Question

Consider the gaseous equation, the rate of which is given by$k[A][B]$. The volume of the reaction containing these gases is suddenly reduced to${{\dfrac{1}{4}}^{th}}$of the initial volume. The rate of the reaction as compared with original rate is
(a) $\dfrac{1}{16}times$
(b)16 times
(c) $\dfrac{1}{8}times$
(d)8 times

Answer 1

Hint: Speed at which a chemical reaction takes place is called rate of the reaction. The rates vary from reaction to reaction. Rate of the reaction will always be positive. If negative sign indicated means the reactant concentration is decreasing. Rate law is the law which shows the relation of the rate of the reaction with its reactant and its products.

Complete answer:
We know that concentration is number of moles divided by volume; i.e.it is inversely proportional to volume ($\dfrac{n}{V}$)
In the equation$k[A][B]$, [A] and [B] are concentrations of A and B respectively in the reaction and k is the rate constant.
It is given that the volume decreases ${{\dfrac{1}{4}}^{th}}$ times the initial volume. So concentration of A and B increases 4 times (as concentration is inversely proportional to volume).
Therefore the equation becomes $k[4A][4B]$
We can also write the equation as $k\times 4\times 4\times [A][B]$
$=k16[A][B]$
Hence, the rate of the reaction increases by 16 times. The correct answer for the question is option (b).

Additional Information: If A and B are reactants of the reaction and k is the rate constant of the reaction, according to rate law
Rate of the reaction =$k[A][B]$
This shows that the rate is dependent on the concentration of the reactants. The other factors which he rate is dependent on are temperature, concentration, and presence of a catalyst.

Note: We should keep in mind that the concentration is inversely proportional to volume. Mistakes are made by taking the value of the volume as such in the equation.