CHEMISTRY COMMUNITY

Sanjana Munagala_1j

The problem says that the reactants are equal to half the partial pressures of the product. Therefore he set up an expression where x would equal the partial pressure of reactants and 2x would be the partial pressure of products.

Hope that helps!

Sanjana Munagala_1j

Thanks again Lyndon! Like everyone has already said you have been very helpful and a part of the reason for why I have been doing well in chemistry. I hope your future endeavors go very well for you, as I am sure they will. Good luck for you and your future! Thanks again!

Sanjana Munagala_1j

That is correct. The units of rate will always be concentration over time and the rate constant will differ based on the order so that units align on both sides of the equation.

Hope that helps!

Sanjana Munagala_1j

So if you are calculating entropy at a constant temperature, use the equation S=q/t. If you have changing temperatures then you would us the integral equation which can be simplified to S=nCln(T2/T1)

Hope that helps!

Sanjana Munagala_1j

No, catalysts will not affect the equilibrium constant since the catalyst affects the rates of both the forward and reverse reactions equally.

Hope that helps!

Sanjana Munagala_1j

You could be able to determine the order of a reaction or the order in respect to a certain reactant by experimental data. In this case you would see how the concentration changes in relation to the change in rate. Another way to find the order of the reaction is through the units of K, the rate con...

Sanjana Munagala_1j

You would use the second equation with Change in U= H -PV, since at constant pressure q=H. However, it makes it easier for me to just memorize the first equation U=q +w since it also resembles the first law of thermodynamics, but also because it is the skeletal equation that other equations can be d...

Sanjana Munagala_1j

Generally these questions are applicable to concentration cells. When I approach such problems, I think about the concentration difference between the two cells. If reducing the concentration of one half cell will reduce the concentration difference between the two half cells, then E cell will decre...

Sanjana Munagala_1j

Electromotive force is an analogous term to standard cell potential. It is considered to max cell potential essentially when there is no current flowing. This is because, before any electrons are transferred, the cathode side has the strongest affinity toward electrons. This "greed" or pul...

Sanjana Munagala_1j

I personally would write out the individual half reactions and balance them separately. Then when I combine then, I consider the least common multiple and the n will be the number of electrons in the half reactions that get cancelled out when balancing the overall reaction.

Hope that helps!

Sanjana Munagala_1j

We need to reverse the reaction in order to get an oxidation half reaction. When you look at the half reactions in the appendix, the reactions are given as reduction reactions. Therefore, whichever half reaction takes place in the anode must be flipped.

Hope that helps!

Sanjana Munagala_1j

The standard electron potential is basically a measure of an atom's affinity for electrons. If the standard electron potential is (+) then a molecule has a high affinity for electrons and is more likely to be reduced. If the electron potential is (-) then the molecule has a low affinity for electron...

Sanjana Munagala_1j

I have seen some cell diagrams on the homework that include Hg(l) as the end without any solid metal to act as the conductor. Is this acceptable and if so why?

Sanjana Munagala_1j

That is correct! You could see these charges in the appendix as well with all the half reactions.

Hope that helps

Sanjana Munagala_1j

D is actually deuterium and the values can be found in the Appendix

You could use that formula at 25 degrees celsius since the values in the appendix are at that temperature. However, for 150 degrees celsius, you'll need to use the equation delta G= delta H-T(delta S)

Hope that helps!

Sanjana Munagala_1j

Would we be incorrect if we wrote OH-(aq) instead of KOH(aq)? Also how do we know that using solid Nickel as a conductor would not interfere with the reaction? Would we be incorrect if we wrote Pt(s) as the metal conductor? And finally, when I balance the reduction reaction I get: Ni(OH)3(s) + 1e- -...

Sanjana Munagala_1j

Because the ratio is really small, we know that the denominator is going to be a higher number. The denominator in the ratio is the protonated drug. Thus the protonated drug would be in larger quantity allowing us to conclude the drug would remain protanated. Alernatively, you can compare the pKa to...

Sanjana Munagala_1j

I believe it depends on the species. Usually when something is reduced or oxidized, it remains the same atom but just has a different charge. Thus you can separate it by the atoms.

