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Question: For the reaction, \[3A+2B\to C+D\], the differential rate law can be written as: A.\[3\dfrac{d[A]...

For the reaction, 3A+2B→C+D3A+2B\to C+D, the differential rate law can be written as:
A.3d[A]dt=d[C]dt=k[A]n[B]m3\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}}
B.−d[A]dt=d[C]dt=k[A]n[B]m-\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}}
C.+13d[A]dt=d[C]dt=k[A]n[B]m+\dfrac{1}{3}\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}}
D.−13d[A]dt=d[C]dt=k[A]n[B]m-\dfrac{1}{3}\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}}

Explanation

Solution

Hint: To solve this question, you need to apply the concepts of chemical kinetics. The differential rate law can be related to the rate of appearance (of product) and the rate of disappearance (of reactant).

Complete step by step answer:
The general formula for appearance or disappearance of a compound is calculated by the formula –
±1stoichiometric coefficientxreactant/producttime\pm \dfrac{\text{1}}{\text{stoichiometric coefficient}}\text{x}\dfrac{\text{reactant/product}}{\text{time}}
The rate of disappearance has a ‘-’ sign, because the rate decreases with time.
Rate of appearance of C =+1stoichiometric coefficientxproducttime=+\dfrac{\text{1}}{\text{stoichiometric coefficient}}\text{x}\dfrac{\text{product}}{\text{time}}
Rate of appearance of C =+11xd !![!! C !!]!! dt=+\dfrac{\text{1}}{1}\text{x}\dfrac{\text{d }\\!\\![\\!\\!\text{ C }\\!\\!]\\!\\!\text{ }}{\text{dt}}
Rate of disappearance of A = −1stoichiometric coefficientxreactanttime-\dfrac{\text{1}}{\text{stoichiometric coefficient}}\text{x}\dfrac{\text{reactant}}{\text{time}}
Rate of disappearance of A =−13xd !![!! A !!]!! dt=-\dfrac{\text{1}}{3}\text{x}\dfrac{\text{d }\\!\\![\\!\\!\text{ A }\\!\\!]\\!\\!\text{ }}{\text{dt}}
The rate of appearance and disappearance is always the same.
Therefore, −13d !![!! A !!]!! dt=d !![!! C !!]!! dt-\dfrac{\text{1}}{3}\dfrac{\text{d }\\!\\![\\!\\!\text{ A }\\!\\!]\\!\\!\text{ }}{\text{dt}}=\dfrac{\text{d }\\!\\![\\!\\!\text{ C }\\!\\!]\\!\\!\text{ }}{\text{dt}}
Now, let us write the rate law –
Rate law = k=[A]n[B]mk={{[A]}^{n}}{{[B]}^{m}}, where, n and m are the stoichiometric coefficients of the reacting elements.
For the given equation, the value of n and m are 3 and 2 respectively.
So, we can combine the two equations and write the rate of the reaction as –
−13d[A]dt=d[C]dt=k[A]n[B]m-\dfrac{1}{3}\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k{{[A]}^{n}}{{[B]}^{m}}
Therefore, the answer is – option (d).

Additional Information: The word kinetic is derived from the Greek word ‘kinesis’ which means movement.

Note: As we can see, the rate of a reaction depends on concentration and time. Therefore, it is expressed in terms of concentration / time. If concentration is in mol/L, the unit of rate becomes molL−1s−1mol{{L}^{-1}}{{s}^{-1}}. However, in case of gases, concentration is expressed in terms of its partial pressure. The rate becomes atm/s.