Question

In the given reaction $aA{{ }} + {{ }}bB{{ }} \to {{ }}Product$ , $dx/dt{{ }} = {{ }}k{\left[ A \right]^a}{\left[ B \right]^b}$ . If concentration of A is doubled, the rate is four times. If concentration of B is made four times, the rate is doubled. What is the relation between the rate of disappearance of A and that of B?
A. - \left\\{ {d\left[ A \right]/dt} \right\\}{{ }} = {{ }} - \left\\{ {d\left[ B \right]/{{ }}dt} \right\\}
B. - \left\\{ {d\left[ A \right]/dt} \right\\}{{ }} = {{ }} - \left\\{ {4d\left[ B \right]/{{ }}dt} \right\\}
C. - \left\\{ {4d\left[ A \right]/dt} \right\\}{{ }} = {{ }} - \left\\{ {d\left[ B \right]/{{ }}dt} \right\\}
D. None of these

Answer 1

In order to solve this we need to find the values of $a$ and $b$ using the set of information which is already provided in the question and then equate in the rate expression of disappearance of reactants.

Formula used:
Rate of disappearance of $A = {{ }} - {{ }}1/a{{ \\{ }}d\left[ A \right]/dt\\}$
And Rate of disappearance of $B{{ }} = {{ }} - 1/b{{ \\{ }}d\left[ B \right]/dt\\}$
And we know that $R =$ Rate of disappearance of $A =$ Rate of disappearance of $B$
Where, $a$ and $b$ are the stoichiometric coefficients of $A$ and $B$ respectively, and $dt$ is the change in time, $R$ is the reaction rate.

Complete step by step answer:
In the question the equation given to us is $aA{{ }} + {{ }}bB{{ }} \to {{ }}Product$
And for this the rate given is
$dx/dt{{ }} = {{ }}k{\left[ A \right]^a}{\left[ B \right]^{b\;}} = {{ }}R{{ }}\left( {rate} \right)$
Now if we double the concentration of $A$ , rate becomes $4$ times, i.e.,
$4R = {{ }}{\left[ {2A} \right]^a}{\left[ B \right]^b}$
This is possible when $a = 2$
Also, rate becomes double when $B$ is increased to four times
$2R = {{ }}{\left[ A \right]^a}{\left[ {4B} \right]^b}$
This is possible only when in the rate equation $b = 1/2$
Now we know that for the given reaction
Rate of disappearance of $A = {{ }} - {{ }}1/a{{ \\{ }}d\left[ A \right]/dt\\}$
And Rate of disappearance of $B{{ }} = {{ }} - 1/b{{ \\{ }}d\left[ B \right]/dt\\}$
And we know that $R =$ Rate of disappearance of $A =$ Rate of disappearance of $B$
Therefore $- {{ }}1/a{{ \\{ }}d\left[ A \right]/dt\\} {{ }} = {{ }} - 1/b{{ \\{ }}d\left[ B \right]/dt\\}$
Putting the values of $a$ and $b$ , we get
$- {{ }}1/2{{ \\{ }}d\left[ A \right]/dt\\} {{ }} = {{ }} - 2{{ \\{ }}d\left[ B \right]/dt\\}$
Hence, $- d\left[ A \right]/dt{{ }} = {{ }} - 4{{ \\{ }}d\left[ B \right]/dt\\}$

So, Option D is correct.

Note: The rate of a reaction generally increases with the increase in the concentrations of the reactants and as per the law of mass action the former is directly proportional to the latter, rate is also highly dependent upon the type and the nature of the particular reaction.
The rate of disappearance is shown with a negative sign because this sign is indicative of decrease in value since as the reaction proceeds the concentration of the reactant decreases.