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 start and after time- t- For a first order reaction-

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Factors Affecting Rate of 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 start and 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 t

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Solution

The correct option is B $\frac{d f}{d t} = k (1 - f)$
We need to substitute the differential rate law:
$f = 1 - \frac{C}{C_{0}}$
$\frac{d f}{d t} = \frac{d}{d t} (1 - \frac{C}{C_{0}})$
$\frac{d f}{d t} = \frac{\frac{d C}{d t}}{C_{0}} = k \frac{C}{C_{0}} = k (1 - f)$

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Similar questions

Q. In a reaction involving one single reactant, the fraction of the reactant consumed may be defined as

f = (1 - \frac{C}{C_{0}})

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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 start and after time, t. For a first order reaction.

The correct option is B dfdt=k(1−f)We need to substitute the differential rate law:f=1−CC0dfdt=ddt(1−CC0)dfdt=dCdtC0=kCC0=k(1−f)

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 start and after time, $t$ . For a first order reaction.

The correct option is B $\frac{d f}{d t} = k (1 - f)$
We need to substitute the differential rate law:
$f = 1 - \frac{C}{C_{0}}$
$\frac{d f}{d t} = \frac{d}{d t} (1 - \frac{C}{C_{0}})$
$\frac{d f}{d t} = \frac{\frac{d C}{d t}}{C_{0}} = k \frac{C}{C_{0}} = k (1 - f)$