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# For Zero order reactions, the linear plot was obtained for $\left[ A \right]$ vs $t$ . The slope of the line is equal to:A) ${k_o}$B) $- {k_o}$C) $\dfrac{{0.693}}{{{k_0}}}$D) $- \dfrac{{{k_0}}}{{2.303}}$

Last updated date: 20th Jun 2024
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Hint:We know that the rate of order is determined by the differential rate law or the integrated rate law, rate of reaction is speed of chemical reaction proceeds measure of the change in concentration product or the change in concentration of the reactants.
If the kinetics of the reaction depends on the concentration of only one reactant then the order of the reaction is first order present, but each reactant will be zeroth-order.

We need to know that the integrated rate law for a zero-order reaction is frequently written in two different ways: one using exponents and one using logarithms. The exponential form is as follows:
$\left[ A \right] = {\left[ A \right]_0}{e^{ - kt}}$
Initial concentration of reactant A at $t = 0$ is denoted as $\left[ {{A_0}} \right]$ ,
K is that the rate constant, and e is that the base of the natural logarithms, which has the worth $2.718$ to $3$ decimal places. Integrating on both sides and finding the value of C we get,
$\left[ A \right] = - {k_0}t + \left[ {{A_0}} \right]$
Where,
$- {k_o}$ is rate constant.
$\left[ {{A_0}} \right]$ is the initial concentration. So the slope of the line is $- {k_o}$ .
Hence option B is correct.

We also know that for chemical reaction $A{\text{ }} + {\text{ }}B{\text{ }}C{\text{ }} \to {\text{ }}E\;$ , the rate of the reaction is doubled when the concentration of B was doubled, if the concentration, of both A and B was doubled rate become doubled and if the concentration of both B and C was doubled rate become quadrupled. The order of the reaction with respect to A, B and C and the total order given as,
The Total sum of order $= 0 + 1 + 1 = 2$