MeadowBubble

Happy Birthday!

MeadowBubble, I'm still just a stranger? That random guy who is only there to help you with chem homework?
I see how it is ;_;

Haha, I jest. It's a pleasure to help. Take care, MeadowBubble!

MeadowBubble, hey, you're tired. Don't worry about it.

Oh, you asked about the Arrhenius equation and I didn't notice!

k = A * e^(-Ea/RT)

k = rate constant
Ea = activation energy
R = gas constant
T = temperature in Kelvin
A = I actually forget what this is, but I think it's usually 1.

It's also written as:

ln k = (-Ea/R * 1/T) + ln A

by taking the ln of both sides.

What's it used for? To find the activation energy of a given reaction.

MeadowBubble, let's see...

Oh! Wow, that's overly simple that the answer went past me. Silly me!

-delta[BrO3-]/delta t=1.5 x 10^-2 M/s
represents the overall rate of the reaction itself, with respect to bromate. The coefficient of bromate is 1, and the coefficient of bromine is 5, so to get -delta[Br-]/delta t, multiply by 5. Your answer should be 7.5 x 10^-2 M/s.

MeadowBubble, can you give me all the information it's giving you (as well as what it wants you to solve for)? Let me see if I can crack it.

MeadowBubble, s does stand for seconds, but is it giving you data in the form of a table like I gave you (with multiple columns, usually 4 or more, with concentrations for all the reactants), or is it giving a different sort of table (with only 2 or so columns, with concentrations for only 1 reactant vs what time it is in seconds)?

If the former, it's probably differential. If the latter, it's probably integral.

MeadowBubble, I don't think you need to know exactly what -Δ[Br] / Δt is. That was just giving you the rate of the reaction itself. -Δ[A] / Δt is just the generic form of the integral rate law; don't worry about it.

You will need to know what order it is, though. Is it giving you data for concentrations of reactions vs reaction rates, or concentrations vs time?

MeadowBubble, you have told me how amazing I am before, and I love you too

Integrated rate law questions are a bit trickier. They will require knowledge of being able to construct a graph of form y = mx + b, that is, a straight line, and for many of them, you'll also be solving for the rate constant k (something that isn't usually done in differential rate law problems), as well as using that to find the concentration of a reactant at time t. Each form of the integral rate law (order 0, 1, or 2) has its own special rate law equation, as well as a half-life equation (to find the time it takes for the reactant concentration to halve).

Zero-order integral rate law:
[A] = -kt + [A]0

where [A] is the concentration of an aqueous reactant, [A]0 is its concentration at time t = 0 (i.e. initial concentration), t is time, and k is the rate constant.
If a reaction is zero-order, you can plot a graph of [A] versus t (data for this will be provided for you), and the result will be a straight line of slope -k. If it doesn't give a straight line, it's not zero-order.

The half-life for a zero-order reaction is t(1/2) = [A]0 / 2k, where t(1/2) is the time of the half-life, and is written as t with 1/2 as a subscript.


First-order integral rate law:
ln[A] = -kt + ln[A]0

Looks the same as zero-order, except it uses the natural logarithm of the concentration instead.
If a reaction is first-order, you can plot a graph of ln[A] versus t, and the result will be a straight line of slope -k. If it doesn't give a straight line, it's not first-order.

Half-life for first-order: t(1/2) = ln 2 / k
(This one is easier. Just plug in k and you're done, since ln 2 is a constant.)


Second-order integral rate law:
1/[A] = kt + 1/[A]0

Takes the reciprocal of the concentration. For second-order reactions, you can plot 1/[A] vs t, and get a straight line of slope k. NOTE!! This is not a slope of -k you will get here; the line for second-order will be a positive slope. Zero-order and first-order lines will have negative slopes.

