The integrated rate equation for first-order reaction is $A$ $\to$ products.

K = \frac{2.303}{t} l o g \frac{[A]_{0}}{[A]_{t}}

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K = - \frac{1}{t} l n \frac{[A]_{t}}{[A]_{0}}

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K = \frac{2.303}{t} l o g_{10} \frac{[A]_{t}}{[A]_{0}}

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K = \frac{1}{t} l n \frac{[A]_{t}}{[A]_{0}}

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Solution

The correct option is A
$K = \frac{2.303}{t} l o g \frac{[A]_{0}}{[A]_{t}}$

For the first reaction, $A$ $\to$ Products, the integrated rate law is given by the equation.

$K = \frac{2.303}{t} l o g \frac{[A]_{0}}{[A]_{t}}$

where,
$[A]_{0}$ is the initial concentration of the reactant at $t = 0$

$[A]_{t}$ is the concentration of the reactant at t $=$ t.

So, the correct option is $A$ .

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Similar questions

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. If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by:

Q. Match the column I with column II and mark the appropriate choice.

Column I	Column II
(A) Zero order	(i) $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$
(B) First order	(ii) $Δ H > 0$
(C) Endothermic reaction	(iii) $k = \frac{2.303}{t} log \frac{[A]_{0}}{[A]}$
(D) Activation energy	(iv) $k = \frac{1}{t} ([A]_{0} - [A])$

Q. For a first order reaction, the plot of ‘t’ against log C gives a straight line with slope equal to:

Q. If a reaction obeys the following equation

k = \frac{2.303}{t} l o g \frac{a}{a - x}

then the order is:

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The integrated rate equation for first-order reaction is A → products.

The correct option is A K=2.303tlog[A]0[A]tFor the first reaction, A → Products, the integrated rate law is given by the equation.K=2.303tlog[A]0[A]twhere, [A]0 is the initial concentration of the reactant at t=0[A]t is the concentration of the reactant at t = t.So, the correct option is A.

The integrated rate equation for first-order reaction is $A$ $\to$ products.