CHEMISTRY COMMUNITY

Isita Tripathi 2E

During a phase change, equilibrium exists between the phases. For example, during the phase change between liquid and gas, the reaction does not "favor" either phase, which means that both liquid and gas molecules will exist. Once you lower the temperature an infinitesimal amount, it will ...

Isita Tripathi 2E

Yes because rate = k[A]
all other factors being equal, a higher rate constant means a higher speed because rate = mol/L/s
this applies to all of the other rate laws too!

Isita Tripathi 2E

Lavelle sets the grading scale at the end of the quarter, so you can't calculate your letter grade right now based on the points you have in the class. In any case, the scale he sets won't end up hurting you.

Isita Tripathi 2E

It's all about the wording. For example, the rate of decomposition is positive (even though its concentration is decreasing) because it is the rate at which it is disappearing. This is why we have to take the negative of the d[A]/dt for the rate in A --> B.

Isita Tripathi 2E

Ar is a catalyst; it appears as a reactant and a product, which means it cancels out and it isn't used in the reaction. Since catalysts are involved in the interactions between reactants, they can be included in the rate law, but not in the overall balanced chemical equation.

Isita Tripathi 2E

Using c as an example: (c) Rate = k[O2][NO] (Products are NO2 and O.) Since you are given the rate law, you know that the reaction depends on the interaction between o2 and NO (you will draw this in your activated complex). You are also given the products, which allows you to visualize which bonds a...

Isita Tripathi 2E

In the problem it says "all other factors being equal," which means that the only thing changing is the rate constant. The concentrations are kept the same. This makes the changes in the rate constant directly proportional to changes in the rate.

Isita Tripathi 2E

You have to identify the rate determining step, as this is the one that will not have a reverse reaction. Write the rate for the forward reaction for each step, and determine which one looks closest to the experimentally determined rate law. This is usually the rate-determining step. Then, you assum...

Isita Tripathi 2E

They're probably referring to the overall order of the reaction. Since there are two reactants, and each is first order, the overall order = 1+1 = 2.
The rate constant's units depend on the overall order.

Isita Tripathi 2E

If you graph the concentration vs time of a zero-order reaction, you get a straight line. Since Δconcentration/Δtime = rate, an increased concentration has no effect on the rate, since the slope is constant, even though the concentration changes over time. However, if you graph a first order reactio...

Isita Tripathi 2E

In class, Lavelle said that you can use any experiment to plug the values in and find the rate constant. I personally used the last reaction, and got the same answer as the solutions manual.

Isita Tripathi 2E

I am confused about this as well. I am thinking it may have something to do with the mmol^4, as this may mean that you are multiplying by a factor of 10^-3 4 times.

Isita Tripathi 2E

You need to know the differential rate laws (rate in terms of concentration) and the integrated rate laws (rate in terms of conc+time). The derivations are really helpful to conceptually understand going from the differential to integrated, but both are provided on the constants and equations sheets...

Isita Tripathi 2E

If you were to graph just concentration vs time, it would curve downwards, like an exponential decay function. However, if you graph ln of the concentration vs time, then it would be a straight line with a negative slope. This is because taking the log of an exponential is essentially cancelling out...

Isita Tripathi 2E

I think the trend loosely follows electron affinity trends, but there are exceptions. Elements with greater electron affinity seem to generally have a more positive reduction potential because they are more likely to accept electrons.

Isita Tripathi 2E

Platinum is the most common, but you can use any inert electrode in this situation (it shouldn't participate in the reaction, it should just conduct).

Isita Tripathi 2E

Since #13 already gives you the reaction in the correct direction, you can break it down into half reactions without needing the appendix. But in #15, you have to determine which half reaction to use on the cathode/anode side, so you need to reference the reduction potentials in the appendix. This w...

Isita Tripathi 2E

Yes, it is asking if the Gibbs free energy of its decomposition (breaking up into its elemental parts) is positive. This means that the compound does not spontaneously decompose at room temperature, as spontaneously decomposing would make it unstable. Stable implies that nothing will happen to the c...

