CHEMISTRY COMMUNITY

Luc Zelissen 1K

For order (m + n), the rate constant has units of mol1−(m+n)·L(m+n)−1·s−1 For order zero, the rate constant has units of mol·L−1·s−1 (or M·s−1) For order one, the rate constant has units of s−1 For order two, the rate constant has units of L·mol−1·s−1 (or M−1·s−1) And for order three, the rate const...

Luc Zelissen 1K

It easiest if you change the t value to the time value of your rate constant. In this way you can easily multiply and get the right answer without changing the rate constant units.

Luc Zelissen 1K

I wouldnt be surprised to see a derivation on the exam. Some of the UAs had a derivation problem on their second test last year. Also lavelle specifically went through the derivations in class, so we should probably know how to do it.

Luc Zelissen 1K

For Zero Order reaction: t½ = [Ao] / 2k
For First Order reaction: t½ = 0.693 / k
For Second Order Reaction: t½ = 1 / k [Ao]

They are given, but they are not labeled on the constants sheet.

Luc Zelissen 1K

For order zero, the rate constant has units of mol·L−1·s−1 (or M·s−1)
For order one, the rate constant has units of s−1
For order two, the rate constant has units of L·mol−1·s−1 (or M−1·s−1)
And for order three, the rate constant has units of L2·mol−2·s−1 (or M−2·s−1)

Luc Zelissen 1K

When K > 1, the reaction is spontaneous because this makes delta G negative. So when the question asks to find if K > 1, it wants you to find the sign of delta G and thus if the reaction is spontaneous. You can easily find the sign of delta G, by finding the E^{o} of the cell. You can use the equat...

Luc Zelissen 1K

You will need to use the nernst equation: $E_{cell}=E^{o}-.0592/n*log(Q)$ . We are given E cell, and you can find the E standard from the half reactions from appendix 2b.

Luc Zelissen 1K

I believe the solutions manual is wrong.

Luc Zelissen 1K

$\Delta G_{cell}=-nFE_{cell}$ is an important equation. You can replace $\Delta G$ with work also.

Luc Zelissen 1K

A Concentration cell, is a like a galvanic cell except the chemical species on both sides is the same. If you wanted to find the E standard, it would be 0 because the 2 rxn are the same, and so the you would add the reduction E standard with the E standard of the oxidation, which is equal but opposi...

Luc Zelissen 1K

Your work is right, I don't know why they rounded; It might just be a solution error

Luc Zelissen 1K

I believe it can be either or. I always right in the way the reaction proceeds, so if Cr4+/Cr3+ are on the anode side, I would write the cell diagram $Cr^{3+} | Cr^{4+}$

Luc Zelissen 1K

This is also part of the reason for the pH scale. We take numbers like .001 and .0001 and say they are 1 pH value apart, while in reality, .001 is ten times greater than .0001. We use ln and log to view very large or very small numbers on an easier scale that we can easily interpret

Luc Zelissen 1K

.075 Molar is part of the oxidation half reaction, but it ends up on the products side of the overall reaction.

Luc Zelissen 1K

A concentration cell is an electrolytic cell that is comprised of two half-cells with the same electrodes, but differing in concentrations. The only work done from a concentration cell occurs from a difference in concentration

Luc Zelissen 1K

Maya Pakulski 1D wrote:What are oxidation numbers used for?

If you know the oxidation numbers of a species on both sides of a chemical reaction, then you will know if it is reduced, stay the same, or oxidized

Luc Zelissen 1K

Your Oxidation half-reaction is correct

Luc Zelissen 1K

According to other forum posts, there is an error in the questions and it should be: $Cl_{2}\rightarrow 2Cl^{-}$

Luc Zelissen 1K

The oxidation state of oxygen in its compounds is usually -2, except for peroxides like H2O2, and Na2O2, in which the oxidation state for O is -1. The oxidation state of hydrogen is +1 in its compounds, except for metal hydrides, such as NaH, LiH, etc., in which the oxidation state for H is -1.

Luc Zelissen 1K

Since we need to usually need to balance the reactions in a solution, we need to add water. By doing this you might need to then balance the Hydrogens, and to do this you add H+ and OH- based on the context of the question

Luc Zelissen 1K

No work is done in free expansion because there is no opposing force. No work is done by pushing if there is nothing to push against, so w=0. Since its isothermal, so \Delta U = 0 . \Delta U = q+w , where 0 = q + 0, so q is also 0, if q is 0 , then \Delta S_{Surrondings} = 0 , because \Delta S = q_{...