Hope that helps!

Sanjana Munagala_1j

You want to remember that cell potential needs to be positive in order to have a favorable redox reaction. For this reason when you have two redox reactions calculate the E cell using the equation: Ecell=Ecathode - Eanode. Use this equation twice putting the individual cell potentials for the half r...

Sanjana Munagala_1j

I believe you can if you wanted but since it will be on both sides of the equation it would cancel out anyway. Not entirely sure about this answer though.

Hope that helps!

Sanjana Munagala_1j

R is a constant that you can find on the equations sheet posted on Dr. Lavelle's website. The one you want to use here is 8.314 J.K^-1.mol^-1. T is the temperature which in this problem is given as 1200. K.

Hope that helps!

Sanjana Munagala_1j

You need to also be cautious when you have a system that has an entropy change due to both a temperature change and a volume change. When you calculate the entropy change for just the temperature change you have to make the assumption that the volume is constant.

Hope that helps!

Sanjana Munagala_1j

You can try to understand this fact in one of two ways. 1) In an isothermal process temperature is constant so the delta T is equal to zero. If you recall the equation, delta U=3/2R(deltaT) if delta T is equal to zero then delta U is equal to zero. 2) If we want to keep the temperature constant for ...

Sanjana Munagala_1j

I want to also point out that you use Cv also when solving for an entropy change in a system that has both a volume change and a temperature change. This is because you have to treat the two changes independently. Therefore when calculating the entropy change due to a temperature change, you have to...

Sanjana Munagala_1j

Is a constant volume bomb calorimeter an isolated system? I remember Professor Lavelle gave us an example of an isolated system being glucose combustion in bomb calorimeter. However, this would imply that internal energy is equal to zero, work is equal to zero, and therefore heat would equal zero. H...

Sanjana Munagala_1j

Usually, if you are looking for a variable that remains constant throughout a change in the system, to calculate it, use the conditions at the final state of the system. Do not use change in any of the variables if you are looking for a variable that is also constant.

Hope that helps!

Sanjana Munagala_1j

If I am not mistaken, if HCl was the limiting reactant, then all of the Zinc would not dissolve which affects the amount of moles of Zinc that are used when determining the standard enthalpy of the reaction. After calculating the enthalpy of the reaction, you get that answer as per mole of the react...

Sanjana Munagala_1j

When we are calculating ΔH vaporization we could calculate it by subtracting ΔHf of the liquid from ΔHf of the gas. (ΔHf gas- ΔHf liquid) or (final - initial). However, usually these values are given to you as per mole values and so when calculating them in problems just multiply the given values by...

Sanjana Munagala_1j

Like the previous commenter said, q is only the heat released or absorbed due to a temperature change. In order to account for a phase change you need to use the enthalpy of vaporization or fusion, depending on what a question is asking you. For example, if a question asked about the enthalpy for a ...

Sanjana Munagala_1j

These values are the enthalpy of fusion and the enthalpy of vaporization used for phase changes. You can find these values on some tables located in Chapter 4C or 4D I believe.

Hope that helps!

Sanjana Munagala_1j

A lower heat capacity implies that not much health is required to change the temperature. In other words that change in temperature for the heat input will be large, making the slope on a heating curve steeper. On the other hand, if a substance had a higher heat capacity, such as water, it means tha...

Sanjana Munagala_1j

I would agree with that reasoning. To add to what you said, we use the integral for reversible expansions because we are only changing the pressure by an infinitesimal amount which requires the use of an integral.

Hope that helps!

Sanjana Munagala_1j

Honestly that is the biggest difference, and because of the difference there are two different equations that can be used. For irreversible reactions, use the equation w=-P(change in V) since pressure would be constant. In a reversible reaction, use the integral equation which can be simplified to w...

Sanjana Munagala_1j

At constant Pressure:
q= change in H
Change in U= q + w
w= -P(change in V)

At constant volume:
Change in u= q
w=0

I am not sure if I hit everything, but I hope this gives you some idea.
Hope this helps!

