A chemical reaction proceeds into the following steps,
Step I, $2A \rightleftarrows X\left( {fast} \right)$
Step II, $X + B \rightleftarrows Y\left( {slow} \right)$
Step III, $Y + BA \rightleftarrows X\left[ {Product} \right]\left( {fast} \right)$
The rate law for the overall reaction is:
A.$Rate = k{\left[ A \right]^2}$
B.$Rate = k{\left[ B \right]^2}$
C.$Rate = k\left[ A \right]\left[ B \right]$
D.$Rate = k{\left[ A \right]^2}\left[ B \right]$
Answer
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Hint: We have to first identify the rate determining step and the rate law expression is written. We have to know that if any concentration term of an intermediate is found, then a suitable expression from the equilibrium step is used to express the intermediate concentration.
Complete answer:
Let us now consider Step I, $2A \rightleftarrows X\left( {fast} \right)$
We can write the expression for the forward reaction rate. The expression is written as,
$r = k{\left[ A \right]^2}$.. (1)
We can write the expression for the reverse reaction rate. The expression is written as,
$r = k\left[ X \right]$…… (2)
When a reaction is in equilibrium, the rate of forward reaction is the same as the rate of reverse reaction. So, let us now equate the equation (1) and equation (2).
$k{\left[ A \right]^2} = {k^{'}}\left[ X \right]$
$\left[ X \right] = \dfrac{k}{{{k^{'}}}}{\left[ A \right]^2}$
$\left[ X \right] = K{\left[ A \right]^2}$ …… (3)
Here, we have to know that K is the equilibrium constant for Step (1).
Let us take the slow step (or) rate determining step, i.e. the step (2).
$X + B \rightleftarrows Y\left( {slow} \right)$
We can express the rate of the reaction as,
$r = {k^{''}}\left[ X \right]\left[ B \right]$…… (4)
Let us now substitute equation (3) in equation (4)
$r = {k^{''}}K{\left[ A \right]^2}\left[ B \right]$
$r = k{\left[ A \right]^2}\left[ B \right]$
Here, $k = {k^{''}}K$ is the rate constant for the overall reaction.
So, we can write the rate law for the overall reaction as $Rate = k{\left[ A \right]^2}\left[ B \right]$.
Option (D) is correct.
Note:
We can say that the rate of chemical reaction indicates the speed at which the formation of products occurs from the reactants. We have to know that when there are different steps found in the reaction, so the rate of overall reaction is similar to the rate of rate determining step. We have to know that the slowest step of reaction is the rate determining step of the reaction.
Complete answer:
Let us now consider Step I, $2A \rightleftarrows X\left( {fast} \right)$
We can write the expression for the forward reaction rate. The expression is written as,
$r = k{\left[ A \right]^2}$.. (1)
We can write the expression for the reverse reaction rate. The expression is written as,
$r = k\left[ X \right]$…… (2)
When a reaction is in equilibrium, the rate of forward reaction is the same as the rate of reverse reaction. So, let us now equate the equation (1) and equation (2).
$k{\left[ A \right]^2} = {k^{'}}\left[ X \right]$
$\left[ X \right] = \dfrac{k}{{{k^{'}}}}{\left[ A \right]^2}$
$\left[ X \right] = K{\left[ A \right]^2}$ …… (3)
Here, we have to know that K is the equilibrium constant for Step (1).
Let us take the slow step (or) rate determining step, i.e. the step (2).
$X + B \rightleftarrows Y\left( {slow} \right)$
We can express the rate of the reaction as,
$r = {k^{''}}\left[ X \right]\left[ B \right]$…… (4)
Let us now substitute equation (3) in equation (4)
$r = {k^{''}}K{\left[ A \right]^2}\left[ B \right]$
$r = k{\left[ A \right]^2}\left[ B \right]$
Here, $k = {k^{''}}K$ is the rate constant for the overall reaction.
So, we can write the rate law for the overall reaction as $Rate = k{\left[ A \right]^2}\left[ B \right]$.
Option (D) is correct.
Note:
We can say that the rate of chemical reaction indicates the speed at which the formation of products occurs from the reactants. We have to know that when there are different steps found in the reaction, so the rate of overall reaction is similar to the rate of rate determining step. We have to know that the slowest step of reaction is the rate determining step of the reaction.
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