CHEMISTRY COMMUNITY

Joshua Chung 2D

Think of it as catalysts are tools used to increase the rate, and so will exist before and after the reaction. On the other hand, intermediates are created and consumed in the reaction, so will not exist before and after.

Joshua Chung 2D

The slowest step of a reaction is the one that determines the rate of the entire reaction. You can identify it by comparing the individual steps' rate laws to the overall rate law.

Joshua Chung 2D

Intermediates are substances produced, then consumed in a reaction. Unlike catalysts, they do not exist in the first step's reactants.

Joshua Chung 2D

Catalysts are not consumed. You may be thinking of intermediates, which are consumed.

Joshua Chung 2D

Hello, so simply use the Arrhenius equation to find the k-values for each reaction (catalyzed and uncatalyzed), keeping all variables constant except the activation energy. Then divide the k of the catalyzed reaction by the k of the uncatalyzed reaction.

Joshua Chung 2D

If the reactants of the slow step are not given, just keep in mind that they will always be a part of the rate law.

Joshua Chung 2D

Lucy explained it very well above. Simply, it is just a way of describing and categorizing reactions using the order system (0-order means that reaction concentration does not affect rate, 1st order means that the concentration of only one reactant affects the rate, and 2nd order means that two part...

Joshua Chung 2D

You can plot out experimental values to find the reaction order. Given that A is the reactant and t is time: If the plot of [A] in the y-axis and t in the x-axis is linear, then the reaction is 0-order. If the plot of ln[A] in the y-axis and t in the x-axis is linear, then the reaction is 1st-order....

Joshua Chung 2D

Simply sum up the orders of each of the reactants, usually denoted by n, m, etc.

Joshua Chung 2D

Think of the saying "weakest link of a chain"--the slowest step determines the rate because it's the one that is, in a sense, holding back the rest of the reaction.

Joshua Chung 2D

In a forward reaction, the average rate will always be negative since the concentration of the reactant(s) decreases as it is converted into products.

Joshua Chung 2D

Equilibrium was just so satisfying to me once i got it down!

Joshua Chung 2D

It goes (from left to right) anode, ion compound, ion, double line, ion, ion compound, anode.

Joshua Chung 2D

Oxidation takes place at the anode, so the substance will have lower reduction potential.
Reduction takes place at the cathode, so the substance will have higher reduction potential.

Joshua Chung 2D

Yes, I personally found it helpful to just ctrl+f the particular element in question.

Joshua Chung 2D

I've tried switching up the order of the elements on either side and get different, unhelpful error messages each time, so if someone did this question right, let me know. Hello, this is what I was able to get hope it helps. Thanks so much, this worked! But can anyone explain why this is correct an...

Joshua Chung 2D

Yes, for some questions (I don't particularly remember how many exactly), the state of matter is required.

Joshua Chung 2D

No. Adiabatic processes have no heat transfer, so q=0 and delta U=w. Since w is the energy transferred through work rather than heat, w is not affected by whether or not a process is adiabatic.

Joshua Chung 2D

You can use either, they just have the negative sign in different places.

Joshua Chung 2D

Delta G naught is simply Delta G under standard conditions (273 K / 0˚C, 1 atm).

Joshua Chung 2D

A reaction moves forward from left to right (as given to you). Since the reactants are on the left and the products are on the right, a forward reaction is R —> P. Thus, the reverse of the same reaction moves from right to left, or P —> R.

Joshua Chung 2D

In this context, K is always the equilibrium constant.

Joshua Chung 2D

You can use the equation delta S = deltaH/T, where delta S is taken from the surroundings and delta H is taken from the system.

Joshua Chung 2D

There isn't really a notable difference, just that qrev is used to specify that the heat value is taken from a reversible system.

Joshua Chung 2D

Reversible work is much slower, more efficient, and takes an infinite number of infinitely small steps. On the contrary, irreversible work is done rapidly and/or in large jumps one direction or another.
W = -Pex delta V is used for irreversible work, and W = -nRTln(V2/V1) is used for reversible work.

Joshua Chung 2D

Molar heat capacity is defined as the amount of energy needed to raise 1 mol of a certain substance by 1 degree Celsius. Enthalpy, on the other hand, is applied to chemical reactions/phase changes. It is defined as the amount of energy absorbed or released as a process takes place. Molar heat capaci...

