CHEMISTRY COMMUNITY

Jordan Young 2J

You would only use the coefficients of the reactants, not the products, because the rate laws only use the concentrations of the reactants in the reaction.

Jordan Young 2J

You plug the concentrations into the first order integrated rate law equation to find k
ln[A]=-kt+ln[Ainitial]

Jordan Young 2J

Dr. Lavelle said that we won't have to develop our own reaction mechanism. We just need to be able to decide if a given proposed mechanism supports a rate law, what a rate law would be given a mechanism, etc. But we won't be ask to develop a full mechanism without given much information.

Jordan Young 2J

You can assume x is small when the equilibrium constant (K) is less than 10^-3 as a general rule of thumb. But to test it, after you calculate the equilibrium concentration, you can do the equilibrium concentration of the conjugate base divided by the initial concentration of the weak acid times 100...

Jordan Young 2J

If there is a catalyst in the slow step of a reaction mechanism, is it included in the rate law even if it cancels in the overall reaction?

Jordan Young 2J

I think it means that you need in ensure that there isn't product, because if there is then the reverse reaction can occur and you are trying to studying just the rate of the forward reaction with the integrated rate law(which only considers reactant). Therefore, you need to either remove product or...

Jordan Young 2J

I think it just means that the redox reaction isn't spontaneous (assuming that the cell potential was calculated correctly). A negative cell potential means that there is no energy to do work and the reaction is not spontaneous.

Jordan Young 2J

The question states: For each reaction that is spontaneous under standard conditions (that is, K > 1), write a cell diagram, determine the standard cell potential, and calculate delta G for the reaction. My question is, are we suppose to know if the reaction is spontaneous before calculating the cel...

Jordan Young 2J

To find the equilibrium constant at 150 Celsius, why can't you use △G=-RTlnK? Why do you need to use the Van't Hoff Equation? For reference: Calculate the equilibrium constant at 25 C and at 150 C for each of the following reactions, using data available in Appendix 2A: NH4Cl(s) --> NH3(g) 1 HCl(g)

Jordan Young 2J

Can you use △G=-RTlnK at any temperature or does it need to be at standard conditions (298K)?

Jordan Young 2J

The solid isn't included in the equilibrium constant. The salt (C6H5CH2CH(CH3)NH3+Br-) is a solid, but the Br- doesn't affect the solution so in the equilibrium constant, it is only C6H5CH2CH(CH3)NH3+, which is not a solid.

Jordan Young 2J

The coefficient matters in terms of redox reactions. For 2Fe2+ you will need 4e- to balance the charge to 0, but for Fe2+ you only need 2e- to balance the charge to 0. However, Fe2+ has a 2+ overall charge regardless of the coefficient and H+ has a +1 charge regardless of the coefficient, etc.

Jordan Young 2J

They aren't included in the half reactions. In this case, the graphite and platinum are used at inert conductors to allow the flow of electrons in the half reactions.

Jordan Young 2J

Write the half-reactions and the balanced equation for the cell reaction for each of the following galvanic cells: Pt(s)|O2(g)|H+(aq)||OH-(aq)|O2(g)|Pt(s) The answer is: 2H2O --> 4H++ O2 + 4e- for the anode and O2 + 2H2O + 4e- --> 4OH- for the cathode How do you know that H+ and O2 are both products...

Jordan Young 2J

The question asks: Write the half-reactions and the balanced equation for the cell reaction for each of the following galvanic cells: b) C(gr)|H2(g)|H+(aq)||Cl-(aq)|Cl2(g)|Pt(s) The answer gives the half rxns a: H2(g)-->H+ + 1e- Cl2(g)+2e- --> 2Cl-(aq) My question is how do you know that Cl2 is the ...

Jordan Young 2J

In some cases, the Cl2 will lose electrons and become oxidized and act as a reducing agent and in other cases the Cl2 will gain electrons and become reduced, acting as an oxidizing agent. It'll just do both in the system since it's the only reactant and it is being both reduced and oxidized in terms...

Jordan Young 2J

n is the number of moles and it comes from the fact that the value of Faraday's constant is the charge per mole so that's why you have to multiply it by the number of moles.

