Question

\[aA+bB\to P\];\[\dfrac{dx}{dt}=k{{[A]}^{a}}{{[B]}^{b}}\]. If conc. of A is doubled, rate is doubled. If B is doubled, rate becomes four times. Which of the following is correct?
(A) \[-\dfrac{d[A]}{dt}=-\dfrac{d[B]}{dt}\]
(B) \[-\dfrac{d[A]}{dt}=-2\dfrac{d[B]}{dt}\]
(C) \[-\dfrac{2d[A]}{dt}=-\dfrac{d[B]}{dt}\]
(D) None of the above

Answer 1

Hint: Here the rate of reaction depends on concentrations. And for finding the relation between rate of A and B we have to find their stoichiometric coefficient by using given information between rate of reaction given in terms of concentration and by using given data.

Step by step solution: The given reaction is an elementary reaction because here sum stoichiometric coefficient is equal to the order of the reaction.
Given expression for rate of reaction:
\[\dfrac{dx}{dt}=k{{[A]}^{a}}{{[B]}^{b}}\] ….(1)
Here, [A] is concentration of A and [B] is concentration of B.
Also given that if,
If concentration of A is doubled then rate also get doubled
\[2\dfrac{dx}{dt}=k{{[2A]}^{a}}{{[B]}^{b}}\] ….(2)
Now the equation (1) / (2)
\[\dfrac{1}{2}=\dfrac{1}{{{2}^{a}}}\]
So, the value of a = 1;
And also given that B is doubled then rate becomes four times.
\[4\dfrac{dx}{dt}=k{{[A]}^{a}}{{[2B]}^{b}}\] …..(3)
Equation (1) / (3):
\[\dfrac{1}{4} = \dfrac{1}{{{2}^{b}}}\]
So, the value of b = 2
We know that:
\[\dfrac{-\dfrac{d[A]}{dt}}{a}\]=\[\dfrac{-\dfrac{d[B]}{dt}}{b}\] …..(4)
Now putting value of a and b in the equation (4):
\[\dfrac{-\dfrac{d[A]}{dt}}{1}\]=\[\dfrac{-\dfrac{d[B]}{dt}}{2}\]
\[-2\dfrac{d[A]}{dt}=-\dfrac{d[B]}{dt}\]

So, from the above derivation and calculation we can say that the correct answer is option “B”.

Note: Here you have to remember the relation between rate of consumption of A, rate of consumption of B and overall rate of reaction. Negative sign is given because of consumption of the reactant.