CHEMISTRY COMMUNITY

Cate Cattano 1L

A sealed beaker of water where the beaker doesn't insulate

Cate Cattano 1L

It lowers the activation energy so it takes less energy for the reaction to occur. I think that would lower the temperature because less energy is required.

Cate Cattano 1L

Catalysts lower the activation energy so they increase the rates of both the forward and reverse reactions. They aren't used up and can be reformed for future reactions. Intermediates are formed in a reaction and are used up.

Cate Cattano 1L

The forward reaction would be spontaneous because K>1 favors the products.

Cate Cattano 1L

I think it would be low because we only really covered 0,1st & 2nd order reactions.

Cate Cattano 1L

The overall order of a reaction is the sum of all the orders involved.

Cate Cattano 1L

C is not included because it is zero order so the overall reaction rate is independent of [C]. For that reason, we just include [A] & [B]/

Cate Cattano 1L

he forward rxn has the higher Ea, so an increase in temperature will increase the forward rate constant, more than it will the reverse rate constant. This is because the higher Ea means it takes more energy for the rxn to occur.

Cate Cattano 1L

Hi! You just use the formula ΔH = Ea(forward) - Ea(reverse) to get your answer. You have ΔH and Ea(forward) so all you need to do is plug in your values.

Cate Cattano 1L

I was confused by this one too. I plugged in my info to the formula and still got the wrong answer. What I did instead was find the difference between the two concentrations and then divided that answer by the rate constant. I am pretty sure this only works for zero-order reactions, though.

Cate Cattano 1L

Hi! Does anyone have any strategies for remembering the formulas for the different reaction orders? And the changes in initiate rate when things are halved, tripled, etc?

Cate Cattano 1L

I did very poorly on the first midterm and I found that going through the textbook problems and attending the step-up sessions super helpful. In the step-up sessions, they walk through the problems and let you ask as many questions as you want which in turn helps with the textbook work. It takes a l...

Cate Cattano 1L

This is my #9 for Achieve. I'm on my 13th attempt now so I thought I'd ask for help. These are the two half-reactions I have been using. Cr3+(aq) + 3e– → Cr(s) –0.74 Au3+(aq) + 3e– → Au(s) +1.498 I have gotten ± 2.238 and ± 0.758 and neither are correct. Am I using the wrong values? Calculating it w...

Cate Cattano 1L

I've been really stuck on #3 for a while now. Can someone help me figure out where I am going wrong? I've watched a few videos but it still isn't clicking.

Cate Cattano 1L

Hi! Do you all have any general tips on how to balance redox reactions? I did well with question 2 because there weren't any intermediates. How would you approach problems like numbers 3 and 4 where intermediates get involved? Thanks!

Cate Cattano 1L

Q: What do you do with a sick chemist?
A: If you can't helium and you can't curium, then you may as well barium.

Cate Cattano 1L

Use (500 W)(3.36 mins)(60 sec) and convert that to kJ
Then divide that value by the molar mass of the respective molecule to get ΔHvap

Cate Cattano 1L

So first I changed Watts into Joules per second in respect to the amount of heat time the question listed. So I would do 550 times 60 seconds times 3.09 minutes. For the first sample (and these steps would be repeated for all samples), I would find the amount of moles of the sample were left after ...

Cate Cattano 1L

Hi! This problem has a lot of steps and information. I looked at the answer key and it sort of makes sense. Can someone walk me through their thought process? It's hard for me to figure out all the equations. Thanks!

Cate Cattano 1L

Since it asks for change in internal energy, we know we're solving for ∆U. I started with the formula ∆U = q + w. Now, we know that our "game plan" will be to solve for q and w and then add those two values together to get ∆U. Because the problem mentions nothing about pressure, I assumed...

Cate Cattano 1L

Can someone explain how they derived the equations for 4D.7? I've looked at the answer key but am a bit confused about where the numbers are coming from. Particularly, they made PΔV = ΔnRT negative, and why they never did anything with the volume.

Thanks :)

Cate Cattano 1L

Hi!

I just wanted to make sure I understand Kirchoff's Law correctly. It is basically that the enthalpies of the reactants and products increase with temperature but the difference between the two respective enthalpies does not change?

Cate Cattano 1L

Exothermic reaction the system loses heat and the sign of q is negative. Endothermic reaction the system gains heat and the sign of q is positive. If you look at the magnitude of the q arrows you can tell which one is which.

Cate Cattano 1L

It is how much disorder there is in a system. The higher the temperature, the more disorder because there are more arrangements of places the particles could be.

Cate Cattano 1L

I had to use all four. Be careful, though! I didn't pay attention to the states of matter in my first two attempts.

Cate Cattano 1L

My answer was 5 sig figs. I used 3-5 sig figs for each number when I was doing the calculations.

Cate Cattano 1L

Hi! How did you get the 0.082057 value?

Cate Cattano 1L

I'm confused why the fraction of the concentrations is set up the way it is. I don't understand why we are subtracting 7.2 x 10^-3 from 0.1 in the denominator if we know the concentration.

Cate Cattano 1L

Standard reaction enthalpy is when the reactants and products are in the standard state.
Standard enthalpy of formation is deltaHº for the formation of 1 mol of a substance from its elements in standard form.

Cate Cattano 1L

When it is less than 10^-3

Cate Cattano 1L

You can't have negative values of concentration. If you have 2 positive values, make sure you pic one that is smaller than the initial concentration.

Cate Cattano 1L

You add up all of the partial pressures at the end to get the total pressure.

Cate Cattano 1L

I think we can only approximate when K is less than 10^-3

Cate Cattano 1L

Basically, you use the table to get the K values and cancel stuff out.

H2 + Cl2 <-> 2HCl
2BrCl <-> Br2 + Cl2 (The Cl2s cancel)
-------------------------
H2 + 2BrCl <-> Br2 + 2HCl

Then you multiply the K values
4 x 10^31 x 377
To get K=1.5 x 10^34

Cate Cattano 1L

The equilibrium constant is set at a certain temperature. It's the baseline that the reaction is trying to stay at / get to. Temperature changes the equilibrium constant because it changes the reaction rates.

Cate Cattano 1L

You need to look at the original chemical equation and find the pieces of it that you need in the other ones. It's a lot like balancing a chemical equation in that you have to multiply equations by numbers to match the equations for your original ones. The equations won't all match up perfectly so y...

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