CHEMISTRY COMMUNITY

Shail Avasthi 2C

Dr. Lavelle has a great instructional sheet on his website on how to balance redox reactions. It explains all of this is detail. Helped me a lot for test 2

Shail Avasthi 2C

This is done using two different symbolic temperature values, plugging them into the Arrhenius Equation, and subtracting these equations to solve for the rate constants. Pretty much the same as what we did for the Van't Hoff equation (relating temperature changes to a change in K)

Shail Avasthi 2C

Catalysts are consumed then regenerated throughout the course of the reaction. Intermediates are produced during one step and consumed in another step of the reaction. Both will cancel out of the final equation.

Shail Avasthi 2C

Catalysts DO speed up reactions. You're right that catalysts lower activation energy by providing an alternate mechanism for the reaction. If you look at the exponential form of the Arrhenius equation, when you decrease activation energy, you are increasing the rate of the reaction exponentially.

Shail Avasthi 2C

Gibbs free energy is the maximum amount of energy released during a reaction that is available to do work (i.e. that energy can be used for other purposes)

Shail Avasthi 2C

The ion in aqueous solution that is growing in concentration is the product (i.e. at the anode). The ion in aqueous that is decreasing in solution (i.e. at the cathode) is the reactant. The pure metal at the anode and cathode is not involved in Q because they are solids. Solids and solvents don't fa...

Shail Avasthi 2C

A concentration essentially uses the -(RT/nF)lnQ term of the Nernst Equation to generate (a small amount) of electricity. Since the half reactions at the cathode and anode are the same, the standard cell potential is zero. Therefore, the difference in concentration (i.e. Q) makes the overall cell po...

Shail Avasthi 2C

Because if there is still potential difference between the cathode and anode, then the reaction is not at equilibrium, and electrons will continue to flow from the cathode to the anode. This generates electricity.

Shail Avasthi 2C

Three reactant molecules are involved to form the products in a single step.

Shail Avasthi 2C

A is the fraction of molecular collisions that occur at the proper angle for a reaction to occur.

Shail Avasthi 2C

The rate law isn't determined by the overall equation. Certain reactions might proceed in multiple steps. In this case, the rate is determined by the slowest step of the reaction, not the overall equation. HI is second order because the rate determining step is bimolecular with respect to HI (two HI...

Shail Avasthi 2C

Since reaction and product concentrations follow smooth curves that are defined by equations, we can easily use the instantaneous rate of change, which will give us a more accurate result than using an average. It also allows us to turn the differential rate laws into the integrated rate laws.

Shail Avasthi 2C

Lyndon's review sessions are always super helpful. Do his packet before attending the session. Other than that, probably do a lot of practice problems from the homeworks and make sure you know how to do everything.

Shail Avasthi 2C

CO is not in the rate law because its concentration does not affect the rate of the reaction (i.e. it is zero order). This implies that CO is not involved in the slowest step of this multi-step reaction. It does not mean that CO is not involved in the reaction at all. Just that it's not significant ...

Shail Avasthi 2C

Reaction rates are determined empirically by changing one reaction concentration at a time and observing the effect on the reaction rate. If you double the initial concentration of A and the rate does not change, A is zero order. If the rate doubles, A is first order. If the rate quadruples, then A ...

Shail Avasthi 2C

Calculate the oxidation number of the compounds before and after the reaction. e.g. if MnO4- becomes Mn2+, then the reactant Mn has an oxidation number of 7+ (each oxygen has a charge of -2, and there are 4 of them. The manganese needs to be +7 to bring the overall charge to -1), and the product Mn ...

Shail Avasthi 2C

Electrons always flow from the reaction that is losing electrons (i.e. oxidation), which occurs at the anode, to the reaction that is gaining electrons (i.e. reduction), which occurs at the cathode.

Shail Avasthi 2C

This is the one I learned for AP chem

the RED CAT LEO says GRR at AN OX
RED CAT: reduction occurs at the cathode
LEO: loss of electrons is oxidation
GRR: gain of electrons is reduction
AN OX: oxidation occurs at the anode

Shail Avasthi 2C

You need a solid metal to conduct the flow of electrons when both compounds of your anode or cathode are in aqueous solution. Platinum is used when it is not involved in the chemistry of the galvanic cell, and therefore will not be oxidized or reduced while acting as a conducting metal.