Isita Tripathi 2E

Since it is isothermal, delta H = 0. Since it is free expansion, there is no work being done, as there is no external pressure, which makes P delta V = 0. Since delta U = delta H + P delta V, delta U = 0 as well.

Isita Tripathi 2E

Water isn't counted into the moles in this situation, since you only plug in the net moles of gas. This is because expansion work is only calculated using moles of gas, since liquids and solids hardly change in volume as a result of expansion.

Isita Tripathi 2E

ΔS = nR ln (V1/V2)
using boyle's law for ideal gases:
P1V1 = P2V2
P1/P2 = V2/V1
replacing this in the equation:
ΔS = nR ln (P1/P2)

Isita Tripathi 2E

The universe is an isolated system because it cannot exchange heat or matter with its surroundings, as technically it has no "boundary" where you could distinguish external surroundings.

Isita Tripathi 2E

I think that absolute entropy is measured relative to temperature in Kelvin, while statistical entropy uses Boltzmann's equations and the existence of microstates. Since entropy based on temperature change is derived using a form of Boltzmann's equation along with the idea of internal energy, I woul...

Isita Tripathi 2E

In lecture, I was wondering why Lavelle replaced
ΔS = kB ln W2/W1 with
ΔS = kB ln V2/V1? I think I may have missed the explanation.

Isita Tripathi 2E

Could someone explain why we can replace PΔV = ΔnRT for P(ex)Δv in the equation for work at a constant pressure? For example, say you had a balloon with an ideal gas expanding to twice its volume at an unstated, constant external pressure. The pressure inside the balloon decreases as volume increase...

Isita Tripathi 2E

In lecture, Lavelle said to ignore the vertical brown line. I think it was drawn to compare the irreversible and reversible reactions. But, the external pressure was always constant, it didn't drop down all of a sudden.

Isita Tripathi 2E

For that problem, since you are using the CHANGE in temperature, your answer will be the same whether you use celsius or kelvin. This is because 0 C = 273.15 K, so its just a transformation of the temperature scale.

Isita Tripathi 2E

Forming a bond creates a more stable molecule than the individual atoms alone, since they have completed their octets. This is why it takes energy to break bonds (you are forcing something from more stable to more unstable) and releases energy to form bonds (you are forming a more stable complex).

Isita Tripathi 2E

For 8.45 part b, why can we use the standard reaction enthalpy to calculate the heat absorbed when the textbook explicitly states that carbon is not in its pure form? I thought that by definition, standard reaction enthalpy means that this is the enthalpy for when reactants in their standard states ...

Isita Tripathi 2E

An isolated system is coffee in a sealed thermos because neither energy nor mass (coffee) can leave/enter. A closed system is coffee in a sealed cup, because it can exchange energy in the form of heat with the surroundings, but it still cannot exchange mass.

Isita Tripathi 2E

Does anyone have an example of a chemical equation for which the reaction enthalpy is not the standard reaction enthalpy because one of the reactants/products isn't in standard state? I know of examples in which the phase is not the standard one at that temperature, but what about a molecule?

Isita Tripathi 2E

I think the pressure is changing for isothermal expansion of an ideal gas. That's only one type of reversible process, so it's not necessarily a rule.

Isita Tripathi 2E

Nitrous acid has a pKa of 3.16, so the pKb of its conjugate base must be 14-3.16 = 10.84 Acetic acid has a pKa of 4.75, so the pKb of its conjugate base must be 14-4.75 = 9.25 Since 10.84 > 9.25, and a large pKb corresponds to a weaker base, the conjugate base of nitrous acid is weaker. This can als...

Isita Tripathi 2E

An oxoacid will more readily lose H+ if the resulting anion is stabilized. This occurs when electronegative atoms that are part of the molecule delocalize the negative charge on one atom, as this makes the anion more stable overall. For this problem, since Cl is more electronegative than Br, it has ...

Isita Tripathi 2E

I think polydentate depends on lone pairs, and whether the molecule is large enough for it to bind at two sites.

Isita Tripathi 2E

You have to convert from kPa to bar, and 100kPa = 1 bar, so each value has to be divided by 100.

