CHEMISTRY COMMUNITY

deepto_mizan1H

We learned in class that a catalyst serves to provide a new pathway for the reaction mechanism to go through to produce the products faster. However some sources say that a catalyst can't be described as lowering the activation energy, instead providing a different pathway with a lower activation en...

deepto_mizan1H

Essentially the equation is just finding the change from initial [A(initial)] to find the value of [A] after a point in the reaction. Since we know the initial concentration of the product [B] is 0, the formation concentration of it can be treated as a decomposition of [A] if we multiply by the mole...

deepto_mizan1H

They can coincidentally be the same value, but we can't set the order of reaction based off stoichiometric coefficients. It takes some work, but calculating and determining the order from the experimental data is the best and only way to get the right order.

deepto_mizan1H

To start, you can use the half life for first order reaction like you said in order to find the rate constant k. Then, you can use another equation that describes the behavior of the reaction summed over time and plug in k and use ratios to find the value of t. You can also use a sample initial conc...

deepto_mizan1H

The graph of a 2nd reaction is graphed on the axes of (x = time) and (y = 1/[concentration]). In order to visualize our reaction in a linear format, we consider it in terms of 1/[Molarity]. This is similar to how we take the natural log of the concentration for the 1st order reaction to get a line. ...

deepto_mizan1H

Trying to keep it in terms of ratios can help to figure out the exponent values easier. If we need to reduce the coefficients a bit in order to do so can help as well, but in essence nothing should specifically change.

deepto_mizan1H

Another aspect of kinetics is that we're considering rates (derivatives), which become time and activity dependent. This helps to give a context in exactly what we're analyzing in terms of the reaction. In thermodynamics, we didn't particularly study any rates of activity.

deepto_mizan1H

In addition to using data tables, it can also be determined through the usage of graphs as the example we saw today. In applying a ln(x) function to the concentration function, we can potentially determine whether it's a first or different order reaction. However, experimentally calculating it is of...

deepto_mizan1H

There aren't really "rules" for redox reactions, but they make use of the typical oxidation states of the elements. When calculating the charge changes, we won't be given them in the question. Instead, we'll easily be able to find them from the periodic table, or memory as we've used these...

deepto_mizan1H

Our spontaneity can help us to conceptually understand what stability means. In essence; the spontaneous reaction which increases the entropy of the universe is most favored. This can be considered a value of stability, as the reaction ends up being universally favorable. The opposite reaction would...

deepto_mizan1H

Using an inert electrode (most often Pt) allows the half of the cell to facilitate reduction/oxidation when the solution is fully aqueous. Electrons will be transferrable on the surface of the electrode. If the electrode is involved in the half cell with the solution (solid copper and aqueous), then...

deepto_mizan1H

The salt bridge essentially helps to prevent charge accumulation on either side (which would nullify the battery system). Since it is made out of tubing with gel, it will not chemically interact with the cell. It allows the charges to move, and to keep the cell going.

deepto_mizan1H

It helps to relate the effect of temperature on the equilibrium constant of a reaction. Back when we were looking at Le Chatelier's principle, we noticed that only a change in temperature changed our equilibrium constant as a whole, and this equation can be used to see how a change in temperature di...

deepto_mizan1H

Exergonic: (-) sign. It releases energy spontaneously, which fits our definition for a negative delta G value. The bonds formed are essentially stronger than the bonds broken. Endergonic: (+) sign. It absorbs energy, which fits the definition of the positive delta G value, being not spontaneous at a...

deepto_mizan1H

To start understanding what our free energy means, we can look at the equation of delta G = delta H - Temp*(delta S). Our delta H (enthalpy) value is in joules or kilojoules, and multiplying our entropy value times the temperature of the reaction will give us a value in joules as well. So, we can se...

deepto_mizan1H

I think that the statements above are correct, as a large error in W (degeneracy) constitutes a small error in the residual entropy. It seems that the residual entropy is flexible to the errors in W. But, as always it doesn't count the thermal entropy we can calculate using the same equation.

deepto_mizan1H

In addition to the responses above, we can see that entropy would not be 0 in any cases as it violates the laws of thermodynamics, where entropy only approaches 0 when it's at ~0 K. (Third Law). In an isothermal system our temperature is constant, so it doesn't approach 0. (Only delta T is 0). When ...

deepto_mizan1H

In addition to the responses above, it's important to draw the structures for the molecules we are considering, as it can help us to see what the resonance structures are and to total the amount of equal energy states (or positions) that we have. From this information it'll allow us to determine deg...

deepto_mizan1H

When the system is expanding, the volume will be a positive value, ultimately making the work value negative. When it's compressed, the volume will be a negative value, which makes the work in total a positive value. So, when the system does work, it is negative. When work is done on the system, it'...

deepto_mizan1H

Since our integral is of integral of P with respect to dV, we are integrating a constant. So our difference is of the difference of V, which is why we aren't necessarily taking an *actual* integral. In essence, we are doing the integral through using the formula of -P* delta V. (This is all based of...

