## Direct Proportions and First Order

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Kyra Dingle 1B
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### Direct Proportions and First Order

Why does a directly proportional relationship between the rate of the reaction and the concentration of the reactants indicate a first order?

Alyssa Pelak 1J
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### Re: Direct Proportions and First Order

It indicates first order because the concentration change of the reactant directly correlates to the change in the rate. So, if the concentration was doubled the rate would double too.

John Huang 1G
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### Re: Direct Proportions and First Order

Then what would it mean for a reaction to have a second order or zero order?

Payton Schwesinger 1J
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### Re: Direct Proportions and First Order

If it is second order, then I believe that means that as concentration of reactants doubled, then the rate would quadruple (concentration =2, n=2---> 2^2=4)

melissa carey 1f
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### Re: Direct Proportions and First Order

First order indicates that doubling the reactant concentration doubles the rate because rate = k [A]^1

Julie Steklof 1A
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### Re: Direct Proportions and First Order

John Huang 1G wrote:Then what would it mean for a reaction to have a second order or zero order?

The rate of a zero order reaction does not depend on the concentration of reactant.