Joshua Chung 2D

delta U = q + w. You can substitute 0 for q if the system does no work, such as if it is isothermal. Thus, you can also substitute 0 for w if no work is being done, such as if there is no external pressure or no change in volume.

Joshua Chung 2D

Thank you so much!

Joshua Chung 2D

I was just wondering, how we are supposed to know if something is a an atom, linear molecule, or nonlinear molecule, like those pertaining to question 20 on the sapling assignment, that were used to fid the heat capacity? For that specific problem, the molecule name was given to you (ex: H2). You c...

Joshua Chung 2D

The standard equation for internal energy is deltaU = q + w. Since w = -PdeltaV, w will be 0 when deltaV is 0. In other words, deltaU=q when the volume is constant.

Joshua Chung 2D

Since the ice is already at 0 degrees celsius, it doesn't need to change temperature to undergo a phase change. Thus, we only use the enthalpy of fusion to account for the shift from solid to liquid, and then use the heat capacity of water afterwards to account for changes in temperature after the p...

Joshua Chung 2D

No, because by definition the heat capacity is the amount of heat required to raise the temperature of a substance by 1 degree. Since you cannot increase temperature by taking away heat, heat capacity will always be positive.

Joshua Chung 2D

Calorimeters are simply used to calculate the transfer of heat in a reaction, whether it moves in or out of the system.

Joshua Chung 2D

You can flip and rearrange the reactions used, just make sure that you keep track of which direction that they end up in so that you use the correct sign for the enthalpy when calculating.

Joshua Chung 2D

Hess's Law states that enthalpy is a state function, meaning that it doesn't take into consideration how the enthalpy arrived at its current value. Thus, enthalpies are additive, making it very straightforward for us to problem solve with them.

Joshua Chung 2D

Since this reaction is a textbook example of a combustion reaction, we can immediately single that out as the most likely answer. If we look at the other possibilities, the bond enthalpies don't line up (as described above). Thus, the enthalpy is equal to the combustion of CH4.

Joshua Chung 2D

Pretty sure we don't have to memorize every standard state besides the diatomic molecules as mentioned above (H2, N2, F2, etc.)

Joshua Chung 2D

A phase change is whenever a substance changes its state of matter (solid, liquid, or gas).

Joshua Chung 2D

I was running into the same issue as you; turns out that I simply wasn't using the right value for B[initial]. Keep in mind that B[initial]=[B] + [x], and you should get an appropriate percentage.

Joshua Chung 2D

If it is a gas, then we include it. If it is in liquid form (as is in most chemical equations dealt with so far), then we exclude it.

Joshua Chung 2D

You can start solving this problem by writing out the chemical equation with the appropriate conjugate acids and bases. In this case, the K+ ions are spectator ions, so you can exclude them from your solving process. Then, set up an ICE box to solve.

Joshua Chung 2D

According to Dr. Lavelle, we can use this approximation when the K-value is smaller than 10^-4.

Joshua Chung 2D

Default to Kelvin here.

Joshua Chung 2D

Since we can choose to use + or - x for each of the compounds in a reaction, we simply have to determine which direction the reaction will shift with the given change. If it will shift to the right, we use -x for the reactants and +x for the products, and vice versa.

Joshua Chung 2D

For this question, I found it helpful to write out the expressions for each chemical equation's K value (using the brackets and such). Then, you can see that you can get the desired K value by multiplying the K of the reciprocal of equation 1 by equation 3.

Joshua Chung 2D

Since any value to 10^-4 is extremely small, it is deemed "close enough" to zero to treat it as zero in your calculations.

Joshua Chung 2D

Yep; they are completely interchangeable in this coursel; just remember to keep the units of measurement consistent throughout the problem.

Joshua Chung 2D

If a reaction is shifting to the left, then it will begin to produce more reactants relative to products. If it is shifting to the right, then it will produce more products relative to reactants.

Joshua Chung 2D

It outlines how reactions adjust to changes in concentration, pressure, and temperature.

Joshua Chung 2D

Since inert gases don't react with anything, all the concentrations will remain proportional.

Joshua Chung 2D

Yes, as adding an inert gas will not change the reaction. Only by changing individual concentrations will the reaction be affected.

Joshua Chung 2D

The principle specifies how reactions will naturally balance themselves out from changes, meeting the same k value.

Joshua Chung 2D

Since endothermic reactions require heat to occur, increasing temperature will aid in meeting the activation energy.