deepto_mizan1H

If we are given the actual steps for each part of the reaction, then that is the most accurate way to find the enthalpy for the reaction (Hess's Law). Then, from the enthalpies of formation we would be able to get another method of finding the total enthalpy, but it would not be as accurate since it...

deepto_mizan1H

Another aspect of this is thinking of the heat as a product or reactant, where in an endothermic reaction (+) it would require heat which can be considered as a reactant. In an exothermic reaction (-), where heat is given off, we can think of it as a product.

deepto_mizan1H

The example of altitude in the lecture from last week is a good example, as for a state function we just see the initial and final and compare it regardless of the path take (the two teams traveled a total different amount of distance but ended up in the same point of altitude). However in terms of ...

deepto_mizan1H

I think it depends on what information we are given for the calculation. If we are given something with a step by step enthalpy for each step, we could use Hess's method, but if we aren't given the enthalpies for each step of the total reaction then we might decide on using the bond enthalpies to su...

deepto_mizan1H

When an inert gas is added to the system, it doesn't participate in the reaction at all, and the reacting gases will retain their same volume and remain unchanged. That's why when we have inert gases we can pretty much disregard it.

deepto_mizan1H

Temperature (as said above) is the only change that will have an effect on the K constant. This is due to a change in energy provided and other conditions related to temperature change. Often raising the temperature will shift it to endothermic direction.

deepto_mizan1H

Le Chatelier's principle tries to explain how the system will try to reduce the stress put on it. If the system loses pressure or (loss of concentration) on one side, the system can push towards the reverse (or whichever direction) it requires in order to equalize. In this case, it'll reduce on the ...

deepto_mizan1H

It's a way to estimate how the moles factor in to the pressure of each part of the system, a "quick" way to understand which direction it'll go. But overall it'll help to check logically each step.

deepto_mizan1H

In addition to the reply above, it is a strong example of how the principle works to rebalance the system (pushing the equilibrium to either direction).

deepto_mizan1H

Since the K constant is a ratio of concentrations and the temperature change does affect the K constant's value, I think it would make sense that the concentrations would shift accordingly to reflect the change in ratio.

deepto_mizan1H

I believe the partial pressure will be used in terms of when all (or possible most) the products or reactants are in the gas phase.

deepto_mizan1H

The reaction quotient also may not always match with the equilibrium constant, which should also be checked for when doing the problem. It'll probably come up later on in the quarter.

deepto_mizan1H

One of the reasons Dr Lavelle subtracted the value of the equilibrium value from the initial value was to determine the change in molarity. Since the values were 0 for the others (ADP and Pi), and the equation was balanced with coefficients of 1, we were able to understand the change was equal to th...

deepto_mizan1H

In addition to the posts above, since it is a cis version of the molecule, it is also a polar molecule since the Cl groups (x2) face opposite of the NH3 groups (x2), which can be thought to let it have unique properties compared to the trans version where it would be nonpolar.

deepto_mizan1H

Strong acids and bases are ones that deprotonate or protonate almost fully in water solutions, and they're the ones I think he said we will mostly use in this class (with exceptions). For calculation we can treat a strong base like KOH in whatever concentration as equal to the OH ion concentration, ...

deepto_mizan1H

Bronsted Acids and Bases don't really go beyond that definition as it's a proton-centric model of how we can understand acid and base compounds. Proton donation and accepting are characteristics of acids and bases in this definition and that is how we can analyze in a reaction what our acid and base...

deepto_mizan1H

In addition to the great explanations above, the lone pairs become very important in differentiating certain compound models due to their repulsive forces, which also directly changes the bond angles. They are tricky but important for understanding the compounds better.

deepto_mizan1H

The chelating compounds have to do with the ligands involved around the metal ion in the center, which forms the rings characteristic of a chelating compound. The reason certain ligands allow the development of the chelating compound has to do with what types of bonds can be formed from those compou...

deepto_mizan1H

One way to do it is since we are labelling the hybridization starting from B to F, we would identify the shape of the atom based on VSEPR (for example Trigonal Planar), then relate the corresponding hybridized orientation value. There is a good table in the book than can help to identify it.

deepto_mizan1H

It wouldn't correspond to a different model in total, but the angles can differ based on which atoms are present and their relative forces of electronic repulsion.

deepto_mizan1H

Shape of molecules acts as a big character of the compound, especially in biological systems where form equals function and other principles can determine the reactivity and action. For example, certain prion diseases can be from benign ones unfolding or folding in a different state, causing lethal ...

deepto_mizan1H

Often we can answer the less than values by determining the differences in repulsion compared to other examples of the same orientation, where greater repulsive force from a pair or atom will push the others accordingly. It depends on each unique compound.

deepto_mizan1H

The number of electrons in an atom do not necessarily determine bond strength, but the electrons involved in the bond do. For example, with more electron involvement in a triple bond, the atoms need to be closer to each other with a greater attraction in order to facilitate more covalent bonding. Be...

deepto_mizan1H

In addition to the responses above: what helps to form our VSEPR models is to be aware of the lone pairs in the model, which will cause repulsion and push the atoms into specific shapes. Even if our number of atoms are the same across different models, it is very possible to have different VSEPR str...

