$r_1 = k[A] [B]$ When volume is reduced to half then concentration gets doubled. $\therefore \:\:\: r_2 = k[2AJ] [2B] = 4 k[A] [B] = 4\, r_1$
When the volume of a gaseous reaction vessel is reduced to half, the concentration of the reactants will double.
The rate equation for the reaction is given by r = k[A][B]. If the concentration of both A and B is doubled, the new rate of the reaction can be calculated as:
r' = k[2A][2B] = k(2A)(2B) = 4kAB
So, the new rate of the reaction (r') is four times the initial rate (r). Therefore, the reaction rate relating to the original rate will be 4.
Hence, the correct answer is 4.
Chemical kinetics is the description of the rate of a chemical reaction. This is the rate at which the reactants are transformed into products. This may take place by abiotic or by biological systems, such as microbial metabolism.
The speed of a reaction or the rate of a reaction can be defined as the change in concentration of a reactant or product in unit time. To be more specific, it can be expressed in terms of: (i) the rate of decrease in the concentration of any one of the reactants, or (ii) the rate of increase in concentration of any one of the products. Consider a hypothetical reaction, assuming that the volume of the system remains constant. R → P
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