Sanjana Munagala_1j

I believe you would treat the kettle and the water as one whole system and everything around it as the surrounding. Because you treat them as one, you can simply add the heat changes for both of them, assuming constant pressure. Heat at constant pressure is enthalpy which is a state function and thu...

Sanjana Munagala_1j

I'm not sure if there are different versions of solution manuals, but my solution manual has the correct specific heat capacity for water and the correct specific heat capacity for copper.

Sanjana Munagala_1j

The first step is to find the heat capacity that is calibrated by the calorimeter. To do this use the eqaution C=q(calorimeter)/change in T. Essentially you would do C=(3.50kJ)/(7.32K) = 0.478kJ/K. Then you can use this heat capacity to calculate the heat released by the reaction. So set up the equa...

Sanjana Munagala_1j

So when dealing with standard enthalpies of formation, we are looking at reactions at standard conditions. Does this mean we can't use this method for reactions that are not in standard conditions. If we can, how can we account for the difference. Please let me know!
Thanks!

Sanjana Munagala_1j

The general idea is that for a weak acid, the Ka value is going to be very small. Let's say you have this reaction for an acid in water: HA(aq) + H2O --> H3O+(aq) + A-(aq) The Ka for this reaction would be: [H3O+][A-] -------------- [HA] Since the reaction is of a weak acid, it won't fully dissociat...

Sanjana Munagala_1j

You could also use an ICE table for this problem:
A B C
I 27.5 0 0
C -2(5) +5 +2(5)
E 17.5 5 10

Where 5 is equal to "x"
Therefore the reaction would be:
2A --> B + 2C

Hope that helps!

Sanjana Munagala_1j

I know in biology we look at delta G to see if a reaction is endergonic or exergonic, but how come in chemistry we look at delta H to see if a reaction is endothermic and exothermic? Are the terms endergonic and endothermic different? Please let me know?
Thanks!

Sanjana Munagala_1j

What is the difference between heat(q) and enthalpy? I know heat is not a state property and enthalpy is a state property. Also are the two related somehow as well? Please let me know!
Thanks!

Sanjana Munagala_1j

Yes it would reflect the number of decimal places of pOH. 14.00 is just a set number, for example when you would use molar mass. You can adjust the number of decimal places in Kw to match with the decimal places in pOH.

Hope that helps!

Sanjana Munagala_1j

Usually when we determine a shift in a reaction by comparing the number of moles on each side of the reaction it is usually when analyzing a change in the reaction given a change in pressure. In such a case, if you increase pressure, a reaction would shift to the side with less moles of gas in order...

Sanjana Munagala_1j

When he says Cl- doesn't react with water, he means that it doesn't exhibit a strong enough pull to withdraw a hydrogen proton from a water molecule and thus doesn't form hydroxide ions. On the other hand, a molecule like CH3COO-, the conjugate base of a weak acid would react with water to remove a ...

Sanjana Munagala_1j

K is usually a given and a fixed value for a given reaction at equilibrium for a certain temperature. For this reason, in order to determine if a reaction is at equilibrium, you have to solve for Q, exactly how you would solve for K, and compare the calculate Q value with the K of the reaction. If Q...

Sanjana Munagala_1j

I believe so, I keep getting the same answer as you well. The only calculations that you would do this for this problem is:
(1.0 x 10^-14)/(3.1) in which you would get 3.2 x 10^-15 M.

Hope that helps!

Sanjana Munagala_1j

When you take away a product, the reaction needs to get back to equilibrium so it will have a net movement toward the products in order to maintain the same ratio of products and reactants at equilibrium.

Hope that helps!

Sanjana Munagala_1j

If you think of a reaction written like this: H2 + H2 + O2 -> H2O + H2O then the way you would write the equilibrium constant equation is like so: K = [H2O][H2O] ---------------- [H2][H2][O2] or simply K = [H2O]^2 ------------ [H2]^2[O2] which aligns with the reaction: 2H2 + O2 -> 2H2O Hope that hel...

Sanjana Munagala_1j

No you would include aqueous molecules in the equilibrium equation since you are able to calculate concentrations from them.

