## zero order in rate law

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

Vanessa Reyes_1K
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Joined: Fri Sep 28, 2018 12:28 am

### zero order in rate law

If a reactant is a zero order reactant, is it included when writing out the overall rate law for the entire reaction?

Jim Brown 14B Lec1
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Joined: Fri Sep 28, 2018 12:28 am

### Re: zero order in rate law

It is included in the unique rate law but in the overall rate law it is only included if it is greater than zero. If the reactant is zero order then it will be to the power of zero and equal to 1.

Mukil_Pari_2I
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### Re: zero order in rate law

In the overall reaction law, the equation would just be the reaction constant.

Kessandra Ng 1K
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### Re: zero order in rate law

Just be sure to know that while the reactant is still in the overall reaction equation, when you write out the rate law, it would be [A]^0 (assuming the reaction is zero order with respect to A) - which equals 1, therefore you wouldn't need to include it in the rate law when you're writing it out.

Lauren Huang 1H
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Joined: Fri Sep 28, 2018 12:16 am

### Re: zero order in rate law

You don't need to include it since anything to the zeroth power is equal to one.

Vicky Lu 1L
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Joined: Fri Sep 28, 2018 12:18 am

### Re: zero order in rate law

If a reactant is a zero order reactant, it would not be included in the overall rate law since anything to the 0th power as an exponent would be one. Also since it is a 0th order, changes to the concentration of that reactant would have no effect on the rate and therefore, it make sense that it would not be included in the rate law.

Anand Narayan 1G
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Joined: Fri Sep 28, 2018 12:15 am

### Re: zero order in rate law

In this case, you would not need to include it since any number to the power of zero is one.

Brian Chang 2H
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Joined: Fri Sep 28, 2018 12:17 am

### Re: zero order in rate law

It's technically included, but since it is raised to the 0th power it just becomes 1.