CHEMISTRY COMMUNITY

Cameron_Greenberg_3C

Gibbs Free Energy is dependent on temperature, entropy, and enthalpy. A high temperature can still be multiplied by a positive or negative value of change in entropy, so just knowing the temperature does not give the sign of the Gibbs Free Energy value.

Cameron_Greenberg_3C

While exothermic reactions release energy, they can't be assumed to be spontaneous because a low entropy value could counter a negative enthalpy value ad make delta G positive, making it non-spontaneous.

Cameron_Greenberg_3C

Yes, this is more of a temporary fix in the presence of a lot of reactant. Eventually once it is all consumed the reaction cannot procees.

Cameron_Greenberg_3C

It depends on the the oxidation and reduction half reactions and how many electrons are transferred. It is necessary to multiply by the correct coefficients to find the least common multiple between the two half reactions to determine the correct number of electrons that are actually transferred.

Cameron_Greenberg_3C

Usually a Van't Hoff question will give you the reaction enthalpy and ask you to either find a K value from an original K and two temperatures, or a T value from and original T and 2 equilibrium constants.

Cameron_Greenberg_3C

Looks like you answered but in addition, make sure to write out both half reactions and multiply by coefficients as needed to find the true number of electrons transferred.

Cameron_Greenberg_3C

At constant temperature and pressure (fusion for example), the standard entropy of fusion is the standard deltaH over the temperature at which the fusion takes place.

Cameron_Greenberg_3C

Just know that a negative Gibbs Free Energy corresponds to a thermodynamically stable reaction and vice versa. Labile and nonlabile refer to reaction rates.

Cameron_Greenberg_3C

When it is necessary to balance hydrogens, H+ is added when the reaction is specified as acidic, and OH- and H2O are added when it is basic.

Cameron_Greenberg_3C

The behavior of redox reactions is similar to a normal reaction in that molecules can cancel on either side of the arrow. So if a basic half reaction has 4 H2O on the left and 2 on the right, it can be simplified to just 2 H2O on the left.

Cameron_Greenberg_3C

For our purposes in redox problems, the only relevant oxidation numbers are O (-2) and H (+1). These are the only elements coupled with metals, and the metals have an unspecified oxidation number that can be deduced from adding up the oxidation numbers of the O and H ions and then subtracting back t...

Cameron_Greenberg_3C

A basic redox problem will not have H+ ions present. Balancing hydrogens is instead done with H2O and OH-. If the left side of the equation needs, for example, two more hydrogens to be balanced, instead add 2 H2O to the left side and 2 OH- on the right side, resulting in a net gain of only two hydro...

Cameron_Greenberg_3C

In addition, a reversible reaction will instead have the equation w=-nrTln(V2/V1).

Cameron_Greenberg_3C

This equation is more used in this class to explain the heat capacity at constant volume/pressure to be 3/2R and 5/2R respectively. The equation itself is unlikely to appear on any exam.

Cameron_Greenberg_3C

Michelle Nwufo 2G wrote:Also, it’s kind of confusing why work and degeneracy have the same constant. Are they somehow related?

Work has a lowercase w, while degeneracy is an uppercase W. They are not related.

Cameron_Greenberg_3C

In our class, (delta)U=0 for all reversible reactions. Irreversible ones will not equal zero if there is any expansion work. So if deltaV is not zero, there is energy transferred in the system.

Cameron_Greenberg_3C

Enthalpy by definition is heat at constant pressure, making it a state variable. So (delta)U= (delta)H + w at constant pressure

Cameron_Greenberg_3C

In addition, reversible and isothermal for our purposes in 14B can be considered synonyms. So any reaction that is reversible is therefore isothermal, and vice versa. These reactions have a (delta)U of 0.

Cameron_Greenberg_3C

In addition to the other answers, the heat capacity at constant volume and the heat capacity at constant temperature are always separated by a factor of R. (C)v,m = R + (C)p,m. So, for single atoms 3/2R is the (C)v,m and 5/2R is the (C)P,m, while linear atoms have a (C)v,m of 5/2R, and more complica...

Cameron_Greenberg_3C

When a phase change occurs in the reaction, there is an additional enthalpy change that must be accounted for on top of the normal reaction enthalpy. The enthalpies of molecules are different in different states.

Cameron_Greenberg_3C

298 K is the standard temperature in an ideal gas equation and it is derived from 25 C + the conversion factor of 273. 298 seems like a random number but is in fact the Kelvin value in standard conditions.