Joshua Chung 2D

Ion-dipole forces are stronger than hydrogen bonds, as hydrogen bonds are simply a strong type of dipole-dipole forces. Since ion-dipole forces are always stronger than dipole-dipole, ion-dipole forces are stronger than hydrogen bonds.

Joshua Chung 2D

Since Iodine is a larger molecule than Fluorine, the bond is stronger due to it being easier to distort its electrons. Thus, the compound with iodine is harder to break, resulting in a higher melting point.

Joshua Chung 2D

Think of it as the third row p-block elements can have expanded octets, and B, Be, Li, He, and H can have less than an octet.

Joshua Chung 2D

Writing out the lewis structure would help here. In this compound, boron has only three bonds and no lone pairs, so can easily accept electrons from a lewis acid.

Joshua Chung 2D

You look at the geometry and the differences in electronegativity. If there are clear differences that do not cancel out, then the bond is polar.

Joshua Chung 2D

The less difference in electronegativity (look at position on periodic table), the more covalent it is.

Joshua Chung 2D

Yep, both are angular. Keep in mind that the angles are slightly less than those for a typical trigonal planar and tetrahedral structure, as lone pairs "push" other regions of electron density more than bonds. Thus, the observed bonds are more squished together.

Joshua Chung 2D

The central atom is bound to two atoms and has two lone pairs, thus holding 4 regions of electron density. This forms a tetrahedral-esque structure, which has a 109.5 degree bond angle.

Joshua Chung 2D

Octahedral structures are formed when 6 bonds orient regularly around a central atom, forming a structure with 8 faces (thus lending the name "octahedron"). I think of bipyramidal structures as two pyramids sharing a base, forming a shape similar to the green thing from Sims.

Joshua Chung 2D

Yes; once you determine the shape of the molecule, you can see if the dipole moments cancel each other out geometrically. If so, then the molecule is nonpolar.

Joshua Chung 2D

First, you gotta determine the differences in electronegativity between the individual atoms. If none exist or if they cancel each other out, then the molecule is nonpolar. If there is a noticeable direction of electronegativity difference, then the molecule is polar.

Joshua Chung 2D

I'm not completely sure if double bonds are commonly referred to as pi bonds, but they include one sigma and one pi bond. Thus, it is distinct from a normal pi bond (ie single bond).

Joshua Chung 2D

You can either look at the number of bonds formed (more bonds=more covalent) or the bond length (shorter=more covalent).

Joshua Chung 2D

More electrons = larger radius and vice versa, as adding additional electrons will increase their repulsion from the nuclei.

Joshua Chung 2D

Sunday night at 12 am.

Joshua Chung 2D

Delocalized simply means that there exist resonance structures for the compound, so the pi bond(s) can exist in multiple different positions.

Joshua Chung 2D

There is no need to memorize specific bond angles for different compounds. However, bond angles can be calculated and/or estimated by playing around with the fact that a circle has an angle of 360 degrees. For example, you can deduce that a trigonal planar molecule has bond angles of about 360/3=120...

Joshua Chung 2D

Although molecule size is often correlated with its ionic character, a better way to determine ionic character is by looking at the difference in electronegativity between the bonded elements. The greater the difference, the stronger ionic character the molecule exhibits.

Joshua Chung 2D

You simply look at the difference in electronegativity between the bonded elements. For example, a molecule with carbon and hydrogen is extremely covalent because they are very similar in electronegativity.

Joshua Chung 2D

Shape can be used to determine if a molecule is polar or nonpolar, but you must consider whether or not it has existing dipole moments. If dipole moments are not present or if they cancel each other out (with respect to shape), then it is nonpolar. If they exist and do not cancel out, then it is pol...

Joshua Chung 2D

Generally, you can determine which molecules accept (acids) or donate (bases) by looking at their respective Lewis Structures. Lone pair electrons usually indicate that the molecule is a base.

Joshua Chung 2D

Personally, I take the number of valence electrons (V) and subtract from it the number of bonds (s/2) combined with the number of non-bonding electrons (L).

Joshua Chung 2D

In general, comparing the given bond lengths to the values on the table provided should give a strong indication of ample or overwhelming characteristics. In other words, on an exam, the difference between the two classifications should be extremely clear, and will allow you to simply make your best...

Joshua Chung 2D

In the case of dispersion forces, larger molecules have stronger attraction due to the distance between the nuclei and the outer electrons. Since the outer electrons in large atoms are more loosely held compared to smaller atoms, they are more easily affected by external charges.

