For a first order reaction, the plot of $^{'} t^{'}$ against $l o g C$ gives a straight line with slope equal to:

(k / 2.2303)

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(- k / 2.303)

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(I n k / 2.303)

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- k

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Solution

The correct option is B $(- k / 2.303)$
For a first-order reaction,
$l o g (a - x) = - \frac{k t}{2.303} + l o g a$

It is similar to straight-line equation $y = m x + c$

where intercept is equal to $l o g a$ and slope equal to $- \frac{k}{2.303}$

Option B is the correct answer.

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Similar questions

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Q. For a first-order reaction, the plot of t against log[a-x] gives a straight line with slope equal to:

If [A] is the concentration of reactant A at any time t and [A_o] is the concentration at t=0, then for the first order reaction the rate equation can be written as:

Q. Rate constant

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of a reaction varies with temperature according to the equation

log k = c o n s t a n t - \frac{E_{a}}{2.303 R} \times \frac{1}{T}

where

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(

R = 8.314 J K^{- 1} {m o l}^{- 1})

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For a first order reaction, the plot of ′t′ against logC gives a straight line with slope equal to:

The correct option is B (−k/2.303)For a first-order reaction, log(a−x)=−kt2.303+logaIt is similar to straight-line equation y=mx+cwhere intercept is equal to loga and slope equal to −k2.303Option B is the correct answer.

For a first order reaction, the plot of $^{'} t^{'}$ against $l o g C$ gives a straight line with slope equal to:

The correct option is B $(- k / 2.303)$
For a first-order reaction,
$l o g (a - x) = - \frac{k t}{2.303} + l o g a$

It is similar to straight-line equation $y = m x + c$

where intercept is equal to $l o g a$ and slope equal to $- \frac{k}{2.303}$

Option B is the correct answer.