Jordan Young 2J

An oxidizing agent does actually gain electrons and becomes reduced. A reducing agent loses electrons and then becomes oxidizes. It is a little weird because it seems backwards but I like to think of it when it's a paired redox reaction and how the oxidizing/reducing agent helps another molecule bec...

Jordan Young 2J

I believe we assume that entropy remains constant because I think we would normally use that equation to find equilibrium constants based on temperature. Also, entropy gets cancelled when you try to find K at two different temperatures if delta H is know

Jordan Young 2J

The salt bridge allows the charges of the two compartments to remain neutral. Without the salt bridge, one compartment would become negatively charged and the other would become positively charged and the battery/galvanic cell would stop working(there would be no flow of electrons)

Jordan Young 2J

Extensive properties depend on the amount of substance (such as heat capacity), but intensive properties don't depend on the amount of substance (such as specific heat capacity)

Jordan Young 2J

Enthalpy of formation of O2 is zero because it's in its most stable form of oxygen

Jordan Young 2J

If the volume increases then there is expansion work.

Jordan Young 2J

For Question 4I.9, it asks: Initially an ideal gas at 323K occupies 1.67L at 4.95atm. The gas is allowed to expand to 7.33 L by two pathways: (a) isothermal, reversible expansion; (b) isothermal, irreversible free expansion. Calculate deltaStot, deltaS, and deltaSsurr for each pathway. I got the cor...

Jordan Young 2J

I think it just means that the difference in entropy between a gas a liquid is much greater than the difference in entropy between a liquid and a solid because the entropy of gas is a lot of larger since a liquid and solid are still contained to some degree and don't have as many possible positions ...

Jordan Young 2J

It says that if and object reaches absolute zero temperature (0 Kelvin), then entropy is equal to zero.

Jordan Young 2J

In questions 4F.1 and 4F.3, they are saying that the change in entropy is less when the object is at a higher temperature, not that entropy is decreasing.

Jordan Young 2J

ΔH of fusion is the change in enthalpy required for a substance to undergo a phase change from liquid to solid (or solid to liquid) and ΔH of vaporization is the change in enthalpy required for a substance to undergo a phase change from liquid to gas (or gas to liquid). You would use these values to...

Jordan Young 2J

If work is done by the system then your work value for the system would be negative because the system is losing energy due to work. If the work is done on the system, you work value for the system would be positive because the system is gaining energy.

Jordan Young 2J

A reversible reaction is when you incrementally increase the volume and allow the pressure to come to equilibrium at each small step so because of the infinite small steps, the process is the maximum possible work.

Jordan Young 2J

In this problem you are using Hess's Law to find the reaction enthalpy. You have to rearrange/multiply the given equations to eventually end up with the equation H2 + Br2 --> 2HBr when you add all of the reactions together. To do this you will need to reverse the first equation and multiply it by a ...

Jordan Young 2J

qp is enthalpy at constant pressure and qv is at constant volume. They are used when finding enthalpy/delta H but also find change in internal energy which is q + w so q can either be qp or qv depending on the given conditions

Jordan Young 2J

I think it might be because of significant figures

Jordan Young 2J

Because the change in thermal energy is equal to q + w and since there isn't any work being done on the system, I believe that the answer is simply delta U = q and q = delta H so the answer is -320 because that is the value of delta H in the reaction for 1 mol of OF2.

Jordan Young 2J

Enthalpy is the change in heat or delta H. Entropy is a measure of randomness/disorder. Enthalpy is used in calculating q, work, and energy due to heat. Entropy can be used to measure delta G which is the change in free energy, but we haven't discussed that yet.

Jordan Young 2J

I think that that conversion factor is only for if the gas is at standard temperature and pressure. Otherwise you can use the ideal gas law to convert from liters to mols at any temperature and pressure.

Jordan Young 2J

You use it to check if you can appropriately approximate x to be a small enough that it would effect the initial concentration that much when using the calculation involving K. You will most likely use this when solving equilibrium equations.

Jordan Young 2J

If delta H is positive (endothermic) and temperature is increased, then the reaction would proceed towards the products. If delta H is negative (exothermic), then the reaction would proceed towards the reactants if temperature is increased.

Jordan Young 2J

5.5*10^-6 is the value you get for x when you solve for the change in concentration using the equilibrium constant and it's multiplied by 2 because of the stoichiometric ratios in the equation.