Half-life for second-order: t(1/2) = 1 / k[A]0


For problems you'll be given, you may need to solve for any or all of the following:
~Whether a reaction is of order 0, 1, or 2 (done by graphing the points out and see if you get a line - don't forget, you don't use [A] unmodified as your y-axis unless it's zero-order; first-order is ln[A] and second-order is 1/[A])
~Solving for k (done after you find out what order it is usually)
~Finding when the half-life is
~Finding the concentration of a reactant at any given time t (to do this, you take the integral rate law and plug all the numbers in; you'll need to know k and [A]0 for this)

And that, I believe, should cover it! I'm free to answer any leftover questions you have.

MeadowBubble,
Rate law = how the rate of a reaction is expressed mathematically. There are 2 types: differential (goes based on concentration) and integral (goes based on time)
Reaction rate = the R, on the left side of the equation, in a rate law.
Rate constant = the k, on the right side of the equation. Different for every reaction and must be calculated by experiment.

(I'm about halfway done with the integral rate law write-up, so don't run away yet!)

MeadowBubble, ah, it looks like you're up to the integrated rate law as well.

Differential rate law (used for finding reaction rates based on reaction concentration):
Rate = k * [A]^m * ^n

Integral rate law (used for finding reaction rates based on time):
Rate = -Δ[A] / Δt = k[A]^n
(n can be 0, 1, or 2; each case has another special integrated rate law that goes with it)

Gimme a bit and I'll get you a write-up for the integrated rate law as well!

(Make sure you read the post below first before reading this one)

When you're given problems to find the order of a reaction, you'll be given data for several experiments, each of which will have the initial concentrations of all reactants (given in M, usually), plus the rates of the reactions themselves (my book gives it in mol/L * s). To solve for the exponents (which will be the unknowns in your problems), you'll need to compare two experiments together via division.

As an example, I'll give you the example my book uses. The reaction it gives is:

NH4+ (aq) + NO2- -> N2 (g) + 2 H2O (l)
(Ammonium and nitrite react to give nitrogen gas and water.)

The data it gives is in 4 columns, with the columns from left to right being Experiment, Initial Concentration of NH4+ (in M), Init. Conc. of NO2- (in M), and Initial Rate (in M * s). Data is as follows:

1 _______ 0.100 M _______ 0.0050 M _______ 1.35 x 10^-7
2 _______ 0.100 M _______ 0.0100 M _______ 2.70 x 10^-7
3 _______ 0.200 M _______ 0.0100 M _______ 5.40 x 10^-7

The rate law for the reaction would be:
Rate = k * [NH4+]^m * [NO2-]^n

We are given the rate as well as initial concentrations, and we're told to find what the overall order of the reaction is. To do so, we solve for m and n.
First step is to compare two reactions. Let's start by comparing 1 and 2, since the NH4+ concentration remains constant and we can cancel it out.

Rate 1 = 1.35 x 10^-7 M * s = k * (0.100 M)^m * (0.0050 M)^n
Rate 2 = 2.70 x 10^-7 M * s = k * (0.100 M)^m * (0.0100 M)^n

I got these figures by just taking data from the table, and plugging it into the rate law. We can actually solve for n right now by taking the ratio of the two rates - dividing them together:

Rate 2 / Rate 1
=
2.70 x 10^-7 M * s / 1.35 x 10^-7 M * s
=
k * (0.100 M)^m * (0.0100 M)^n /
k * (0.100 M)^m * (0.0050 M)^n

You can do a lot of cancelling out here. The two M * s terms divide nicely to give you 2.00, the ks drop out, the [NH4+]^m both drop out, and the two [NO2-]^n terms can be combined, leaving you with:

2.00 = (2.00)^n

Now we know n = 1. Note that result, and put it aside for later while we solve for m. This time we're going to compare experiments 2 and 3, since the two [NO2-]^n terms will cancel out so we can solve for m. Here's how that will look:

Rate 3 / Rate 2
=
5.40 x 10^-7 M * s / 2.70 x 10^-7 M * s
=
k * (0.200 M)^m * (0.0100 M)^n /
k * (0.100 M)^m * (0.0100 M)^n

After cancellation and combining terms, we get:
2.00 = (2.00)^m

So m = 1 as well! With that, we can safely say that this reaction is first-order with respect to ammonium (since m = 1), is first-order with respect to nitrite (since n = 1), and is second-order overall (since m + n = 2).