Luc Zelissen 1K

If a system is reversible that means that any $\Delta State functions$ are equal to zero

Luc Zelissen 1K

Also remember that for Ideal gasses you can use the equation, Cpm = Cvm + R

Luc Zelissen 1K

My final equation for number 10 is (25_{grams}*2.01_{C_{Ice} }*T_{f}-0) + (6010_{\Delta Hfus_{Jules}}*1.38_{moles})= -(265_{grams}*4.184_{C_{Liquid}}*(T_{f}-25)) , but I keep getting the wrong degrees. What am I doing wrong? I realized that after the dHfus occurs, th...

Luc Zelissen 1K

For Question 5, I have $\Delta S=moles*8.314*ln(v2/v1)$ for both gasses, but when I add them up I dont get the right answer. Is there another equation to find the change of Entropy of the 2 gasses combined, if so what is it?

Luc Zelissen 1K

My final equation for number 10 is $(25_{grams}*2.01_{C_{Ice} }*T_{f}-0) + (6010_{\Delta Hfus_{Jules}}*1.38_{moles})= -(265_{grams}*4.184_{C_{Liquid}}*(T_{f}-25))$ , but I keep getting the wrong degrees. What am I doing wrong?

Luc Zelissen 1K

First you need to heat up the liquid water to 100 degrees and calculate entropy, this is the heating part. Then you need to vaporize, and use the given number in the question. Then you need to cool it back down to 85 degrees in order to correctly answer the question.

Luc Zelissen 1K

$\Delta U$ is a function of N and T. If $\Delta T$ does not change, $\Delta U$ will also not change

Luc Zelissen 1K

It always depends on the system's pressure and volume. In some situations You can use the equation $\Delta S=q/t$ . If there is no heat transfer then q is 0, so delta S would be zero

Luc Zelissen 1K

005384106 wrote:will we ever need to do any conversions for units? do these problems always need to be in kJ?

I assume that you will need to do unit conversions on the test to get the answer right. Its always easier to use the units that are in the questions.

Luc Zelissen 1K

Also remember that degeneracy, Capital W, has no units. So for the equation $\Delta S=k_{b}*ln(W)$ , the units of boltzmann constant is the same as entropy

Luc Zelissen 1K

\Sigma \Delta H_(Bonds broken) - \Sigma \Delta H(Bondsformed) . You add up the bond enthalpies from the product side and subtract it from the sum of the bond enthalpies of the reactant side. Also remember to take into account the stoichometric values in your math. You are right that...

Luc Zelissen 1K

I would say that you should be able to use all 3, since the problem could give you constants and values that only one of the methods can utilize

Luc Zelissen 1K

For 4.E.5, Looking at the lewis structure for C2H2 gives us 1 C-C triple bond, which has an enthalpy of 837, and also 2 C-H bonds which have a combined value of 824. Add these 2 numbers and multiply by the moles, which is 3, and you get the total enthalpy for C2H2, 4983. For benzene, the table of bo...

Luc Zelissen 1K

You can think of this in another way: imagine putting heat as a reactant or product based on if the reaction is endothermic or exothermic. Then you can use le chatelier's principle to determine how K will change. For an exothermic reaction, the heat is on the products side because it is released. If...

Luc Zelissen 1K

Do solids and liquids only use C? Or do they use Cv? I'm pretty sure they don't use Cp. They are the same for solids and liquids. For gasses, they are easily compressible, meaning their specific volume can easily change, while solids and liquids cannot change volume that easily so we just say that ...

Luc Zelissen 1K

I believe that anytime the the concentration of H30+ = concentration OH- is considered neutral even if the ph isn't 7, which is neutral for water at room temperature. If kw changes because of a change in temperature then we should expect the ph scale and what is acidic, neutral, and basic should al...

Luc Zelissen 1K

Yes, 500 cal/g is the middle point of a phase change from liquid to vapor. Half of the water would be in the liquid state and the other half would be in the vapor state. It is also important to understand that the temperature is not changing during a phase change because all of the heat is going to...

Luc Zelissen 1K

A strong Acid(high Ka) has a weak conjugate base(Low Kb). A strong Base(High Kb has a weak conjugate acid(Low Ka).

