
Derive an integrated rate equation for the rate constant of a zero-order reaction.
Answer
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Hint: An equation that shows the dependence of the rate of reaction on the concentration of reactants is known as differential rate equation and its integration is known as an integrated rate equation. Here in this question, one has to find out for zero-order reaction in which the rate of the reaction depends upon the zeroth power of concentration of the reactant species. By using this data one can elaborate and derive the equation for the zero-order reaction.
Complete step by step answer: 1) First of all let's understand the concept of integrated rate equation where the equation shows the dependence of the rate of reaction on the concentration of reactants is known as differential rate equation and its integration is known as integrated rate equation.
2) Now let's calculate the integrated rate equation for the rate constant of a zero-order reaction by starting with the general rate equation as follows,
$A\xrightarrow{{}}B$
${\text{Rate of reaction = - }}\dfrac{{d\left[ A \right]}}{{dt}} = k{\left[ A \right]^0}$
As ${\left[ A \right]^0} = 1$ the equation becomes,
$ - \dfrac{{d\left[ A \right]}}{{dt}} = k$
Which we can write as follows,
$d\left[ A \right] = - kdt$
When we integrate both sides to get the integrated equation,
$\int {d\left[ A \right]} = - k\int {dt} $
$\left[ A \right] = - kt + I$ ……..equation$(1)$
Where, $I$ is used as an integration constant.
When the $t = 0$ then the value of $\left[ A \right] = {\left[ A \right]^0}$
Let's calculate ${\left[ A \right]^0}$ by putting it in the equation $(1)$,
${\left[ A \right]^0} = - k \times 0 \times I$
${\left[ A \right]^0} = I$
Let's substitute the value of $I$ in equation $(1)$,
$\left[ A \right] = - kt + {\left[ A \right]^0}$
$kt = {\left[ A \right]^0} - \left[ A \right]$
$k = \dfrac{{{{\left[ A \right]}^0} - \left[ A \right]}}{t}$
This is an integrated rate equation for the rate constant for a zero-order reaction.
Note: The proved equation above is called an integrated rate equation for the zero-order reactions. We can observe the above equation as in a graph plotted with the concentration of reactant on $Y - $ axis and time on $X - $ axis which gives a straight line for zero-order reaction. The slope drawn on the straight line gives us the value of the rate constant i.e. $k$
Complete step by step answer: 1) First of all let's understand the concept of integrated rate equation where the equation shows the dependence of the rate of reaction on the concentration of reactants is known as differential rate equation and its integration is known as integrated rate equation.
2) Now let's calculate the integrated rate equation for the rate constant of a zero-order reaction by starting with the general rate equation as follows,
$A\xrightarrow{{}}B$
${\text{Rate of reaction = - }}\dfrac{{d\left[ A \right]}}{{dt}} = k{\left[ A \right]^0}$
As ${\left[ A \right]^0} = 1$ the equation becomes,
$ - \dfrac{{d\left[ A \right]}}{{dt}} = k$
Which we can write as follows,
$d\left[ A \right] = - kdt$
When we integrate both sides to get the integrated equation,
$\int {d\left[ A \right]} = - k\int {dt} $
$\left[ A \right] = - kt + I$ ……..equation$(1)$
Where, $I$ is used as an integration constant.
When the $t = 0$ then the value of $\left[ A \right] = {\left[ A \right]^0}$
Let's calculate ${\left[ A \right]^0}$ by putting it in the equation $(1)$,
${\left[ A \right]^0} = - k \times 0 \times I$
${\left[ A \right]^0} = I$
Let's substitute the value of $I$ in equation $(1)$,
$\left[ A \right] = - kt + {\left[ A \right]^0}$
$kt = {\left[ A \right]^0} - \left[ A \right]$
$k = \dfrac{{{{\left[ A \right]}^0} - \left[ A \right]}}{t}$
This is an integrated rate equation for the rate constant for a zero-order reaction.
Note: The proved equation above is called an integrated rate equation for the zero-order reactions. We can observe the above equation as in a graph plotted with the concentration of reactant on $Y - $ axis and time on $X - $ axis which gives a straight line for zero-order reaction. The slope drawn on the straight line gives us the value of the rate constant i.e. $k$
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