## Units of Reactions. [ENDORSED]

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

Jose Avelar
Posts: 10
Joined: Fri Sep 25, 2015 3:00 am

### Units of Reactions.

Can someone please explain to me how to know which unit of k to use for zero order, first order, and second order reactions? Why are they different for each order?

Richmund Tan 1L
Posts: 25
Joined: Fri Sep 25, 2015 3:00 am

### Re: Units of Reactions.

Units of k for a zero order reaction is just M*s^-1 because the Rate = k. There is no concentration that affects the equation.

Units of k for a first order reaction is s^-1 because Rate = k[A]. Since there is one concentration value, s^-1 must multiply with M to give us a rate of M*s^-1.

Finally, units of k for a second order reaction is M^-1 * s^-1 because Rate = k[A]2. It makes sense that M^-1 must cancel out M^2 to give us a rate of M*s^-1.

Michael Lee 2I
Posts: 51
Joined: Fri Sep 29, 2017 7:05 am

### Re: Units of Reactions.

Basically whatever the exponent of [A] is the order of reaction.

Esin Gumustekin 2J
Posts: 57
Joined: Thu Jul 27, 2017 3:01 am

### Re: Units of Reactions.

If you forget the units, plug in mol/Ls for the rate and mol/L for the concentration of A and find k according to what power A is raised to (so that the units cancel on both sides of the equation). I usually don't memorize the units for the rate constant but find it like this when I need to.

Posts: 53
Joined: Fri Sep 29, 2017 7:03 am

### Re: Units of Reactions.

What helps most is just knowing that rate must be defined as mol/L*S so whatever reactants you have and what they multiply out to, just convert K to make sure the final units are in mol/L*S

Ilan Shavolian 1K
Posts: 58
Joined: Fri Sep 29, 2017 7:03 am

### Re: Units of Reactions.  [ENDORSED]

Just subtract 1 from the total order of the reactants (x), and than put Lx / molx * s