## Dependence on concentration

$\frac{d[R]}{dt}=-k; [R]=-kt + [R]_{0}; t_{\frac{1}{2}}=\frac{[R]_{0}}{2k}$

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zanekoch1A
Posts: 25
Joined: Thu Jul 13, 2017 3:00 am

### Dependence on concentration

Why is it said that 0th order reactions do not depend on concentration while first and second do. In the equations it looks like they all depend on concentrations.

Isita Tripathi 2E
Posts: 52
Joined: Sat Jul 22, 2017 3:00 am
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### Re: Dependence on concentration

If you graph the concentration vs time of a zero-order reaction, you get a straight line. Since Δconcentration/Δtime = rate, an increased concentration has no effect on the rate, since the slope is constant, even though the concentration changes over time. However, if you graph a first order reaction of concentration vs time, you get an exponential decay function. In this, the slope is decreasing, which means that the rate is changing as the concentration changes over time.

All the equations have concentration in them because using rate laws can allow us to calculate the concentrations at different times. This does not mean that the rate depends on concentration, as the rate is implicit in these equations. You have to analyze the slope to determine dependency on concentration.

Ridhi Ravichandran 1E
Posts: 35
Joined: Sat Jul 22, 2017 3:01 am

### Re: Dependence on concentration

The rate law for a zero-order reaction is: rate=k[A]^0. However, anything to the power of 0 is just one, so the rate is simply equal to k. k is a constant that doesn't change throughout the reaction, and concentration is not even part of the equation, so zero-order reactions do not depend on the concentration of the reactant. They have constant rates that vary only with temperature.

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