Shail Avasthi 2C

The oxidizing agent is the compound that becomes reduced. The reducing agent is the compound that becomes oxidized. I remember it this way: The reducing agent is the compound that starts off reduced and causes the other compound to become oxidized. The oxidizing agent is the compound that starts off...

Shail Avasthi 2C

Just because ΔΗ is constant doesn't mean it's not an important part of the equation. We are able to factor it out because it's not changing, but since its value can be something other than 1, we cannot eliminate it from the equation.

Shail Avasthi 2C

When ΔG is positive, the reaction is non-spontaneous and requires an input of energy to proceed towards the products. When ΔG is negative, the reaction is spontaneous and releases energy as it proceeds towards the products.

Shail Avasthi 2C

The released free energy is the maximum amount of energy that can be used as work. In reality, it can't all be perfectly utilized, but we calculate it that way.

Shail Avasthi 2C

In this problem, delta S and delta U are equal to 0 when referring to the change in entropy/internal energy over the course of the entire process, which the question is asking for. Entropy and internal energy are state functions, meaning we can look at the initial and final states directly without ...

Shail Avasthi 2C

In this problem, delta S and delta U are equal to 0 when referring to the change in entropy/internal energy over the course of the entire process, which the question is asking for. Entropy and internal energy are state functions, meaning we can look at the initial and final states directly without ...

Shail Avasthi 2C

I did it the way you advised and still got 22.85C as my answer for Tf. For your Step 3, shouldn't the right side of the equation be negative (e.g. -265 g*specific heat of water...)? Why do you not use the specific heat of ice on the right side? (Using the specific heat of ice gives me 20.93C for my...

Shail Avasthi 2C

Because for infinitesimally small expansion, work must be done and equilibrium must be regained. This repeats over and over and this creates a decreasing curve where as irreversible expansion occurs immediately so work occurs instantly only once and equilibrium is only regained once (this creates a...

Shail Avasthi 2C

A reversible reaction is where products and reactants are being formed at the same time, and once the forward and reverse reactions occur at the same rate, they reach equilibrium. Not sure this applies in a thermodynamic setting. In my understanding, a reversible expansion is when the external pres...

Shail Avasthi 2C

Degeneracy is the number of all possible states of the particles in a system. It is used to calculate entropy.

Shail Avasthi 2C

In an irreversible expansion, the pressure of a system suddenly changes. In a reversible expansion, the pressure is slowly changed in infinitesimal increments (requires an integral).

Shail Avasthi 2C

When ΔG is negative, the reaction is spontaneous.

Shail Avasthi 2C

Wait, but I assumed that heat would still technically be able to escape from a hydroflask to the surroundings. Are there any other examples that would be an isolated system? In reality, a hydroflask is a closed system. In theory, perfectly insulated thermos bottles would be isolated systems because...

Shail Avasthi 2C

Specific Heat Capacity = Heat Capacity per gram of substance

Molar Heat Capacity = Heat Capacity per mole of substance

Shail Avasthi 2C

ΔU = q + w, so to change the energy of a closed system, you would heat it or do work on it.

Shail Avasthi 2C

Closed systems don't allow the exchange of matter, but they do allow the exchange of energy. An isolated system is perfectly insulated, and does not allow the exchange of matter or energy.

Shail Avasthi 2C

One example would be an expanding gas pushing a piston.

Shail Avasthi 2C

Because at a certain point, the energy input is not being used to raise the temperature, and is instead being used to change the phase of the molecules (i.e. break intermolecular interactions)

Shail Avasthi 2C

Enthalpy is a state function because you can use the final and initial values as your only information to calculate the change in enthalpy. The path taken is not required information.

Shail Avasthi 2C

You calculate the percent ionization by taking the concentration of H3O+ and dividing it by the initial concentration of the acid.

Shail Avasthi 2C

First, set up an ICE table to determine the concentration of H3O+. Then take the negative log of that value to find the pH. Then subtract that value of pH from 14 to find the pOH.

Shail Avasthi 2C

The value of R you use depends on the units given to you in the problem and the units which your answer needs to be in. The units of R will cancel out with all of the given terms to give you your final answer's units.