Hope that helps!

Sanjana Munagala_1j

At equilibrium the forward reaction rate and the reverse reaction rate are equal. This however, does not imply that the concentrations of products and the concentrations of reactants are equal. However, what is true at equilibrium is that concentration of reactants remain the same over time and the ...

Sanjana Munagala_1j

The 5.5 x 10^-6 is the value of x that you can calculate after setting up your ICE table: Cl2 Cl I 0.001 0 C -x +2x E 0.001-x 2x Then set this up with Kc so you get: 1.2x10^-7 = (2x)^2/0.001-x Use the quadratic equation to solve for x and plug it back in to to get the equilibrium concentrations. Hop...

Sanjana Munagala_1j

No the values would not be the same. If they were the same then this conversion P=n/V has to be true for all the gases which is highly unlikely. Kp and Kc values are different but all Kc values are the same for the same reaction at a certain temperature and the same goes for Kp values. Hope that hel...

Sanjana Munagala_1j

These molecules make the solution more acidic because of the strong highly charged cations, Al 3+ and Cu2+, If you recall these form coordinate complexes with water. Because the pulling power of the cations are really strong, it weakens the bond between the O-H bond in the water molecule and it allo...

Sanjana Munagala_1j

F- is that conjugate base of a weak acid and thus it will result in the whole solution being basic. This is the reaction: F- + H20 = HF + OH- please note the equal sign is representative of the two arrows Since this reaction makes more hydroxide ions, it increases the pOH and thus decreases the pH. ...

Sanjana Munagala_1j

Polydentate is when a single ligand has multiple sites where it can bind to a metal cation. Chelate refers to the whole complex which contains a ligand that forms a ring of atoms that include the central metal atom. When talking about the chelating ligand, it essentially is a polydentate ligand that...

Sanjana Munagala_1j

Though it is different from what polydentate means, essentially if a ligand is polydentate then it can form a ring with the metal cation and thus also be considered a chelating ligand.

Hope that helps!

Sanjana Munagala_1j

There should be common examples of polydentate ligands that you should be familiar with that can help you with this problem. For example, knowing edta, en, dien, oxalate are the most common ones. If you don't have these molecules, for example in water and carbonate ion, there are two things I look f...

Sanjana Munagala_1j

First you want to look at the charges of all the atoms other than the cation within the brackets. After determining that, figure out what the charge of the cation has to be so that when it is added to the rest of the charges of the atoms inside the bracket, it yields the total charge of the whole co...

Sanjana Munagala_1j

When the complex is a cation, naming rules remain the same. However, when the complex is an anion, then at the end of the name where the transition metal is stated, the ending of the metal becomes "-ate".

Hope that helps!

Sanjana Munagala_1j

There are also common polydentate compounds that might be useful to know. They are mentioned in the textbook. Oxalate, edta, ethylenediamine, and diethylenetriamine are probably the ones that you should be familiar with.

Hope that helps!

Sanjana Munagala_1j

If you get a question asked like this, I would probably include both answers just to be safe.

Hope this helps!

Sanjana Munagala_1j

Since a coordinate bond is formed when a lewis base donates both electrons to a lewis acid, why is it still considered a covalent bond?
Thanks!

Sanjana Munagala_1j

The reason Neon has weaker London Dispersion forces is because it has a lower amount of electrons compared to Argon that can fluctuate and cause instantaneous dipole interactions. You can also consider how Neon has a lower polarizabilty and thus has a lower potential energy making the LDFs weaker. H...

Sanjana Munagala_1j

Be sure to also consider VSEPR shape, or how the molecule would look in 3D. Sometimes when you draw the 2D Lewis structure, it seems as if the dipoles should cancel out however, this is not the case for certain shapes, specifically tetrahedral molecules.

Hope that helps!

Sanjana Munagala_1j

You should also take into consideration the VSEPR shapes. For example, if you have a tetrahedral, CH4, you know it is a nonpolar molecule because of symmetry. However, if you had H2Cl2C, you might draw a lewis structure where the two chlorine atoms are opposite from each other and the same for the h...