Jordan Young 2J

A buffer has a weak acid/base and its conjugate so that it can thus resist a chance in pH. You want to have large enough concentrations of both so that it can react if an acid or base is added to the solution

Jordan Young 2J

If you need to find the equilibrium concentration of a reactant of product or need to use the equilibrium concentration in a calculation you generally will need an ICE table. Also if you have to find the pH of the solution given the concentration of reactant, you will also most likely need an ICE ta...

Jordan Young 2J

Yes you can. You just have to make sure that all of the values in your ICE tables have the same units (initial, change, and equilibrium) and that if you are going to calculate a Kc or Kp value, that you convert to the proper units.

Jordan Young 2J

Yes the statement would be true because if you start with a higher concentration of reactants, you will end up with a higher equilibrium concentration of products since the K stays the same.

Jordan Young 2J

Temperature just has an affect on the K constant. So in this case, at that temperature, the equilibrium constant Kw is equal to 2.1*10^-14 instead of 1*10^-14 which is Kw at 25 degrees Celsius. From the question you know that the concentration of H3O equals the concentration of OH- because it is a s...

Jordan Young 2J

Solids don't have concentrations because it doesn't make sense for it to have one. Liquids/solvents in reaction generally do not change much in terms of concentration so we assume that the value stays the same and will be on both sides of the reaction so it has no effect on the equilibrium constant.

Jordan Young 2J

Solids don't have concentrations because it doesn't make sense for solids to have a concentration. The concentration of liquids (which are usually solvents) don't change that much during a reaction so it is therefore not included in the equilibrium expression.

Jordan Young 2J

It doesn't matter which unit you use, but you have to make sure the R constant value/units corresponds with that pressure unit to make sure that units cancel.

Jordan Young 2J

I haven't done the exact problem yet but if you know the equilibrium concentrations of dichloride and chlorine or difluorine and fluorine, the one with the higher equilibrium concentration is more stable because the reactions proceeds to the more stable molecules.

Jordan Young 2J

You do include molecules that are in an aqueous solution (aq)

Jordan Young 2J

K is general term for the equilibrium constant of products over reactants. Kp means that the values in the equilibrium constant are the partial pressures of usually gases. Kc means that the values in the equilibrium constant are concentrations of aqueous solutions

Jordan Young 2J

Increasing the concentration of the products will cause the reaction to proceed towards the reactants in order to reach equilibrium

Jordan Young 2J

I don't think so because the cations and anions of strong acids and bases are stable so they won't act as acids or bases (ex: Cl- from the dissociation HCl is a stable ion)

Jordan Young 2J

I think it is because HBrO is a weak acid, which makes BrO- a relatively stronger base than morphine. It also may have to do with nitrogen on the morphine, but I'm not 100% sure.

Jordan Young 2J

Firstly, you would the equation En= -hR/n^2 and the equation △E = Ef-Ei where Ef is for the final state/energy level of the atom and Ei is the initial state where the atom started. Once you find the change in energy, you can use E=hv to find the frequency as the change in energy is the energy of the...

Jordan Young 2J

Each lone pair on an oxygen, nitrogen, or fluorine is a potential hydrogen bonding site and each hydrogen that is attached to an oxygen, nitrogen, or fluorine is a potential hydrogen bonding site. For example, water has 4 hydrogen bonding sites (2 lone pairs on oxygen and 2 hydrogens attached to the...

Jordan Young 2J

I think for 14A we only need to know how to calculate pH for strong acids as they 100% dissociate so calculated the pH is straight forward if you're given the concentration. You would just use the pH formula of -log [H30]

Jordan Young 2J

I believe you can tell if an acid or base is strong based on the K value, but we haven't learned this. I think we just have to know the main strong acids (HI, HCl, HBr, HClO4, HClO3, etc.) and main strong bases (NaOH, KOH, etc.). The rest are weak that don't 100% deprotonate/protonate. Also, I think...

Jordan Young 2J

I think that when it comes to the ligands in the coordination complex, the order doesn't matter as much. I think it matters more in actual name.

Jordan Young 2J

The Roman Numeral (III) is for the chromate. There are 3 K so that the charge is neutral. The 6 CN have a total of a 6- charge, the Cr has a 3+ charge, and since each K has a 1- charge, there are 3 of them to make the coordination compound neutral.

