15.9 [ENDORSED]

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

Palmquist_Sierra_2N
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15.9

Express the units for rate constants when the concentrations are in miles per liter and time is in seconds for a) zero-order reactions; b) first-order reactions; c) second- order reactions

alondra_1D
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Re: 15.9

In the textbook page 620, the units for the rate constants are given.

Rochelle Ellison 2H
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Re: 15.9  [ENDORSED]

Rates of reactions are given in Molarity per second (M/s) therefore, depending on the order of the reaction, the rate constant will have a different units. For example when the reaction is zero order the rate law is in the form rate=k so in order for the units of the rate to be M/s the rate constant must have the units M/s.

But when the reaction is first order, the rate law is rate=k[A] where [A] has units of of M so in order to the units of the rate to be M/s, the rate constant must have units of 1/s.

Finally, if the reaction is second order rate=k[A]^2 and the units of [A]^2 are M^2 the rate constant must have units of 1/(M*s) because (M^2) X (1/(M*s)) = M/s

ERIKTORRESDisc3C
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Re: 15.9

So does this mean that overall rate of a reaction is given in mol.L-1.s-1 and the rate constant K will depend on the given concentration to cancel out units and obtain the rate given in a problem? Or is it more so just dependent on the order of the reaction and the rate law used? I am still confused on the conversion of units when calculating rate law and obtaining the proper units mol.L-1.s-1.

Pauline Tze 3B
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Re: 15.9

Hi Erik,
The overall rate of reaction has the units mol.L-1.s-1. The rate constant k depends on the reaction's activation energy and temperature. Problems that ask you to find rate constant k will require you to find its numerical value and units given the chemical reaction and initial rate.

Say you're given a reaction A + B --> C
The general rate law is: rate = k [concentration of reactant A]n[concentration of reaction B]m
From the experimental data (given as a chart), you figure out that n = 1 and m = 1, giving you an overall reaction order of 1+1 = 2.
Now you have
rate = k [concentration of reactant A]1[concentration of reaction B]1

Disregarding the numerical values, in units you can write this as
mol.L-1.s-1 = k*[mol.L-1]1.*[mol.L-1]1

Simplify to:
mol.L-1.s-1 = k*[mol2.L-2]
Therefore, you can conclude that the units of k for this (and any) 2nd order reaction is:
mol-1.L1.s-1

I hope that helps.