Question

What do you call a rate law that expresses how the rate depends on the gradient (concentration) of species being studied?

Answer 1

integrated

Answer 2

are you SURE?

Answer 3

something like below ?
\[\large \dfrac{dy}{dx} = ky\]

Answer 4

\[Rate = \frac{ \Delta[A] }{ \Delta t }=k[A]\]

Answer 5

yeah i think we need to put concentration function on right hand side instead of the original function itself *

Answer 6

then we can just change the Delta into derivative
\[Rate = \frac{ d[A] }{ dt } = k[A]\]
\[Rate = d[A]=k[A]dt\]

Answer 7

is that correct?
d[A] = k[A]dt?

Answer 8

I think it should be \[Rate = d[A] = kdt \]
assuming it's a zero-order

Answer 9

slight change -
A = amount of species, not the concentration right ?

Answer 10

i guess we cannot put A on right hand side directly.. it becomes a growth/decay problem..

Answer 11

the amount of species would dictate the gradient

Answer 12

oh

Answer 13

an example problem may help model the problem better...

Answer 14

\[Rate = [A]^0=k \rightarrow \frac{ \Delta[A] }{ Deltat }=k \rightarrow \frac{ d[A] }{ dt }=k\]
\[d[A] = kdt\]
if expressed as a function of time,
\[\int\limits_{[A]_0}^{[A]_t}d[A]=k \int\limits_{0}^{t}dt\]
FTC
\[[A]_t-[A]_0=kt\]

Answer 15

aah okay :) so the concentration of NO2 depends on time and the amount of NO2 present

Answer 16

we're on right track... its a typical dy/dt = ky problem

Answer 17

as concentration

Answer 18

I am trying to create a widget in my calculator that helps determine the order.
Based on the zero-order, I can just derive higher order integrated rate laws, and then perhaps with the help of a graphing calculator, do a graphical analysis of the data and do a linear-regression (zeroth, first, second ... etc) and IF it showed a straight line then it would be that order. Then the value of k will be determined also because that would be the slope of the line.

Answer 19

chemistry gives me fevers lol, im not entirely sure of this particular problem. but in general, the solutions for these problems are never linear/polynomial type... we get an exponential decay function after solving the differential equation

Answer 20

you know how when we plot everything in physics and make it linear? LAUGHING OUT LOUD

Answer 21

yes but not really good with this regression stuff :/

Answer 22

in the form of:
y= mx+b

zeroth
\[[A]_t=-kt+[A]_0\]
first
\[\ln[A]_t=-kt+\ln[A]_0\]
second
\[\frac{ 1 }{ [A]_t }=kt+[A]_0\]

Answer 23

it's difficult to evaluate curved plots, so we express them linearly

Answer 24

y = ae^kt

lny = kt + lna

Answer 25

you're just changing the coordinate system for plotting convenience right ?

Answer 26

yes! haha

Answer 27

the original function hiding behind the plot is exponential or whatever it is... got you now :)

Answer 28

YEAH. That's why we derived them first if we wanted to solve for the order and constants, since those are to be determined experimentally.

Then the graph will help analyze and determine appropriate constants and order visually. Calculus is making my world brighter and wider each day.

Answer 29

cool @nincompoop :)