Sanjana Munagala_1j

You can also have a t-shaped molecule when there are 6 total electron density regions. 3 of them being atoms and 3 being lone pairs.

Hope that helps!

Sanjana Munagala_1j

The relationship is not necessarily between polarity and melting point. The relationship is between intermolecular forces and melting point. If a molecule is polar, then it has dipole dipole forces which can contribute to a molecule having a higher melting point or boiling point.

Hope that helps!

Sanjana Munagala_1j

It would also be helpful to know the general bond angles since there might be a question where you have a molecule with lone pairs in which case the bond angles are slightly lower than the normal angles we learn in class. You won't need to know the exact bond angle but just compare it with the commo...

Sanjana Munagala_1j

For test 2, the T.A.'s usually make the test and they make it where it is very straightforward. For this reason I would assume that it is more likely where a question would give us a compound and we would be asked to either draw the lewis structure or name the VSEPR shape.
Hope that helps!

Sanjana Munagala_1j

As long as you are able to identify the VSEPR shape you should be more than fine. Professor Lavelle jokingly repeated in lecture that he was not going to test us on our artistic abilities.
Hope that helps!

Sanjana Munagala_1j

Think of the subscripts like exponents if we had X^0 this would yield one, similarly when there are no lone pairs you could think of it as E^0 or simply 1.
Hope that helps!

Sanjana Munagala_1j

I think for these type of molecules you just have to focus on the central atom and any other atoms or lone pairs solely surrounding the central atom. So in this problem Be is only surrounded by two C atoms and therefore it has two regions of electron density meaning that the VSEPR shape would probab...

Sanjana Munagala_1j

Wait this does not make sense. Do you by chance mean BaO instead of BO, because in that case MgO would be more polar since it has more covalent character due to a lower electronegativity difference. Furthermore, Mg2+ has more polarizing power than Ba2+ because it is a smaller cation and thus the eff...

Sanjana Munagala_1j

Another important rule to note is that if an atom does have to a formal charge, it is more stable for a negative formal charge to be placed on the atom with the higher electronegativity, since that is the atom that pulls on electrons more and thus achieves that negative charge.

Hope that helps!

Sanjana Munagala_1j

Normally we are not given electronegative values. Because of this, I recommend using periodic trends. Cations with lower electronegativity, which are located in the bottom left of the periodic table paired with anions with higher electronegativity, which are located in the upper right of the periodi...

Sanjana Munagala_1j

Yes polarizabilty does increase with atomic radius and charge, but it also increases with atoms that have lower electronegativity, which means the outer electrons feel less of a pull from the nucleus and thus are more likely to get distorted and form covalent bonds with cations.

Hope this helps!

Sanjana Munagala_1j

It might also be helpful to know that electronegativity refers to an atom's electron pulling power when bonded with another atom. For ionization energy, it is just the energy required to remove an electron from the single atom itself.

Hope this helps to distinguish the two!

Sanjana Munagala_1j

In general the central atom is the atom with the lowest ionization energy. In some cases, this is not the case for example, in HOCl where the central atom is oxygen and not chlorine or hydrogen. This is mostly due to the formal charge. So for formal charge it is ideal to have all the atoms have a fo...

Sanjana Munagala_1j

Yes there are several exceptions. The first exception is for any elements after Period 3, which could have expanded octets since they have a d-subshell to fill electrons. The other exception are any of the first 5 elements that can have less than an octet. It might be easier to remember that the 4 e...

Sanjana Munagala_1j

The equation to use is:
E(photon)= work function(or threshold energy) + Kinetic energy of electron
Honestly I would write down this equation whenever you have a question involving electrons being ejected from a metal.
Hope that helps!

Sanjana Munagala_1j

Actually sorry, I understand why I can't do all double bonds for a and b because of wanting the formal charge on the atom with the highest electron affinity. However, for part c why can't we bond the atom h and atom cl and then double bond the rest of the oxygen atoms with cl? Wouldn't this reduce t...

