Question

# For the reaction of the Zeroth order, variation of log ${T_{50}}$ with log a (where ${T_{50}}$ is half-life period and a is the initial concentration) is given by:A)B)C)D)

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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.

Complete step by step answer:
We have to remember 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. The half-life of zero order reaction is given by,
${t_{1/2}} = \log \dfrac{{\left[ {{A_o}} \right]}}{{2k}} = \log \left( {\dfrac{1}{{2k}}} \right) + \log \left( {{{\left[ A \right]}_0}} \right)$
Thus the option A is correct.

In a specific response 10% of the reactant decays in a single hour, 20% in two hours, and 30% in three hours and proceeds. Since rate isn't reliant on focus, we can say the response follows zero order.
A response wherein contains that autonomous of the reactant(s) fixation is known as zero request response. At the point when we increment (or) lessen the responding species focus, the pace of the response would either back off (or) accelerate.
$Rate = k{\left[ A \right]^ \circ }$
Where, the rate constant is given as k and A is the concentration of the reactant species.
For the question we can understand that ${t_{1/2}} \propto a$. The dimension of the reaction will be,
$k = mo{l^{ - 1}}{L^{ - 1}}h{r^{ - 1}}$
For a zero order reaction, we can say $t = \dfrac{x}{k}$ (or) $k = \dfrac{x}{t}$
If take the value of t as 10%, ${t_{10\% }} = 10\min$, x=10
Then k the value of is equal to, $k = \dfrac{{10}}{{10}} = 1mol{L^{ - 1}}{t^{ - 1}}$
If take the value of t as 20%, ${t_{20\% }} = 20\min$, x=20
Then k the value of is equal to, $k = \dfrac{{20}}{{20}} = 1mol{L^{ - 1}}{t^{ - 1}}$
If take the value of t as 30%, ${t_{30\% }} = 30\min$, x=30
Then k the value of is equal to, $k = \dfrac{{30}}{{30}} = 1mol{L^{ - 1}}{t^{ - 1}}$ and it goes on.
Thus, the reaction follows zero-order kinetics.

Note:
A substitute strategy to decide the rate of the reaction is graphical technique. We can plot a diagram of percent disintegration to time, and can get a straight line plot. The straight line plot shows the response is zero order. The unit of zero request response can likewise be $mol{L^{ - 1}}\,{\sec ^{ - 1}}$ (or)$M/\sec$. We can discover zero-request responses when a substance that is required for the response to continue, similar to a surface or an impetus, is immersed by the reactants.