CHEMISTRY COMMUNITY

Christina Bedrosian 1B

no rate laws don't include products you are thinking of the equilibrium constant

Christina Bedrosian 1B

It is the peak because it has the highest energy since the carbons have 5 bonding groups around them during that small amount of time, making it highly unstable

Christina Bedrosian 1B

k' stands for k*the reactants that you don't use in the rate law

Christina Bedrosian 1B

it is 1/(1/16) making it 16/1 (it flips); just plug it into the equation and you should be fine

Christina Bedrosian 1B

part a increases because catalysts increase the rate of the reaction in general

Christina Bedrosian 1B

its just an observed trait

Christina Bedrosian 1B

for multiple reactions, see the effect of each reactant on the rate. so if one reactant is 1st order and another is 2nd order, the 1st would double the rate if the concentration is doubled and the 2nd one would quadruple the rate if the concentration was doubled. you usually view them separately lik...

Christina Bedrosian 1B

depending on order, raise the effect on the concentration to that power to see the effect on the rate. for example if we double the concentration of a 1st order than you do 2^1 (1 because 1st order) which equals 2 meaning the rate will also be doubled. for 2nd order it would be 2^2 meaning the rate ...

Christina Bedrosian 1B

you can also usually tell by what is given in the problem. for example if it doesn't give an order but gives you a rate then its most likely average rate instead of instantaneous

Christina Bedrosian 1B

we dont want to be talking about "negative" rates so we add the negative when talking about the rates of the reactants since they are decreasing and their rate would normally be negative (adding the negative makes it positive, which we want, this also represents rate of products)

Christina Bedrosian 1B

theres a pattern to it as you can see in the previous responses so for 2nd reaction the L and mol gets raised to 2nd power and 3rd reaction gets raised to 3rd power and so on; seconds remain just seconds

Christina Bedrosian 1B

We need the longest time taken for products to be formed or else that slowest step would not be able to be completed; if the slowest step takes 10 seconds but every other step only takes 1, you still need to give the reaction 10 seconds to complete that slow step

Christina Bedrosian 1B

they are inert so that they will not change the reaction but will act as electron transporters to allow the redox reaction to occur

Christina Bedrosian 1B

E values never change, even if the chemical reaction is being multiplied by a factor

Christina Bedrosian 1B

The only time they aren't going to be separated by lines is when they are both the same state such as two aqueous solutions

Christina Bedrosian 1B

you need to balance O first because you will only know how much H you need once you get the right amount of O

Christina Bedrosian 1B

an agent is just the thing that does the action so for an oxidizing agent it is the thing that does the oxidizing

Christina Bedrosian 1B

I get confused with this a lot to but you just have to know that under 0 degrees Celsius will be ice and above will be water and if its at 0 degrees it will usually tell you what current state its in

Christina Bedrosian 1B

O3 is a neutral substance so it's oxidation number is 0 (there is no negative or positive charge on the compound)

Christina Bedrosian 1B

you basically need to rely on the other element in the compound as Audrey said to determine the oxidation state in that situation

Christina Bedrosian 1B

you essentially look at the elements location in the periodic table. if it is in the first two groups, it will be +1 and +2 (the group number) and if it is in the last couple of groups, it will be -3, -2, -1 (corresponding to group 15, 16, 17)

Christina Bedrosian 1B

when deltaG is negative, that means deltaS of the universe (total) is positive since deltaS(total)=-(deltaG)/T. deltaS being positive means the reaction is spontaneous (thus deltaG needs to be negative)

Christina Bedrosian 1B

doing work while maintaining a constant temp needs to include a heat transfer in order to maintain the constant temperature (q=-w or w=-q)

Christina Bedrosian 1B

This formula is true for instances of isothermal expansions since having deltaT=0 (no change in temp), then deltaU (internal energy) would be 0 and since deltaU=q+w, then q would equal -w

Christina Bedrosian 1B

The important part is isothermal, meaning deltaT would be 0 making deltaU 0 (since anything *0=0).

Christina Bedrosian 1B

Also spontaneous reactions with gas can't occur while going to a smaller volume because that would require energy but larger volume wouldn't need anything to happen

Christina Bedrosian 1B

I also don't believe we will actually have to derive anything on the tests because it isn't the point of the equation. We only derived it to help us understand where it came from.

Christina Bedrosian 1B

I think when we are given the question, there will be only one method that fits the question so I don't think you would have much of a choice to not use this method but it is just worth knowing that this method is the least accurate.

Christina Bedrosian 1B

In response to your latest question, you would just notice that a bond in the reactants had been broken and a new bond in the products is formed because there are different combinations. For example, if we have CH4 + HBr --- CH4Br and we drew the lewis structures, we would see that the double carbon...

Christina Bedrosian 1B

I am confused about this as well because in both questions 59 and 63 they are asking for the standard reaction enthalpy for the reactions and 59 uses kJ/mol while 63 only uses kJ. The questions are basically identical except for the reactions so how do we know when to use one or the other?

Christina Bedrosian 1B

I've heard that if the answer ends in .5, you would round to the nearest even number, which in this case would be 448. I am not sure if this is the official rule to go by but I remember reading it in 14A so maybe that would explain it.

Christina Bedrosian 1B

An easy way to think about this is that exothermic releases heat, meaning usually involving cooling, so the phase changes being exothermic would be from gas to liquid (condensation), liquid to solid (freezing), or gas to solid.

Christina Bedrosian 1B

Does the length of the sloped lines indicate anything? Or is it just based on the steepness?

CHEMISTRY COMMUNITY

Search found 33 matches

Re: rate law

Re: Transition State

Re: k' in Pseudo-First-Order Rate Laws

Re: Half life

Re: How does a homogeneous catalyst affect (a) the rate law; (b) the equilibrium constant? [ENDORSED]

Re: Homogeneous Catalysts

Re: Concentration and Rate Relationship

Re: How the concentration of a reactant affects the rate of the reaction

Re: When to Use Avg. vs Instantaneous

Re: signs

Re: units of the rate of reaction

Re: Slowest Elementary Step

Re: Cell Diagrams [ENDORSED]

Re: Reaction E [ENDORSED]

Re: Cell Diagram [ENDORSED]

Re: Balancing Redox reactions

Re: agents

Re: Specific heat of water or ice?

Re: Ozone oxidation number

Re: Transition Metals

Re: oxidation

Re: The sign of the change in Gibb's Free Energy

Re: Test Number 1

Re: q=-w [ENDORSED]

Re: isothermal system

Re: Spontineity

Re: The integral equation

Re: Bong Enthalpy accuracy [ENDORSED]

Re: Bond Breaking or forming

Re: HW Ch.8 #59 and 63

Re: Sig Figs for Question 8.45 (a)

Re: Phase Changes

Re: Heating Curve