Jordan Young 2J

The question asks which of the following ligands can be polydentate? CO3 2- was said to be mono or bidentate and oxalate was said to be bidentate. For CO3 2-, why can it be mono or bidentate? For oxalate, where are the two places that the ligand can bind to the central metal? Is it the two oxygens t...

Jordan Young 2J

I think it will be helpful at least to know the common ones such as HCl, HBr, and NaOH because I don't think we will be told if the acid or base is strong and weak unless it's one we haven't talked about.

Jordan Young 2J

I'm pretty sure 2F.3 is about figuring out the number of sigma and pi bonds!

Jordan Young 2J

For atoms that have a seesaw molecular shape, I know that you select one of the equatorial angles to be the lone pair. So does that mean that the bond angles for seesaw shaped molecules is 90 degrees, 180 degrees, and <120 degrees? If not, what are the bond angles for a seesaw shaped molecule?

Jordan Young 2J

I believe they are both the same intermolecular force so they're both the same thing.

Jordan Young 2J

Daria Azizad 1K wrote:I think that both the structure w/ two double bonds and the structure w/ one double bond and one single bond are acceptable, so they are just giving answers for both.

But wouldn't the structure with one double bond and one single bond not be ideal because it doesn't minimize formal charge?

Jordan Young 2J

Yes because when we hybridize orbitals, all of the orbitals become the same level energy, instead of separated by s,p, and d, so they are degenerate.

Jordan Young 2J

I think ground state refers to the lowest possible energy state of an atom or ion which means that there are no excited electrons in the atom or ion. Charge doesn't impact whether electrons are excited or not so that is why ions have a ground state electron configuration. The ground state configurat...

Jordan Young 2J

Yeah, for the ones we learned today, they all had no lone pairs on the central atom. But there are different shapes when the central atom has lone pairs, which we will probably learn soon.

Jordan Young 2J

Yeah there is. It's on his website under Exam Information and its called "Midterm Solutions". Its on the right hand side if you scroll down.

Jordan Young 2J

The principle quantum number is which shell the electron is in. This is determined by which period the electron of the atom is in. Period 1 has a quantum number of 1, period 2 has a quantum number of 2, etc. For example, an electron is the 2s shell has a principle quantum number of 2.

Jordan Young 2J

The Debroglie equation is used to calculate values for things with a mass (ex: electron, but not a photon). The equation is wavelength = h/mv where h is Planck's constant, m is mass in kg, and and v if speed in m/s.

Jordan Young 2J

Van der Waals are an example of intermolecular forces and during lecture he said it won't be on the midterm.

Jordan Young 2J

A more highly polarizable anion means that the anion's electrons are more likely to be pulled towards the cation because the anion has a relatively low electronegativity. So the bond is considered more covalent because electrons are being distorted and pulled away. A greater difference in electroneg...

Jordan Young 2J

I feel like we won't be asked about the specific experiment, but a lot of the experiments relate to concepts or equations that we learned in class like the Photoelectric Experiment and Photoelectric Effect so I think it would be good to know how the concepts and equations were derived from the exper...

Jordan Young 2J

It's because Be 2+ has a smaller atomic radius so its polarizing power is greater because of more effective nuclear charge so its electronegativity/pulling power is higher than Sr 2+

Jordan Young 2J

It just means to give the configuration for the outer most shell of electrons (valence electrons) as opposed to the entire electron configuration, in which you would start with 1s2 2s2 etc. Another way of thinking about it is that the valence electron configuration is the configuration from the last...

Jordan Young 2J

They have to do with double and triple bonds and how many of each are a sigma or pi bond, but I don't think we need to know this for the midterm because we haven't talked about it in class.

Jordan Young 2J

The last slide of the lecture we had today won't be on the midterm. Bond lengths and energies will probably be on the midterm. The values with either be given to you and you have to use them in a question or you'll have to understand the concept of them to answer a question.

Jordan Young 2J

I'm pretty sure it will be the same table! On the midterm review packet, it said to use the Constants and Equations sheet on his website, which I think is similar or exactly the same as the first test.

