Chemistry, asked by akingqueen56, 10 months ago

When concentration of a reactant is doubled, the rate of reaction becomes four times. The order of reaction is?

Answers

Answered by qwtiger
10

Answer:



As we know that, The Rate of reaction is directly proportional to the concentration of the reactant raised to its order.

=>Rate=k * [A] o

(Here k is the rate constant,

[A] is the concentration

o is the order of the reaction.)

According to the question, New rate = 2 * Old rate,

And

new concentration = 4 * old concentration.

=>k∗(4∗[A])o = 2∗k∗[A]o

=>4o = 2

ans: o = 0.5

Therefore the order is 0.5.

Answered by AbdulHafeezAhmed
4

As we know that, The Rate of reaction is directly proportional to the concentration of the reactant raised to its order.

=>Rate=k * [A] o

(Here k is the rate constant,

[A] is the concentration

o is the order of the reaction.)

According to the question, New rate = 2 * Old rate,

And

new concentration = 4 * old concentration.

=>k∗(4∗[A])o = 2∗k∗[A]o

=>4o = 2

ans: o = 0.5

Therefore the order is 0.5

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