Sanjana Munagala_1j

For this same question since the elements after period 3 can have expanded octets, they can hold up to 18 electrons right? So how come for part a, we couldn't double bond all the oxygen atoms to iodine? This problem is consistent with the other parts of the problem as well.
Thanks

Sanjana Munagala_1j

The atomic radius is half the distance between the centers of two neighboring atoms. We have not really needed to calculate atomic radius as of now, and I don't think we will need to know how to calculate it for this class.
Hope that helps!

Sanjana Munagala_1j

Normally, I look at the electron configurations. For example if we had chromium whose configuration is [Ar]3d^54s^1, you know there are 6 valence electrons. For any element's electron configuration, the number of electrons after the preceding noble gas is the number of valence electrons for that ele...

Sanjana Munagala_1j

So when we were drawing the Lewis structure for methane (CH4) in lecture, we drew the structure with carbon in the middle and the 4 hydrogens equally spread around the carbon. However since carbon has two valence electrons that are paired, does this not make a difference on the structure of the mole...

Sanjana Munagala_1j

So this problem asks to write out the valence-shell configuration using quantum numbers. I got a and d but am confused about b and c. B asks for the notation for Group 15 elements and the answer is ns^2np^3, but I don't understand why they did not include the d-orbitals. And C asks for Group 5 trans...

Sanjana Munagala_1j

So I know Professor Lavelle said that the bond length for a double bond is shorter than the length for a single bond due to more electron interaction. However, does electron repulsion not affect the length, implying that a double bond would be longer than a single bond?
Thanks!

Sanjana Munagala_1j

For part a it is ok to use De Broglie's equation since you are solving for the wavelength of an electron which has mass and velocity. Just keep in mind for part c you have to use the equation we learned when discussing the photoelectric effect because it asks for the wavelength of the radiation whic...

Sanjana Munagala_1j

When writing electron configurations, you want to describe where electrons would be located when they are in their ground state or exhibit the lowest energy. In these two exceptions, having a half full or a full d-subshell is more stable than having a full s-subshell, which also makes sense, since i...

Sanjana Munagala_1j

So its easiest to understand the concepts when you are able to understand the quantum numbers. So n is the principle quantum number and it represents the energy level and a shell that an electron in is. For example if an electron is said to be in the 6px orbital, then n is equal to 6, or the electro...

Sanjana Munagala_1j

According to a video that was shown to me in discussion, manipulating equations causes there to be a discrepancy when solving for the mass of a photon. This is why we assume photon's have a different type of momentum, one that is entirely dependent on energy and not mass.

Hope that helps!

Sanjana Munagala_1j

From what I heard from my TA. It won't really cover fundamentals but will cover the quantum world and chemical bonding!

Hope that helps!

Sanjana Munagala_1j

7.00 x 10^2, has three sig figures because any numbers after a decimal point counts. If the number was just 700 then it would only be 1 sig fig and can be written as just 7 x 10^2. If it was 700. (followed by a period), then there are three sig figs.

Hope that helps!

Sanjana Munagala_1j

1/2mv^2 is the equation that you would use. Just note that since the units for the energy is given in kJ/mol, divide the answer by avogadro's constant so that you get energy per photon.

Hope that helps!

Sanjana Munagala_1j

As of now, we haven't really used amplitude, maybe it will be useful later on. All I know is that higher amplitude symbolizes higher intensity, but don't get that confused with higher energy, which is determined by frequency.

Hope that helps!!

Sanjana Munagala_1j

Is there anyway to solve this problem without using the Rydberg equation, similar to how Professor Lavelle wants us to solve such questions?

Sanjana Munagala_1j

How do we get the name of the compound tin(IV) oxide from the empirical formula SnO2?

Sanjana Munagala_1j

when using the first equation, n=cV, you are solving for moles:
moles = concentration * volume
the second equation, m=cMV should be used when mass (grams) is involved:
mass = concentration * molar mass * volume

Hope that helps!

Sanjana Munagala_1j

Are you asking about how to write ionic equations or individual ionic compounds? If the latter maybe this link will be better:
http://www.occc.edu/kmbailey/Chem1115Tu ... _Ionic.htm

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