CHEMISTRY COMMUNITY

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this scenario occurs in a reversible reaction! Since reversible reactions in real life occur slowly, the system has time to regain the heat lost by work, so the overall change in entropy of the universe is = to 0.

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Delta G by itself is the free energy of a particular system; Delta G naught is the standard conditions of free energy for a system.

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Oxidation is loss of electrons in a half-reaction, so they should be on the product side.

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Heat and work are not considered state properties because they are affected by the paths that they take and are changed by them.

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Yes- there are more limited ways to change an isolated system but it is still possible!

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The questions are usually pretty clear about which delta G to look for or calculate, but remember that delta G naught is change in gibbs free energy of a reaction under standard conditions, but delta G can change with each different reaction you are looking at.

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David Jen 1J wrote:Another question, why is it that the greater the Svap, the more ordered it is?

According to Sapling, "A larger molar entropy of vaporization, which correlates to a greater increase in disorder, indicates a more ordered arrangement of the molecules in the liquid state".

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Divide enthalpy (delta h) by the temperature given

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When delta h is positive, heat is being absorbed by the system (endothermic), and when something is absorbing heat it is going from solid to liquid, liquid to gas, or solid to gas. Gases have the highest entropies since they have the most room for disorder and the most possible microstate positions,...

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Larger molecules have more room for higher disorder, which means the entropy would be higher.

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Standard entropy is usually in Joules, but especially when calculating gibbs free energy make sure you convert that to kJ so that the units align with enthalpy.

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If Pext (external pressure) and Pint (internal pressure) are the same, the system is most likely under irreversible expansion (constant pressure). And if they are different, the system is most likely under reversible (isothermal) expansion.

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All you have to do is get all units converted to kj, and then subtract the heat value (q, which is negative since heat is released) and add the work value (which is a positive since work is done ON the system) in order to get delta E, or U (change in internal energy).

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There were some questions similar to 19 in previous questions of this assignment. You have to divide the heat released by the calorimeter (q) by the change in temperature to find the heat capacity (c). Then, you take that c and use it for the next given values. Since delta T is given, you are lookin...

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Since the system is held at constant volume, we know that the system is unable to perform work, so the w term is equal to zero. Therefore, delta U (change in internal energy) is equal to the same value of q (heat).

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Positional/residual entropy also ignores thermal motion and entropy, which is confusing to me because I thought that thermal motion was important to entropy of any system. Why are they different?

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If the system does work by itself, the value will be more negative since energy is required for it to do so. However, when the surroundings do work on the system, they expend their own energy so the work value of the system is positive because it is gaining that energy.

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It is important for #20 that you draw out the molecular geometry (with electron lone pairs included!) so that you know which values in the table given to use. For example, in a question similar to this, if you are given something like a Carbon atom (C), you would have to use the molar heat capacity ...

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Q of the reaction can be positive or negative, depending on whether the reaction is endothermic (+) or exothermic (-). So whatever amount of heat that the reaction absorbs or releases is taken away from or added to the heat of the calorimeter.

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N is the total number of molecules or particles in the system, so even though there are only two components of each CO molecule, there are 2 CO molecules meaning that there are 4 total components to the system.

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N is the total number of molecules or particles in the system, so even though there are only two components of each CO molecule, there are 2 CO molecules meaning that there are 4 total components to the system.

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State functions do not rely on the path taken to get from an initial to final value.

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I think we’re supposed to just assume temperature is 25 degrees Celsius because our calculations don’t really involve temperature right now.

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They don’t allow for matter to be exchanged inside and outside of the system!

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Use when given info on a weak acid or base to compare initial and final concentrations.

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When pH<pKa for an acid that acid is neutral. In an acid when pH>pKa it is charged.

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Assume that x is irrelevant in an ICE table when the equilibrium constant is less than 10^-4

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Q is the reaction quotient and K is the equillibrium constant. If Q>K the reaction will shift left, and if Q<K the reaction will shift right.

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yes- in some tricky problems liquids and aqueous components may be included, but just emit them since partial pressure can only apply to gases.

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Weak acids consist of all acids besides the 7 strong acids. Additionally, it may help to remember that the conjugate acid of a strong base is a weak acid in some cases.

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It is safe to assume that the Kw value will = 10^-14 in all equilibrium systems at 25 degrees Celsius.

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Le Chatelier's principle states that a system at equilibrium reacts in a way/shifts in the direction that minimizes the stress applied to that system.

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P= pressure (atm); V= volume (Liters); n=moles; R=gas constant (no units needed); T=temperature (Kelvin)

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k must be a very small quantity, as mentioned above, in order to disregard x. If K is greater than 10^-4 and you disregard x, it can throw off your answers and your values of final and initial concentrations of x may be incorrect.

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When there is an increase in pressure, the equilibrium will shift to whichever side of the equation has fewer moles of gas.

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molarity, usually in brackets [] like when we try to calculate K or Q by dividing concentration (molarity) of products / concentration (molarity) of reactants, is a measure of moles per liter of chemicals in these reactions. So for the purpose of K and Q problems, the molarity and moles are usually ...

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Yes. Q is used as a tool to decipher whether the system is at equilibrium, tends towards the product side of the reaction (Q>K), or tends towards the reactant side of the reaction (Q<K).

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Delocalized electrons lower the overall energy of the molecule in a resonance structure making it more stable

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It depends on the situation- it can act as either wherever necessary because it is made up of an acidic component (H+) and a basic component (OH-)!

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That's a great question! don't think too much on this one- just know that one of the 6 H2Os from the left side of the equation has to give up an H to the lone water molecule on the left in order to create the H3O+ on the right side (since this is an acid-base reaction). So, you just remove one of th...