Isita Tripathi 2E

Kp is calculated using the unit "bar", so you have to convert kPa to bar.
100 kPa = 1 bar, so you have to divide each one by 100.

Isita Tripathi 2E

In the "change" row of your ICE chart, you have to use the stoichiometric coefficients of the chemical equation to represent the change. For example, if your equation was 3H2 + N2 --> 2NH3 and you had the following initial values, your table would look like this: I 0.4 0.4 0 C -3x -x +2x E...

Isita Tripathi 2E

Usually, a lone pair means that one of the hybrid orbitals is filled with a pair of paired electrons, while the others have unpaired electrons.

Isita Tripathi 2E

You would expect the bond angle to increase. This is because of the trend in bond angles in these hybrids: sp - 180 sp2 - 120 sp3 - 109.5 As the number of p orbitals increases, the "p-character" of the hybrids increases, and therefore the "s-character" decreases. So as the s char...

Isita Tripathi 2E

In lecture, Dr. Lavelle said that we should draw the dipole moment pointing towards the negative charge, but the book says the opposite. Which one is the proper convention?

Isita Tripathi 2E

Usually, if the best structure is still unable to make all the atoms have a formal charge of 0, you draw the "next best" possible Lewis Structures, since these will contribute to the structure. Most of the time, if all atoms have a formal charge of 0, you only need to draw that one. There ...

Isita Tripathi 2E

Just to clarify, the reason you cross the charges is because in order to obtain the chemical formula for the neutral compound, the charges have to cancel. So for c for example, it takes 3 Li + ions to cancel the 3- charge of one Nitrogen atom. The chemical formula is always the lowest amount of atom...

Isita Tripathi 2E

Since it takes less energy to remove electrons from the atom with the lower ionization energy, it has a weaker hold on its electrons than the other atoms. Thus, it is more "willing" to share its electrons. Since the central atom forms the most bonds in a molecule, it makes sense for the at...

Isita Tripathi 2E

The excited state of an atom is a state in which all the electrons are NOT in their lowest possible energy levels. Most commonly, this means that the electron has jumped from its ground state to a higher energy level. But it can also include differing spins, or different placement of the electrons. ...

Isita Tripathi 2E

Trends in ionization energy: 1. increases down a period (as the atomic number increases, the effective nuclear charge increases on each electron, making them more difficult to remove) 2. decreases down a group (as n increases, the valence electrons get farther away from the nucleus, making the elect...

Isita Tripathi 2E

As you move away from the nucleus, the energy level increases, so the electrons in orbitals farther away from the nucleus have greater energy. However, I'm not able to understand what "farther away" really means, since all of the diagrams of the orbitals show that they are centered around ...

Isita Tripathi 2E

The book mentions that the probability of finding an electron is greatest where the undulations of the wave are the greatest. I understand this mathematically (wave function squared in this regions has the highest value), but conceptually, can someone explain why this occurs?

Isita Tripathi 2E

Yes, De Broglie's equation applies to everything that has momentum (so your particle must have mass and velocity). I think it was derived by relating quantum mechanics to classical mechanics. So, E = mc^2 and E = hf E = mc^2 = hf **set equations equal to one another** E = mc^2 = h(c/λ) **substitute ...

Isita Tripathi 2E

Lavelle mentioned that the Balmer Series is for changes in energy levels from or to n = 2, and that the change between n = 2 and n = 1 emits/absorbs UV light (since this is Lyman Series and requires more energy). Does this mean that if you were to take a Hydrogen atom in its ground state and subject...

Isita Tripathi 2E

I think the problem with combining both equations is that you end up deciding between P4 or O2 as the limiting reactants for the production of P4O10, when really you should be deciding between P4O6 and O2. Also, you assume that all of the given amount of oxygen is used in the production of P4O10, so...

Isita Tripathi 2E

I think that the problems are computed using whichever values are in the given periodic table. So if your periodic table says 14.007, the answer key probably uses that value to find their final answer. This means that if you were to round, you might end up with a slightly different value for the las...

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