Cameron_Greenberg_3C

There are three main methods. 1. Standard Enthalpy of Formation, given by the enthalpy of formation of each reactant and product 2. Hess's Law, given by several equations that need to be manipulated to add up to the given equation 3. Bond Enthalpy, given by the enthalpies of each bond in the reactan...

Cameron_Greenberg_3C

It depends on what the equation gives you. Usually the C(delta)T equation is only used in calorimeter problems. If you feel more comfortable with one or the other, you can easily convert between grams and moles with molecular weight.

Cameron_Greenberg_3C

In this type of problem, the total enthalpy is split into 3 parts: the enthalpy of bringing the reactants from their original temperature to the reaction temperature (modeled by mcdeltaT), the reaction enthalpy at the temperature given, and the enthalpy of bringing the products back to the original ...

Cameron_Greenberg_3C

The chemistry definition of enthalpy is the amount of energy held in chemical bonds. Bond enthalpy problems prove this fact as the breaking and forming of bonds sum up the change in enthalpy.

Cameron_Greenberg_3C

Significant figures are not emphasized in 14B. Step sizes between C and K are the same so a change in temperature can be expressed in either unit.

Cameron_Greenberg_3C

You could draw a simple Lewis structure, but most of the time the structure is given or is relatively easy to figure out so that you may focus strictly on adding and subtracting the bond enthalpies.

Cameron_Greenberg_3C

Calorimeter problems are designed to evaluate reactions in an isolated system so it is understood that there is no interaction with the surroundings. This makes the heat gained/released of the elements inside equal to the heat gained/released by the calorimeter.

Cameron_Greenberg_3C

"Open beaker" implies constant pressure of 1 atm.

Cameron_Greenberg_3C

The terms open/closed/isolated describe a system, not a reaction. So it has nothing to do with whether a reaction is exo/endothermic. Instead, it describes the system's ability to transfer energy and matter to its surroundings.

Cameron_Greenberg_3C

The atom with the lowest ionization energy is in the center. I usually look for which atom is furthest to the bottom left of the periodic table. If there is only a single atom bonded to multiple of a different atom, then the single atom will be central. ex: NH3 has nitrogen in the center

Cameron_Greenberg_3C

Lone pairs will always have a stronger repulsion force than atoms, so it is unlikely that the bond angles would be exactly 90 or 120. The lone pair makes each angle slightly smaller.

Cameron_Greenberg_3C

The axes are not relevant for an octahedral shaped molecule because the shape is the same no matter how the molecule is rotated. So any of the planes can have the four atoms with the two lone pairs above and below.

Cameron_Greenberg_3C

Anjali_Kumar3E wrote:How do you know HCN is polar or non polar?

HCN is polar. Although the molecule is linear, the electronegativity difference makes the hydrogen side slightly positive and the nitrogen side slightly negative.

Cameron_Greenberg_3C

Most of our practice problems feature molecules with an even number of total valence electrons so that every electron is paired. But, radicals are different in that there can be a single unpaired electron in special cases. Usually in the real world, a radical is an incomplete molecule that finds an ...

Cameron_Greenberg_3C

If there are no lone pairs present, then the atoms will be symmetrically arranged around the center atom. If they are all the same element, then yes, the molecule will be non-polar. But if there is a large electronegativity difference, then the molecule could be polar.

Cameron_Greenberg_3C

Bonds themselves are not polar or non-polar, only entire molecules have polarity. Hydrogen bonding is a dispersion force that causes molecules to be slightly attracted to each other if they have N, O, or F. But this force between separate molecules has no effect on the polarity of an individual mole...

Cameron_Greenberg_3C

Maybe I'm just reading this wrong but are you referring to see saw? I don't think it's see saw, I think it is T-shape because I believe the structure being described has 3 bonds and 2 lone pairs, but maybe I also read it wrong. Yes, trigonal bipyramidal has five bonds and no lone pairs, while T-sha...

Cameron_Greenberg_3C

As everyone else has said, they are synonyms. But, the book says angular and that seems to be the more technical term.

Cameron_Greenberg_3C

The VSEPR format allows you to be able to tell what the molecular geometry is. If the geometry is asymmetrical or the atoms have a large difference in electronegativity, the molecule is polar. Generally, any lone pairs will cause areas of asymmetry.