Shail Avasthi 2C

An ICE table should be used when you are given initial concentrations and are asked to find concentrations, pH, or pOH at equilibrium. You should memorize the strong acids and strong bases.

Shail Avasthi 2C

You calculate pH and pOH by taking the negative log (-log) of [H3O+] or [OH-]. You typically calculate pH for acids and pOH for bases, but questions will often ask you for pH no matter what. Use pH+pOH=14 to convert from pOH to pH.

Shail Avasthi 2C

Ka x Kb = Kw
Kw is a constant (10^-14)
If Ka is high for a strong acid, Kb must therefore be low because their product is equal to a constant. And vice versa: if Kb is high for a strong base, then Ka must therefore be low.

Shail Avasthi 2C

@AMahadi, if your Ka or Kb is small (<10^-3), then you are allowed to eliminate all x values in the equilibrium expression which follows the form (n-x). This makes calculating the value of x more simple. If the percent ionization of the acid or base is <5%, then this approximation is valid.

Shail Avasthi 2C

Ka x Kb = Kw (10^-14)
pKa + pKb = 14

Shail Avasthi 2C

Don't include solids and liquids in equilibrium calculations.

Shail Avasthi 2C

If you're looking at aqueous equilibrium, use Kc. If looking at gasses, you can either use Kc or Kp. Use the question to determine the proper answer.

Shail Avasthi 2C

You can either use partial pressure or concentration when calculating equilibrium for gasses.

Shail Avasthi 2C

Create an ICE table using the initial and final values for A B and C in the table. Then plug these values into the K expression.

Shail Avasthi 2C

K is the ratio of the concentrations of the products and reactants when the reaction is at equilibrium. Q is the ratio of the concentrations of the products and reactants at any point in the reaction. Comparing K to Q will tell you which way the reaction will proceed.

Shail Avasthi 2C

The +/- delta symbols represent partial positive and partial negative. The +-----> arrow points in the direction of the negative end of the dipole, with the "cross" end of the arrow towards the positive end of the dipole.

Shail Avasthi 2C

No. Take the amphoteric oxides as an example (e.g. Al2O3). If you react Al2O3 with a strong acid, it will act as a base by producing hydroxide ions. If you react Al2O3 with a strong base, it will act as an acid by producing hydronium ions. None of the amphoteric oxides contain hydrogen.

Shail Avasthi 2C

VSEPR predicts the shapes of molecules by using the principle that regions of electron density around a central atom will tend to maximize their distance from one another. This principle leads to the various VSEPR geometries, such as trigonal bipyramidal, tetrahedral, etc. These shapes maximize the ...

Shail Avasthi 2C

Bent molecules will have lone pairs on the central atom in addition to the two bound atoms. These lone pairs will distort the shape of the molecule to bent instead of linear.

Shail Avasthi 2C

T shaped molecules have 3 bonded atoms and 2 lone pairs. The 5 regions of electron density have trigonal bipyramidal geometry, and the two lone pairs are placed equatorially to maximize distance; this forms a T shaped molecule.

Shail Avasthi 2C

For the following acid equilibrium :

HA <----> H+ + A-

Ka = [H+][A-]/[HA]

pKa = -log[Ka]

Shail Avasthi 2C

If an acid has more than one hydrogen which it can dissociate, it is polyprotic.

Shail Avasthi 2C

For the following reaction:

HA ---> H+ + A-

Ka = [H+][A-]/[HA]
pKa = -log(Ka)

For stronger acids, Ka will be high and pKa will be low, and vice versa.

Shail Avasthi 2C

Organic = Carbon. Therefore, organic acids have carbon while inorganic acids do not.

Shail Avasthi 2C

Strong acids fully ionize (i.e. dissociate into products), while weak acids only partially dissociate and require equilibrium calculations to determine concentrations etc.

Shail Avasthi 2C

The Ka is equal to the product of the concentrations of the products divided by the product of the concentrations of the reactants. e.g.: HA ---> H+ + A- Ka = [H+][A-]/[HA] Stronger acids more readily dissociate into their products. Therefore, the stronger the acid, the higher the Ka because the con...

Shail Avasthi 2C

Acids typically lose H+ while bases typically gain them.

Shail Avasthi 2C

CCl3COOH and CH3COOH have the anions CCl3COO- and CH3COO- respectively. The former anion is more stable than the latter because the 3 chlorines delocalize the negative charge on the oxygen. Oxoacids will be more acidic if the resulting anion is stabilized by electronegative atoms which delocalize th...

Shail Avasthi 2C

The resulting anion of trichloroacetic acid is more stable than that of acetic acid because the three chlorines delocalize the negative charge on the oxygen, whereas in acetic acid, the charge is not as delocalized. Oxoacids which have more stable anions will be more acidic.

Shail Avasthi 2C

Amphoteric compounds can have both acidic and basic character. (e.g. H2O can become H3O+ or OH-)

Dr. Lavelle pointed out in class the diagonal band of amphoteric oxides (Be, Al, Ga, Sn, Sb, Pb), quite similar to the band of metalloids. All of these oxides (e.g. Al2O3) are amphoteric.

Shail Avasthi 2C

Neutralization reactions form water and a salt as their products.

Shail Avasthi 2C

Compare the acids HClO and HBrO. Their resulting anions will be ClO- and BrO- respectively. ClO- will be more stable than BrO for the following reason: Chlorine and Bromine are both considered "electron-withdrawing atoms" because of their electronegativity, and they will have the effect of...

Shail Avasthi 2C

Oxoacids will be more acidic if the anion formed is stabilized by electronegative atoms which delocalize the negative charge. Chlorine is more electronegative than bromine, which means it will more effectively delocalize the negative charge on the oxygen, yielding a more stable anion – this causes i...

Shail Avasthi 2C

Oxoacids will be more acidic if the anion formed is stabilized by electronegative atoms which delocalize the negative charge. Chlorine is more electronegative than bromine, which means it will more effectively delocalize the negative charge on the oxygen, yielding a more stable anion – this causes i...

Shail Avasthi 2C

https://lh6.googleusercontent.com/fH1A2_-wM47qFR8X69w6002x-Iw75lx2Lxh6WiY4oTErTr2YyOOluabl5HuJ0usijqiz1QxAQSuuXIDl67mF1AwouYEiNtxQgShsJWJkDqEhFOoSuTIM5ORfbP9O You could memorize, the 5 "parent" shapes (linear, trigonal pyramidal, tetrahedral, trigonal bipyramidal, octahedral), and then in...

Shail Avasthi 2C

Not always. For example, in square planar molecules, the bond angles are still 90 degrees because the axial lone pairs directly oppose each other. Another example would be linear molecules with three equatorial lone pairs: the bond angle is still 180 degrees.

Shail Avasthi 2C

The electron geometry is the arrangement of the regions of electron density (lone pairs + bonding regions) around the central atom. The molecular geometry is the shape of the molecule, which is only determined using the bonding regions. For example, the electron geometry of H2O is tetrahedral becaus...

Shail Avasthi 2C

Lone pairs have greater repulsion strength than bonding regions do. So lone pairs will tend to push bonding regions closer together, making the bond angles "a little less then" their typical value. For example, if you have a tetrahedral with 1 lone pair, the three bonding regions will have...

Shail Avasthi 2C

All three have the 6 regions of electron density arranged as an octahedral. The difference is in the number of lone pairs and the number of bonding regions. If the molecular shape is an octahedral, then it has 6 bonding regions (i.e. 6 atoms bound to the central atom). If the molecular shape is squa...

Shail Avasthi 2C

Every bond after the first is a pi bond. (i.e. single bond = 1 sigma ; double bond = 1 sigma & 1 pi ; triple bond = 1 sigma & 2 pi)

Shail Avasthi 2C

As others have said, formal charges must add up to the overall charge of the molecule you are drawing. However, when it comes to determining which atom(s) gets the non-zero formal charge(s), use electronegativity. For example, if you draw the lewis structure for the chlorate ion (ClO3-), when decidi...

Shail Avasthi 2C

It depends on the molecule. When you arrange 5 identical atoms around a central atom (e.g. PF5), this molecule will be non-polar because of the shape of the molecule. You can also have a symmetrical molecule that is polar (e.g. CH2Cl2) due to dipole moments etc.. Symmetry is not a reliable way to de...

Shail Avasthi 2C

Yes. Lavelle said in lecture that this is the best way of going about drawing lewis structures for this class.

Shail Avasthi 2C

Hydrogen bonds are stronger than dipole/dipole or induced dipole/induced dipole interactions. H2O has the ability to hydrogen bond with itself because of the H-O bond, while H2S cannot. This accounts for H2O's higher melting point.

Shail Avasthi 2C

CHI3 is a much larger molecule than CHF3, because the size of Iodine's electron cloud is much larger than the size of Fluorine's. The larger the molecule (by molar mass and by size), the greater its polarizability and therefore the greater its LDFs (induced dipole/induced dipole)

Shail Avasthi 2C

Cations have polarizing power because their positive charge distorts the electron cloud of the anion. Likewise, anions have polarizability because their electron cloud can be distorted by positively charged cations. The greater the charge on the cation (Al3+ is a good example), the higher its polari...

Shail Avasthi 2C

Try and minimize your formal charges whenever you draw a Lewis structure, because this means that you have likely found the most stable structure.

Shail Avasthi 2C

Formal charges are a useful quick check to see if your Lewis Structure is stable or still needs reworking. I would recommend quickly checking formal charges on most lewis structures you draw for this reason, unless it's ridiculously easy or it's a structure you know already.

Shail Avasthi 2C

Actually I figured it out, but I'm going to keep this here in case anyone else is confused. The power of 10 is off by 3 because I forgot to convert grams to kilograms...

Shail Avasthi 2C

(Problem: A new element GarBreadium is discovered to have molar mass 3.157 g∙mol-1. The GarBreadium atom travels at an average speed of 1.35x103 m∙s-1 at room temperature. What is the average wavelength of the GarBreadium atom at room temperature?) My answer for this problem comes out to 9.36x10^-14...

Shail Avasthi 2C

The atom with the lowest ionization energy should be the central atom. When a molecule contains carbon, it is usually central in some way. Hydrogen should never be chosen as the central atom.

Shail Avasthi 2C

Formal charges also allow you to compare different possible lewis structures for the same molecule. For example, if the oxygen atoms in one lewis structure have positive formal charges and negative formal charges in another, the second lewis structure is probably preferable (assuming that other fact...

Shail Avasthi 2C

Add double bonds to a lewis structure when you still need to satisfy the octet rule after adding your valence electrons, or if it gives the molecule better formal charges.

Shail Avasthi 2C

The atom with the lowest ionization energy should be the central atom.

Shail Avasthi 2C

An anion is an ion with a negative charge and a cation is an ion with a positive charge. (The way I remember this is that the "t" in cation looks like a plus sign which makes it positively charged.)

Shail Avasthi 2C

Different subscripts for different orbitals (e.g. px, py, pz, dxy, dyz, dxy, etc.) are just different labels for the possible orientations of the orbitals of a certain shape, based on the coordinate system you are using. In other words, for ml = -1, 0, 1, it is arbitrary which one you label as px, p...

Shail Avasthi 2C

No two electrons may have the same 4 quantum numbers. i.e. two electrons in the same orbital must have opposite spins. It is not possible for them to have the same spin.

Shail Avasthi 2C

Electron affinity tends to increase as atoms get more electronegative, i.e. moving up and to the right on the periodic table.

Shail Avasthi 2C

You should only compare atomic radii of atoms in their ground state.

Shail Avasthi 2C

As you move right across a row in the periodic table, the positive charge of the nucleus becomes greater due to the increasing number of protons. This causes the nucleus to have a greater pull on the electrons, thereby decreasing the atomic radius. This is the effect of effective nuclear charge.

Shail Avasthi 2C

Yes, understanding conservation of mass is key to solving this problem. The amount of Cobalt is unchanged in the reactants and products, so to find the mass of Fluorine used, you need to subtract the mass of Cobalt (339.20 g) from the final mass of the products (996.08), which you can use to solve t...

Shail Avasthi 2C

Don't worry about significant figures until it's time to write your final answer to the problem. While solving the problem, numbers should remain as exact as possible so that your final answer is as accurate as it can be. Then apply significant figure rules to your final answer.

Shail Avasthi 2C

Try to avoid rounding the numbers you use for your calculation, and only round at the end of the problem. Excess rounding while solving a problem leads to significant discrepancies between your answer and the correct answer.

Shail Avasthi 2C

In addition, many new graphing calculators have systems of equations solvers which significantly cuts down the time required to solve certain problems later on in the course.

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