Solving for order of a reaction and stuff follows much the same pattern. First, note the reaction itself, and draw up your rate law based on the aqueous reactants (sorry, I should have mentioned that earlier!), Rate = k * [A]^m * ^n * [C]^o * [...]. Next, look at the table you're given. To solve for an exponent (m, n, o...), you want to compare 2 experiments for which the concentration of only one reactant differs, and all the others will cancel out. Do that for each one, then add up the exponents and you'll have your answer!

(I'm checking to see if there's more stuff to say on the matter. I haven't gotten into the integral rate law yet, but that should cover most of the basic rate law problems, if you're solving things in respect to concentration!)

MeadowBubble, reaction rates and such have to do with how fast a reaction takes place. It's usually dependent on concentration and a few other things, but to find exactly how fast a reaction takes place needs to be determined by experiment. (Fortunately, for solving problems, that data is usually already provided for you.)

The differential rate law (or simply the 'rate law') of a reaction is generally written like this:

Rate = k * [A]^m * ^n

where k is an arbitrary constant (depends on the reaction itself, but you usually don't need to worry about it),
[A] and are the concentrations of the reactants in the reaction (generally given in M, moles per litre),
and m and n are integers. (They're in the powers for [A] and respectively, in case you didn't notice.) There can sometimes be more terms, e.g. * [C]^o, but you'll rarely (if ever) have to work with more than 3.

The order of a reaction is found by determining how the rate of the reaction is affected, with respect to the concentration. If a reaction is order 0, 1, or 2 with respect to one of the reactants, that's what m or n will be.

Zero-order: The reaction will proceed at the same rate, regardless of the concentration. e.g. if I double the concentration of [A], the reaction will go at 1x the normal rate (unchanged), and m will be 0.
First-order: If the concentration of a reactant is increased (or reduced), the reaction will change speeds, based on how much it changed. e.g. if I double the concentration of [A], the reaction will go at 2x the normal speed (double), and m will be 1.
Second-order: If the concentration of a reactant is increased (or reduced), the reaction will change speeds, based on twice of how much it changed. e.g. if I double the concentration of [A], the reaction will go at 4x the normal speed (quadruple), and m will be 2.

To determine the overall rate of a reaction, add all the exponents up. Say I found out from the data that m and n are both 1. The overall reaction would be order 2 (second-order).

(That's part of it; I'll have more to come in the next post I make on your wall)

MeadowBubble, lemme go dig up my old chem textbook and I'll be right with you with some answers!
Do you want me to tell you on Skype or here?

MeadowBubble, no, because just coming here for chem help would be silly
Heh, it's cool; I don't mind. (If you're on Skype more often than here, though, you could find me there too...except on the weekends.)

"Chemical kinetics"? Does that have to do with exothermic vs endothermic reactions, and whether they're zero, first, or second order?

No problem; glad I could help!

MeadowBubble, regular would be:

sp = bonds at 180-degree angles apart
sp2 = bonds at 120-degree angles apart
sp3 = 107.5 degree angles
dsp3 = a planar ring with 3 bonds at 120-degree angles, then 2 other bonds perpendicular (90-degree) to that ring.
d2sp3 = six bonds at 90-degree angles to each other (4 in a ring, plus 2 perpendicular)

Ammonia and water don't have 107.5-degree angles between their bonds; ammonia is 104.5 (1 lone pair) and water I think is something like 102 (2 lone pairs). I think hydrogen sulfide (H2S) might also be an example.

And now I have to go since the library is closing; if that explanation wasn't good enough, I can give you more when I get home tonight.

MeadowBubble, does this have to do with sp2 / sp3 / tetrahedral, and so on?
because if it is, I know exactly what this is and have even explained it before to someone.

MeadowBubble, I totally remember you! I usually don't forget people who ask for chem hw help. (Then again, there aren't very many that have asked me...)

Help with a chem lab, you say? What's it about?

I saw it. Ellsie has to much free time >__>