Luc Zelissen 1K

Jessica Li 4F wrote:I actually got 8.6 - I think you forgot to convert the given pKa value to pKb, as the molecule you are dealing with is actually the base.

I got the same answer as you!

Luc Zelissen 1K

Any [H30+] over 1 will have a negative pH, because the -log(1)= 0, because 10^0=1. For example if the molarity is 2, to find the pH you should do -log(2), which is equal to -.3

Luc Zelissen 1K

https://opentextbc.ca/physicstestbook2/wp-content/uploads/sites/211/2017/10/Figure_15_03_03a.jpg
This is a graph similar to what we saw in class today.

At ~500 $\Delta Q/m$ does that mean around 50% of the water is in liquid state and the other half in vapor state?

Luc Zelissen 1K

You can do the homework on any content that we discussed in class. For the first 2 weeks, We've gone over the first 2 sections, and in Week 3 we will most likely move on to Outline 3, so you could potentially do problems from that set. If you want practice before the test you could do your homework ...

Luc Zelissen 1K

The strong acids have a lower pH because they fully dissociate and have a lower hydronium concentration. Weak acids are the opposite. You are right that strong acids has a lower pH because they fully dissociate, but they have a higher hydronium concentration. The more H3O+ you have, the lower on th...

Luc Zelissen 1K

Is the TI-36X an acceptable calculator for this course?

Luc Zelissen 1K

Calculate the molar concentration of OH- in solutions with the following molar concentrations of H3O+. c: 3.1mol/L First I calculate pH of it, using -log(3.1)=-.49. Then i use the equation pKw = pH+pOH, also written as 14= -.49 + x. I solve for x and get 14.49. Then i do 10^(-14.49) to find the conc...

Luc Zelissen 1K

I believe that anytime the the concentration of H30+ = concentration OH- is considered neutral even if the ph isn't 7, which is neutral for water at room temperature. If kw changes because of a change in temperature then we should expect the ph scale and what is acidic, neutral, and basic should all...

Luc Zelissen 1K

You should definitly know some strong acids, and even further you should know what the difference between strong acids and weak acids.

Luc Zelissen 1K

If you had Cl- as an inert gas it would be Cl- on the reactant side and Cl- on the product side. If you add this into the chemical equilibrium constant equation, the concentration of Cl- will be on the top and bottom of the division sign, so they will simply cancel each other out.

Luc Zelissen 1K

If a question gives you concentration you can still use the equation, because (n/V)= concentration, where n is number of moles and V is volume. moles/volume is molarity and that is concentration

Luc Zelissen 1K

If the questions asks for just K, and not Kc or Kp, then you need to look at the equation. If the reactants and products are gases then you should use Kp, because it is implied. If the reactants and products of the equation are aqueous then you should use Kc because it is implied. However if it aske...

Luc Zelissen 1K

Hi,
I was wondering if I had access to the solutions manual for my textbook. I bought the sapling learning pack that included the textbook, but I cannot seem to find the solutions manual. Please help.

Luc Zelissen 1K

I'm not sure how you got x=0.0114, but if you plug in all the values into the K c ratio so that it's 1.00*10 -5 =(2x) 2 /((0.114-x)(0.114-x)) and solve for x using the quadratic formula, you should get x=1.8*10 -4 . I did 1.00x10^-5=(2x)^2/(0.114)^2 instead of (0.114-x)(0.114-x) for the denominator...

Luc Zelissen 1K

I believe your question will be adressed in thermodynamics, with equations like $\Delta G=\Delta G^{0} + RTLnQ$ , where we are able to calculate the effect of temperature on a reaction

Luc Zelissen 1K

A sample of ozone,O3, amounting to 0.10 mol, is placed in a sealed container of volume 1.0 L and the reaction 2O_{3}(g)\rightarrow 3O_{2}(g) is allowed to reach equilibrium.Then 0.50 mol O3 is placed in a second container of volume 1.0 L at the same temperature and allowed to reach e...

Luc Zelissen 1K

A quick note, this problem is not listed as a homework problem. For 5G it says Problems 5G: 1, 3, 7, 9, 11;

Luc Zelissen 1K

A sample of ozone,O3, amounting to 0.10 mol, is placed in a sealed container of volume 1.0 L and the reaction 2O_{3}(g)\rightarrow 3O_{2} is allowed to reach equilibrium.Then 0.50 mol O3 is placed in a second container of volume 1.0 L at the same temperature and allowed to reach equilibrium....

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