## Concentration independent of the rate

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

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Melody P 2B
Posts: 35
Joined: Fri Sep 29, 2017 7:06 am

### Concentration independent of the rate

When the concentration of a reactant is independent of the rate does that mean it's zero order? And how would we be able to tell given a table of initial rates/concentrations?

Brian Chhoy 4I
Posts: 66
Joined: Fri Sep 28, 2018 12:16 am

### Re: Concentration independent of the rate

Do you mean the rate of a reaction is independent of the concentration of the reactant? If so, just look at the table and see if that specific reactant concentration is changed ( for example it is doubled), then you check the rate of the reaction. If the rate of the reaction did not change due to the change in that reactant concentration, then that reactant is zeroth order.

gwynlu1L
Posts: 62
Joined: Fri Sep 28, 2018 12:19 am

### Re: Concentration independent of the rate

Also, if you're given a graph and it's comparing time vs concentration (not ln [a] or 1/[a]), then it is a zero order rxn

BenJohnson1H
Posts: 68
Joined: Fri Sep 28, 2018 12:17 am

### Re: Concentration independent of the rate

It might be better put as the rate is independent of the concentration of a zeroth order reactant but yes, this is true

katie_sutton1B
Posts: 61
Joined: Fri Sep 28, 2018 12:17 am

### Re: Concentration independent of the rate

Yes, zero order is independent of the concentration of reactants. This has a graph that has concentration as its y-axis, and time as the x-axis.

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