Cameron_Greenberg_3C

The molecule is linear, but since nitrogen is the least electronegative, it is the central atom. So the molecules are lined up N-N-O. Oxygen is much more electronegative than nitrogen making the molecule negatively charged on the oxygen side and therefore polar. Why are the molecules lined up N-N-O...

Cameron_Greenberg_3C

The molecule is linear, but since nitrogen is the least electronegative, it is the central atom. So the molecules are lined up N-N-O. Oxygen is much more electronegative than nitrogen making the molecule negatively charged on the oxygen side and therefore polar. Why is it lined up that way though. ...

Cameron_Greenberg_3C

I am confused on how to convert the normal orbitals into hybrid orbitals on parts b, c, and d? How do you incorporate the electrons in the d block?

Cameron_Greenberg_3C

It is based upon electronegativity, which is highest on the upper right of the periodic table. Non-polar molecules typically have a difference of 1.5 or less between the two atom's electronegativity values. Polar molecules typically have a 2.0 or greater difference. Symmetry and shape also play a pa...

Cameron_Greenberg_3C

Tetrahedral is typically the angle that trips people up because it is assumed that it would be 90 degrees since there are 4 molecules, when in fact it is 109.5 degrees. All the others are fairly straightforward (ex: trigonal planar, 120, or linear, 180) or are estimated instead of exact values (ex: ...

Cameron_Greenberg_3C

Bigger molecules and more polarizable molecules both have stronger of London forces. A bigger molecule has more electrons and therefore a higher chance of causing an instantaneous dipole moment (like you asked), so with more of these moments, the force exuded will be stronger.

Cameron_Greenberg_3C

The molecule is linear, but since nitrogen is the least electronegative, it is the central atom. So the molecules are lined up N-N-O. Oxygen is much more electronegative than nitrogen making the molecule negatively charged on the oxygen side and therefore polar.

Cameron_Greenberg_3C

Dipole-dipole forces are specifically between two polar molecules. The negatively charged side of one molecule will be attracted to the positive side of another. The force is strictly between two separate molecules. London forces, like dipole-dipole forces, are between two separate molecules. But Lo...

Cameron_Greenberg_3C

Elements that have a d orbital can have expanded octets. However, a few elements on the right side of period 3 can expand with the extra electrons. S and Cl are common examples of central atoms with more than 8 electrons.

Cameron_Greenberg_3C

You must draw out each Lewis structure to determine how many bonds are between the molecules in each specific example.

Cameron_Greenberg_3C

Yes, the range of 120-140 would be correct. Most likely, any number between would be counted correct also.

Cameron_Greenberg_3C

Degeneracy is orbitals with the same energy.

Cameron_Greenberg_3C

The charge of individual atoms that make up the ion should always be zero or as close to zero as possible. Atoms are most stable at that point and they always tend towards their lowest energy state. Even in an ion with 2- charge, the formal charge of each atom should be close to 0, not -2.

Cameron_Greenberg_3C

How can you tell when a central atom is allowed to have more electrons than the octet? I know the extra electrons go to the d-orbital, but it seems like some of the given atoms like P and S are in period three and have no d-orbital? It just seems a bit random as to when to put 5 or even 6 bonds.

Cameron_Greenberg_3C

I started the same way you did and found the frequency using v=c/wavelength. I then used this frequency in the E=hv equation, which showed the energy per photon to be 4.7x10^-19. The lamp gives off 32 J/s so in 2 seconds, it released 64 J of energy. Dividing the 64 J of total energy by the 4.7x10^-1...

Cameron_Greenberg_3C

SydBenedict3H wrote:Does it matter if you use 2.99x10^8 or 3.00x10^8? or are they close enough that the difference is indiscernible?

3.00x10^8 should be accurate enough. The constant will most likely be given on a test on the front page equation sheet. But either number works.

Cameron_Greenberg_3C

Knowing the general range of the visible spectrum is a good idea. But we will not need to identify colors by their wavelengths, as most questions on the topic give both pieces of information.

Cameron_Greenberg_3C

The question states in the form of a ratio, so in this case, I wrote it in C:H:O form. However if it doesn't specify, I would stick with the empirical formula.

Cameron_Greenberg_3C

I'm in a similar place having not taken Chemistry in 4 years. I worried about this as well but I have come to realize that this class is lower division and therefore Lavelle is accustomed to teaching rusty students. He gave many useful resources including study sessions both on the Hill and on campu...

Cameron_Greenberg_3C

The standard SI units should be fine (m, kg, s, as well as L) and the prefixes should be converted back to their bases.

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