Question

For a gaseous reaction the rate=k[A][B] .the volume of the vessel containing the gas is reduced to 1/4 th of the initial volume .The rate of reaction relative to initial rate would be what?​

Answer 1

Answer:

For any reaction, reactant concentration= no. of moles/ volume. Therefore, if volume of vessel reduces by 4 times, concentration of both reactants will increase by 4 times each. Rate will become r'= k[4A] [4B]= 16k[A] [B]=16r

Explanation:

NOTE: If the same question asked how the rate constant changed, the answer would be no change, as rate constant is independent of concentration of reactants in most cases.

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