deepto_mizan1H

Since it shows a structure that isn't fully linear, we can deduce that there must be some sort of electron density on the central atom that is causing repulsion of the other atoms joined to the central atom. Therefore, we can be sure there is a lone pair causing repulsion, "bending" the sh...

deepto_mizan1H

In addition to the response above, the way we can tell the character of the bond has thresholds around 1.5, where less than that it will be covalent while greater it can be ionic character. There's a very nice diagram in that section that graphs out some specific examples and the character identific...

deepto_mizan1H

I think depending on the energy levels and ultimately the distance between each atom, the bond can be weaker even if the size is larger. But, it ultimately becomes the ratios between distance and energy level that will determine the attractiveness.

deepto_mizan1H

Since the bond in the H2O atom is a single covalent bond between Oxygen and Hydrogen and the Hydrogen bond between two water molecules is due to lone pair pull, they exhibit different levels of attraction. A covalent bond is a true physical sharing and distortion of the electron cloud between the mo...

deepto_mizan1H

Another aspect of finding the formal charge and its usefulness comes from being able to understand the overall charge of the ion, which we should be able to derive by calculating charges of the individual atoms. This can help when we have an ion that we are not sure of what the exact charge is, but ...

deepto_mizan1H

The experimentally observed bond lengths were shown in lecture to probably display how our model cannot fully grasp the true nature of the molecule's arrangement, as we have to find an average to find the true length and arrangement. A double bond is shorter than a single bond, but it does not mean ...

deepto_mizan1H

As said above either is fine, but lines are useful in counting the electrons and also being able to keep track of charges when calculating formal charges as well. This continues in to the fact that when we come to resonance structures it makes it easier to visualize the different electron placements...

deepto_mizan1H

Since it has the lowest ionization energy, you can think of it in terms of is it more willing to gain or lose an electron (become an ion) or to ultimately share or gain the electron in a bond. C is generally a "backbone" for structures because it is direct in the middle with 4 valence elec...

deepto_mizan1H

There is a slight difference between psi and psi^2, which helps to understand Schrodinger's Equation conceptually. Psi itself is a wavefunction, which varies with position. Once we square this value it becomes only positive, and provides a dimensional model of the area we can probabilistically find ...

deepto_mizan1H

I think we should be able to convert the values if needed (same as any unit based question). The conversion value is on the sheet so we should be just fine if the question is presented in eVs.

deepto_mizan1H

It helps to understand how the wave functions (psi) in Schrodinger's equation are a mathematical representation of where electrons may be found. When we square psi to find the probability density, we are essentially able to understand which regions there is a greater probability of finding an electr...

deepto_mizan1H

Absorption and Emission are two aspects of how energy is transferred among particles and compounds. For example in reactions energy is absorbed to make bonds, while emitted when breaking. In particles we see light from the excitement of electrons, where energy is absorbed when excited, and given off...

deepto_mizan1H

I think for the purpose of the class and for learning about the "Quantum World", knowing the Schrodinger Wave Function Equation is essential for a conceptual understanding. From my understanding the equation allows us to gain a probable location of the particle (probability density) by sol...

deepto_mizan1H

It is the principle that (in addition to the fact that on an atomic level measurement influences outcome) we cannot accurately define both the momentum and position of a particle at the same time. For example, we can estimate a probable area of where the electron may be in an atom, but we would be u...

deepto_mizan1H

Using this equation (C= Lambda x V) we can analyze specific waves of light against the constant that is the speed of light, so we can glean some information on its wavelength, frequency, and additionally from that where it belongs on the electromagnetic spectrum. It's a type of linear equation that ...

deepto_mizan1H

I had this question too and still do, but I have come to some understanding that our model for light is still debated on. In a large scale sense, our wave model of light works for our use in physics and larger chemistry, however when coming down to particle side such unique phenomena where intensity...

deepto_mizan1H

Thanks for the clarification up there, it is very helpful. I found this interesting because it helps to glean some insight on the importance of "quanta" in the particle model of light, where increasing the intensity of light will by no means change the energy in the photons. If it is alrea...

deepto_mizan1H

When we have percentages for elemental composition, it makes it much easier to determine the grams in the sample (especially if we do not have a sample given), then we can use molar masses to determine the moles and the empirical formula. In essence, it makes it much easier to glean important inform...

deepto_mizan1H

Another interesting fact that can help conceptualize the importance of isotopes is that the molar mass of the atoms we have on the Periodic Table are in fact representative of the most commonly (or long-lasting) isotope of the atom we can observe. These will definitely be important as we get to deep...

deepto_mizan1H

Definitely check out videos regarding concepts of moles and matter, but I can try to give a short description. A mole is the concept of a certain amount of a compound or element, usually defined by Avogadro's number (6.022*10^23), which tells us about how many particles (atoms, molecules) of whateve...

deepto_mizan1H

Since aqueous solutions have been discussed already, the most important thing about keeping track of states of matter during balancing equations is to get familiar with certain reaction patterns and probable products that may be formed. For combustion (for example) knowing the fact O2 gas is a fuel ...

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