## Doubling Concentration in Second-Order Reaction

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

Jeremiah Samaniego 2C
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Joined: Fri Sep 29, 2017 7:05 am

### Doubling Concentration in Second-Order Reaction

Hi, can someone explain what the book means when it states "Doubling the concentration of the reactant in any second-order reaction increases
the reaction rate by a factor of 22 = 4" on page 619 of the textbook?

joycelee1A
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

### Re: Doubling Concentration in Second-Order Reaction

It means that doubling the concentration for a second order reaction should increase the reaction rate by four times the original reaction rate. So mathematically, since the concentration is being raised to the second power, doubling it would cause it to be 2^2 or times 4.

Hannah Chew 2A
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### Re: Doubling Concentration in Second-Order Reaction

The definition of a second-order reaction is that rate is proportional to the second power (ie. squaring) of the concentration. By doubling the reactant concentration, you would be actually quadrupling the reaction rate since 2 squared is 4.

If it was a first-order reaction, rate is proportional to the first power, so doubling the reactant concentration doubles the reaction rate since 2 to the power of 1 is still 2.

If it was a zero-order reaction, rate is independent of concentration, so doubling the reactant concentration does not change the reaction rate since 2 to the power of 0 is 1.

So in conclusion, the order of the reaction just tells you how proportionate a reaction rate changes with a change in concentration.