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Yes I would definitely recommend memorizing the amphoteric oxide compounds (oxides and the other molecules)

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No- we shouldn't have to know that for this course! If anything, assume normally that the diatomic molecules N2 and O2 and common molecules like CO2 are in their gaseous state, and that metals are solids, but other than that I don't believe we are responsible for knowing states of molecules.

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There would be two pi bonds in CH3CN since there is a triple bond between the second carbon and the nitrogen atom in the lewis structure. You can find unhybridized orbitals from an electron configuration first by drawing the normal electron configuration, then combining (hybridizing) the highest ene...

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The oxidation states of groups 1 and 2 are usually the cations they create,and the oxidation states of groups 3-8 are the anions they create (ex: Oxygen is -2) but transition metals can very so we have to find them based on the oxidation states of elements that we already know.

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I usually draw out the lewis structure first, and if it seems to not make sense in any other way than adding lone pairs to the octet of the central atoms, that is how I know it can have an expanded octet (so long as it is in group 3a or after)

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Nonpolar molecules can have polar bonds and still be nonpolar; polar molecules must have polar bonds to be polar.

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From what I understood, whichever molecules/bonds have the strongest dipole moments are where the poles of the molecule are the most negative.

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The two lone pairs have strong repulsion forces and cause the two hydrogen atoms attached to the oxygen central atom to be bent at an angle, so it is not linear molecular geometry.

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No- only lone pairs will affect molecular shape/geometry

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More spread out charges and preferably charges that are not on the center (least electronegative) atom are the most energetically favorable resonance structures.

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the greater the covalent character, the smaller the difference in electronegativity which means the atoms are closer to equally sharing those electrons!

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bond length generally increases as atomic radius increases (down the periodic table and towards the left)

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the equation that I use is "valence electrons on the periodic table - lines - dots"

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I like to watch pop pilates videos on youtube and do long distance running!

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I think the best prep for this midterm is definitely all of the lecture notes and textbook problems because it is less math and more practicing structures to be prepared for what we may see.

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Don't worry about the transition metals as much, for now. There are a lot of exceptions and we are mainly focusing on the s and p shells when looking at bonding for right now.

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Cl can expand its octet, but it actually only has 7 valence electrons normally. It has 17 electrons total, and to complete its octet it tries to obtain one more electron normally.

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What are the benefits to changing it to pass/no pass or does it hurt our GPA in any way?

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no- all bond lengths in specific numbers should be given to us. If we ever have to figure out a bond length, just use a general periodic trend to make an estimate.

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The reason the Oxygen bonds and stays with the Hydrogen atom through the different resonance structures is that the oxygen is meant to bind to the hydrogen atom and have it instead of another lone pair of electrons. There are also intramolecular forces at play having to do with electronegativity tha...

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There are higher subshells past f, but we will not be learning about them in this intro to chemistry class!

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Me too! my email is suttonsellars22@ucla.edu

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The exam was really stressful so it would be great to study with a group before the next one to avoid any test anxiety- support really helps.

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Go in order from lowest energy level up!

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I think that for now, we need to assume that every element follow the octet rule.

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The above statements are true, but remember to look at the element's location on the periodic table because as you go closer and closer to the bottom left corner, the radius gets bigger, then you must remember to consider whether or not it is an ion (positive or negative).

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The electrons must bend in order to go through the slits in this experiment because they cannot go straight through, as particles would. This means they must have wavelike properties as well as particle-like properties.

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Make sure to read the question several times to make 100% sure that you know if it's giving you N initial or N final. One way you may find out if you use the wrong one is if you get a non-whole number or a negative number.

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Tip: if something says Hertz or cycles per second, it is talking about frequency. Velocity is also a regular lowercase v and frequency is a cursive-looking v in equations.

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In order to check yourself and determine the correct spin state, draw out the diagram for each quantum number

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Always remember: Lyman series N1 is 1 and Balmer series N1 is 2.

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Does anyone know if the next midterm and final exam will be multiple choice?

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As valence electrons in a shell get higher and closer to 8, they want that last electron very strongly, making the electronegativity of that element strong. For example, Chlorine has a very high electronegativity because it has 7 valence electrons.

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Make sure to be familiar with the De Broglie wavelength equation and the other equations for frequency and wavelength using the speed of light and Planck's constant so that you can interchange them easily.

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The periodic trend for ionization and electron affinity is the same: the closer to the top right of the periodic table, the stronger. For atomic radius: the closer to the bottom left, the larger the radius

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Is anyone interested in forming a study group? We could find different practice problems from each section before midterms and exams and share them/work together on them!

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In the past, my teachers have told me not to memorize equations but sometimes it is easier just to memorize them to save time on assessments, especially when you are timed. It also helps to feel more confident when answering questions.

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I wish I could help you with that- I didn't realize he removed the negative sign. Maybe the next equation had a negative in front of that variable and they cancelled out?

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I'm struggling with studying because although I have learned about all of these topics previously, they can become confusing and sometimes questions are worded difficultly. So, I am looking up a bunch of old practice problems for these topics that I used in AP Chemistry in high school just to make s...

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If the energy coming into the atom (electron) has a "higher energy," it means that the energy of each photon is higher, and also has a higher frequency, and therefore has a shorter wavelength of light energy.

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Threshold energy is the energy that an electron must overcome to break away with the incoming energy of a photon, and binding energy is the energy that is keeping that electron in its place orbitting around its atom.

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I am on campus too, and I would love to join a study group.

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The same thing happened to me! I attempted the problem several different ways, but I could not reach the correct answer.

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You actually don't need to write out the equation- just assume based on the "image" in the problem that the molar ratio for each molecule is 1, and you can use the values given to answer the question.

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Does anyone know how to know whether to use use the Ephoton equation versus the other energy/wavelength equations?

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I would love to be a part of a virtual study group!

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