Joshua Chung 2D

When an H atom is bonded to a N, O, or F atom, it causes a notable electronegativity difference due to the strong electronegativity of N, O, and F. Thus, the present partial charges allow for the formation of a hydrogen bond.

Joshua Chung 2D

Yep, that's exactly right. Just to clarify for dispersion forces, larger atomic radii cause weaker attraction between the outer electrons and the nuclei, thus allowing for easier polarization.

Joshua Chung 2D

A way to think about hybrid resonance is by taking the "average" of the known resonance structures for the molecule. All the bonds that differ between the structures are "averaged out" to form the hybrid structure.

Joshua Chung 2D

You're spot on! Since ms is the variable that specifies the spin of the electrons, there are two possible electron states for each ml value.

Joshua Chung 2D

As others have already said, it refers to which direction that the electron is spinning on the axis. In short, it's just another level of specificity when it comes to describing electrons/electron behavior.

Joshua Chung 2D

Yes, but we don't have to worry about them for this class.

Joshua Chung 2D

This is considered an "exception" to the general rule of increasing energy through orbitals, as the 4s is lower energy than 3d. Thus, it is filled first, specifically with d-block elements. However, 3d is still written before 4s when notating electron configuration.

Joshua Chung 2D

Since p-electrons are further away from the nucleus of the atom, they have higher energy, as is required to keep them on the atom.

Joshua Chung 2D

N specifies which shell is being isolated, l specifies which subshell of that shell, and ml specifies which orbital of that subshell. Thus m, l, and ml are increasing in terms of specificity.

Joshua Chung 2D

Generally, Hydrogen is placed in group 1 on the Periodic Table. However, it shares some properties with both group 1 and group 17, so some arguments can be made on its classification. But overall, it's just a unique element.

Joshua Chung 2D

As anions have extra electrons and cations have less electrons than the "normal" state of the element, anions are larger and cations are smaller.

Joshua Chung 2D

As other people seem to have already answered the first part of your question quite thoroughly, I thought I'd add that by convention, noble gases are the only elements used as shorthand reference elements (due to their convenience both in location on the table and for having a complete outer shell a...

Joshua Chung 2D

K and Ca are simply exceptions to the rules applied to the previous elements, as 4s is lower energy than 3d. Thus, for these two elements, 4s is written before 3d in order to keep with the convention that configurations are written with ascending energy.

Joshua Chung 2D

The third is written before the fourth simply because it notates a lower n-value (shell). Since these are written in ascending order by convention, 3d will always come before 4s.

Joshua Chung 2D

The third is written before the fourth simply because it notates a lower n-value (shell). Since these are written in ascending order by convention, 3d will always come before 4s.

Joshua Chung 2D

Other people have pretty much covered it, but just make sure that your SI units align in your equations, and when in doubt double check the conversions.

Joshua Chung 2D

For most problems, you just have to know that the Balmer series covers jumps starting or ending at n=2, and the Lyman series with n=1. Also, the Balmer covers the visible regions of the EM spectrum, and the Lyman covers UV regions. I'm pretty sure you don't need to know the specific frequency limits...

Joshua Chung 2D

As other people have said, n1 is the lower energy level and n2 is the higher one (in emission equations). Another way to remember it is to make sure that the value of (1/n1^2)-(1/n2^2) is always positive.

Joshua Chung 2D

Though 3.0 will probably be fine in most questions, you can round to math the sig figs of the given values just to be safe.

Joshua Chung 2D

Yes, amplitude is specifically a feature of wave models. However, increasing the amplitude of a light source will not increase its energy - only its intensity. The only way (that's been mentioned in class so far) to change energy is to change frequency.

Joshua Chung 2D

This equation is used to find the kinetic energy of the electron, which is essentially the leftover energy after the threshold energy is used up to eject the electron in the first place.

Joshua Chung 2D

Essentially, the wavelength determines whether or not the light will be able to eject electrons, as higher wavelength = higher energy of the photons. Since intensity refers only to the number of photons of light emitted, increasing intensity while not changing wavelength will not affect whether or n...

Joshua Chung 2D

The light must meet a certain frequency in order to eject electrons from the material. Since the intensity only refers to the number of photons and not their energy level, increasing the intensity but not the frequency will not affect whether or not electrons are ejected.

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