t} log \frac{[A]_{1}}{[A]_{2}}$
Where $[A]_{1}$ is the initial concentration and $[A]_{2}$ is the concentration of A at time t
Given initial concentration is $C_{0}$ and concentration at time t is C
Integrating option A we get
$ln (1 - f) = - k t$
given $f = (1 - \frac{C}{C_{0}})$
$l n \frac{[C_{0}]}{[C]} = k t$
$K = \frac{2.303}{t} log \frac{[C_{0}]}{[C]}$

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In a reaction involving one single reactant, the fraction of the reactant consumed may be defined as f=(1−CC0) where C0 and C are the concentrations of the reactant at the after time, t. For a first order reaction:

In a reaction involving one single reactant, the fraction of the reactant consumed may be defined as $f = (1 - \frac{C}{C_{0}})$ where $C_{0}$ and $C$ are the concentrations of the reactant at the after time, t. For a first order reaction: