CHEMISTRY COMMUNITY

Sarah Rutzick 1L

Yes I think so.

Sarah Rutzick 1L

Delta U in an isolated system when there is no heat change or work being done.

Sarah Rutzick 1L

Why is it that when K<1, G is + and when K>1, G is -? I know that this is because of the equation G = -RTlnK, but what does it mean conceptually? When K is less than 1, reactants are more favored in the reaction than products. Therefore, G is not spontaneous (positive) because spontaneous forward re...

Sarah Rutzick 1L

Lavelle has something on his website to help with integrating if you still need clarification!

Sarah Rutzick 1L

If you are having trouble figuring out the units of k, write out the entire rate law with the units. You know that the rate must be in mol/Ls, so the units of k must cancel out the units of concentration of the reactants until you get mol/Ls.

Sarah Rutzick 1L

Also, when you have a zero order reaction, the rate is equal to the equilibrium constant k, so you don't need to include the reactant.

Sarah Rutzick 1L

In a first order reaction, there will be one reactant present in the rate law. For a second order reaction, you can either have a rate law with one reactant to the second order, or with two reactants both to the first order. Two examples of second order reactions are rate=k[CO2]^2 or rate=k[CO2][H2O...

Sarah Rutzick 1L

When the OH- concentration doubles, the rate increased by the same factor, and when the CH3Br concentration increases by a factor or 1.2, the rate increased by the same factor as well. Therefore, both reactants form a 1:1 ratio with the rate law, yielding the rate above. If, for example, when the co...

Sarah Rutzick 1L

You would first want to calculate the equilibrium constant (k) for the reaction, and then you plug that into the equation for half life.

Sarah Rutzick 1L

In 14.43, why are we able to mix the concentrations and the pressures given to us when calculating Q?

Sarah Rutzick 1L

K is equal to 1 at equilibrium, which would make Ecell 0 because there are no more electrons being transferred in the cell. I'm not sure whether Q is equal to 1 at equilibrium though.

Sarah Rutzick 1L

Eo is at standard state (25 degrees C and 1 atm) while E is the cell potential at anything other than standard state.

Sarah Rutzick 1L

In questions like 23 part a and c, why is H+ included in the cell diagram on the reduction side?

Sarah Rutzick 1L

Delta S total is zero when a process is reversible.

Sarah Rutzick 1L

Its true that the entropy of a perfect crystal is 0 at T=0 K. This is because perfect crystals only have one micro state, so using Boltzmann's equation and plugging in 1 for W, you get zero. However, most substances are not perfect crystals, so for these substances entropy is not 0 at T=0 K.

Sarah Rutzick 1L

You want to look at the change as a whole and see if there are phase changes between reactants and products to determine the entropy.

Sarah Rutzick 1L

I believe it is a typo and I would trust the formula sheet.

Sarah Rutzick 1L

For 9.15 part a, the answer book gives the heat of fusion as -6.01 kJ/mol. How do we know that the heat of fusion should be negative?

Sarah Rutzick 1L

Larger molecules have more atoms, and therefore have more ways in which they can be arranged. Since entropy measures disorder, larger molecules have larger entropy.

Sarah Rutzick 1L

It doesn't matter what unit you use as long as you use the same unit for any other steps in the problem

Sarah Rutzick 1L

Dr. Lavelle gave us guidelines to help determine when a reaction is spontaneous (deltaG is negative): Positive deltaS and negative deltaH = spontaneous at all temperatures Positive deltaS and positive deltaH = spontaneous at high temperatures Negative deltaS and negative deltaH = spontaneous at low ...

Sarah Rutzick 1L

On the class website, there is a correction to this problem that used Cv (5/2R)

Sarah Rutzick 1L

The enthalpies of formation for the reactants and products are in kJ/mol because they depend on the number of moles of the reactant or product in the balanced chemical equation. When calculating enthalpy of a reaction from the enthalpies of formation of the reactants and products, you multiply the e...

Sarah Rutzick 1L

When work is done by the system, the sign for work is negative. Therefore, by subtracting the negative amount of work from the change in internal energy (so basically adding), the enthalpy will be greater than the change in internal energy. When work is done on the system, the sign for work is posit...

Sarah Rutzick 1L

In lecture on Wednesday of week 2, Dr. Lavelle made the comment that tables of bond enthalpies refer to breaking bonds in gases and therefore we need to add enthalpy of phase change for liquids and solids. I'm confused about what this means/how you would do this, can someone explain or give an examp...

Sarah Rutzick 1L

Molar heat capacities and specific heat capacities are virtually the same thing, just with different units. If you are given moles of a substance, you can easily convert to grams of the substance and vice versa.

Sarah Rutzick 1L

If you are finding the difference from the final to the initial temperature, you can use C and K interchangeably because both will give the same difference. However, if you are using the ideal gas equation (PV=nRT) you want to make sure that your temperature is in K because using C instead of K will...

Sarah Rutzick 1L

If you are finding the difference from the final to the initial temperature, you can use C and K interchangeably because both will give the same difference. However, if you are using the ideal gas equation (PV=nRT) you want to make sure that your temperature is in K because using C instead of K will...

Sarah Rutzick 1L

Going off of what everyone else said, there are several different gas constants so its important to use the one that ensures the correct units cancel out.

Sarah Rutzick 1L

Calorimeters are used to find unknown variables in the q=mCΔT equation, such as specific heat or energy of the reaction.

Sarah Rutzick 1L

Yes, since the equation to find work is w=-PdeltaV, we know we need to use the volume of the pump which we can assume to be a cylinder. The equation we use for volume would vary by problem and the container we're working with.

Sarah Rutzick 1L

Does electronegativity have anything to do with the strength of acids or bases?

Sarah Rutzick 1L

Ionic bonds have higher dissociation energy that covalent bond because the polarity of ionic compounds makes it easier for them to break apart in solution. For example, NaCl is highly dissociative in water because of its polarity.

Sarah Rutzick 1L

State whether the following oxides are acidic, basic, or amphoteric: a) BaO b) SO3 c) As2O3 d) Bi2O3

How do you tell what these are, particularly how do you know when they're amphoteric?

Sarah Rutzick 1L

There is a temperature associated with every K value, which is what allows us to use the K value for all reactions at a given temperature. But, when the temperature changes, we must use the K value associated with that new temperature.

Sarah Rutzick 1L

Sometimes it helps me to think of heat as a reactant or product. If heat is added to the reactant side (endothermic) then more products will be produces, and if heat is added to the product side (exothermic) then more reactants will be produced.

Sarah Rutzick 1L

Sometimes it helps me to think of heat as a reactant or product. If heat is added to the reactant side (endothermic) then more products will be produces, and if heat is added to the product side (exothermic) then more reactants will be produced.

Sarah Rutzick 1L

When there are more reactants than products, the equilibrium sits to the left. This is because the denominator of the K expression is larger than the numerator, which leads to a K value that is less than one meaning that the equilibrium sits to the left.

Sarah Rutzick 1L

hydrogen cyanide, because CN is a compound known as cyanide.

Sarah Rutzick 1L

To add to this, if the shape of the molecule is linear, the number of hybrid orbitals is 2 (sp) since there are 2 regions of electron density. If a molecule is trigonal planar, the number of hybrid orbitals is 3 (sp^2) since there are 3 regions of electron density. It helps to draw the Lewis structu...

Sarah Rutzick 1L

In the answer book, there is a not with this problem that says that other Lewis structures are possible and more stable, like the ones you are talking about, but that they don't change the shape of the molecule. You're right about the double bonded structures being more stable, but I think for the s...

Sarah Rutzick 1L

To add to this, you can tell that beryllium will now have a full octet because of formal charge. In BeCl2, Be has 2 shared bonds, making its formal charge 0, and Cl has 6 unshared electrons and 1 shared bond, making the formal charge of both Cl atoms in BeCl2 0. This proves to be the most stable str...

Sarah Rutzick 1L

According to my notes from lecture, removing the second electron is always harder (second ionization energy is larger), but what about with elements like Mg? Wouldn't the first ionization energy for Mg be larger than the second since Mg has a full s orbital, therefore making it harder to remove the ...

Sarah Rutzick 1L

When writing electron configuration, should the d orbital electrons of lower energy come before the s and p orbital electrons of higher energy? For example, in the electron configuration for gallium, should the configuration be [Ar]3d^10 4s^2 4p^1 or [Ar]4s^2 3d^10 4p^1 ?

Sarah Rutzick 1L

This exception is like the one we talked about in class when we discussed radical exceptions to the octet rule, which is when there is a compound with an unpaired electron.

Sarah Rutzick 1L

I think you are correct that you want to use λ=c/v when dealing with a photon of light, and λ=h/mv for other particles. What also helps is that when you are given a velocity in the problem, you know you must use λ=h/mv, but if you are not given a velocity, you can assume that the velocity is the spe...

Sarah Rutzick 1L

You want n1 to be the energy level that the electron drops to, and n2 to be the energy level that the electron drops from, so n1 will be the smaller n value and n2 will be the larger n value.

Sarah Rutzick 1L

In this problem it makes sense to convert the wavelength to picometers because one picometer is equal to 1x10^-12 meters. Since the wavelength was calculated as 8.8237x10^-12 m, 8.8237 pm expresses this wavelength in the simplest form.

Sarah Rutzick 1L

To add to that, diffraction occurs because of the different wavelengths of the waves passing through the crystal. Since electrons passing through a crystal show diffraction, this proves that instead of being a uniform stream of energy, electrons behave like waves.

Sarah Rutzick 1L

I think that an excited electron in a hydrogen atom will always drop back down to the first energy level because that is where electrons are most stable, especially since hydrogen only has one electron and that electron exists naturally in the first energy level.

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