Jordan Young 2J

I don't think you need to show every single step, but I think it might be good so you don't mess up and your units match up. However, I don't think it's absolute necessary to show every step to manipulate an equation, as long as you manipulate it correctly.

Jordan Young 2J

Formal charge is used to determine which resonance structure is "best" or the most stable. By calculating the formal charge, you can determine which resonance structure is most likely to occur. Ideally, you want the higher electronegative atoms to have the negative charges and you want to ...

Jordan Young 2J

They are related in a sense. Shielding is when electrons in a lower energy level "shield" electrons in an outer energy level causing them to be less attracted or "pulled" by the nucleus. The example Dr. Lavelle gave in class was your friend blocking you from the heat of a fire by...

Jordan Young 2J

Boron also is an exception at times and there are some elements in the periodic table that can hold more than 8 electrons, but I think we'll talk about these in class at some point.

Jordan Young 2J

I think it means that electronegativity is value that is calculated through a formula based on other values, whereas other values such as IE, electron affinity, and atomic radius are already found and don't require a formula for calculation.

Jordan Young 2J

Yes, Na has a lower first ionization energy than Mg because of the fact that it has a weaker nuclear pull because it has less protons. Then if both lose one electron, Na+ has a much higher ionization energy than Mg+ because it now has a noble gas electron configuration

Jordan Young 2J

You would remove from the 4s state first and then from the 3d because the 4s becomes higher in energy once the 3d orbital starts being filled.

Jordan Young 2J

I think it just means to use the De Broglie equation because the De Broglie equation is derived from the equations E=pc, c= wavelength*frequency, and E=hv which we have learned before. However, this equation only works for any particle with a rest mass, momentum, and wavelength so if it asks for the...

Jordan Young 2J

I think you need to multiply the kinetic energy value you found by Avogadro's number to get the energy per mole of electrons, but I'm not 100% sure.

Jordan Young 2J

1 Angstrom is equal to 10^-10m so if you have a wavelength, for example, you can convert it from meters to angstroms with that conversion factor.

Jordan Young 2J

Yes, you write them in order from least to greatest amount of energy

Jordan Young 2J

Photons are particles of light and they do have energy, which you can calculate using the equation E=h*v. And thus, light is a source of energy.

Jordan Young 2J

For me, I think it's better to round for sig figs at the end of the problem because if you round as you are calculating, your answer is more likely to be more off than if you just round at the end of the problem. Generally, if you are fairly close in terms of sig figs and rounding I think you'll be ...

Jordan Young 2J

Rydberg is the constant R and it's equal to 3.29x10e-15 Hz. You use this constant in the E= -hR/n^2. The Lyman and Balmer Series are just names for the spectral lines for atomic spectrum for a H atom. Balmer is the visible section and Lyman is the UV section.

Jordan Young 2J

Why does the kinetic energy stay the same, in the photoelectric effect, even with a higher intensity of light?

Jordan Young 2J

When discussing the photoelectric effect, is "high intensity" the same as "increasing the number of photons"? Are they synonymous? If so, does higher intensity just mean more/brighter light?

Jordan Young 2J

The E that you plug into E=hv is the energy for the photon emitted which is why it is positive. The E you find from the problem was the energy of the electron when it moved from n=4 to n=2 which is negative because the energy decreased. The values are equal because the law of conservation of energy,...

Jordan Young 2J

Yes you're right. Mr. Lavelle's website says that you round to the nearest even digit if the last number if 5 so you are correct for rounding 1.65 to 1.6 and 1.35 to 1.4

Jordan Young 2J

There is also a document on Mr. Lavelle's website called "Everything you wanted to know about Sig Fig" if you want more clarification, but I think the above replies explains it pretty well. Zeros are only significant if they are between two numbers (ex: 705) or at the end of a number with ...

Jordan Young 2J

The first five homework problems from the textbook are due this Friday (Oct 4th) in lecture. And then starting next week, the five homework problems are due during our discussion sections each week. The five online posts on Chemistry Community are due by Sunday (Oct 6th) and then it's five every week.

Jordan Young 2J

It doesn't affect the equation itself but it may be good to know so you can visualize the experiment and sometimes elements change states if you have multiple equations that are linked together. Also, as we learn more it may be important when we expand balancing equations into greater concepts that ...

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