CHEMISTRY COMMUNITY

Justin Sarquiz 2F

If this showed up on the final, the question would tell you that a precipitate would form and ask why. And you would need to know that there are other reactions taking place. But the question is not going to ask you to determine if a precipitate will form.

Justin Sarquiz 2F

I think the free energy of activation refers to delta G. The activation energy is how much energy must be supplied to break the bonds of the reactants.

Justin Sarquiz 2F

You pretty much write out the rate equations for two equations. You then divide the equations from each other. You want to pick two equations where the concentration of one reactant will be left over. The rate constant will cancel and you will have only one variable left over, which is the order of ...

Justin Sarquiz 2F

A species in this case refers to the number of reactants. In the example you gave, this is bimolecular. The number of moles of reactants determines the molecularity.

Justin Sarquiz 2F

Rate = k[R]
The equilibrium constant accounts for the concentrations as well as the rate constant because K=k/k'. Therefore, the equilibrium contains both variables involved in the rate equation.

Justin Sarquiz 2F

I saw that too, and I believe that there should be a platinum electrode on the anode side. Maybe it is just a mistake in the solutions manual.

Justin Sarquiz 2F

If delta G is negative, then that means that free energy is released. That released energy can be used to do work.

Justin Sarquiz 2F

E= E* - RT/nF ln Q
E = E* - 0.0592/n log Q
0.1=0.1 - 0.0592/12 log Q
0= log Q
Q=1
I think you messed up in calculating your E*, and there should be 12 electrons transferred.

Justin Sarquiz 2F

delta G= delta G* + RT ln Q #substitute delta G for -nFE
-nFE = -nFE* - RT ln Q #divide both sides by -nF
E= E* - RT/nF ln Q

Justin Sarquiz 2F

I think the book made a mistake because I got K=107. The book said that 2 electrons were transferred, but it should be that one electron was transferred.

Justin Sarquiz 2F

To calculate the standard potential energy, you do Ecathode-Eanode. It is not always right minus left because someone could draw the anode on the right if he or she wanted to.

Justin Sarquiz 2F

I found delta G* using delta G*=delta H* - Tdelta S*
You have to solve for delta H* and delta S* and then use those values with the temperatures to find delta G*
deltaG*= -RT ln K
You know everything except for K
Then do this same procedure for the other temperature

Justin Sarquiz 2F

Cobalt alone is always a solid because it is a transition metal, which is why Cobalt is the electrode in this example.

Justin Sarquiz 2F

The R constant should be 8.314 J/(mol*K) instead of 0.0831.

Justin Sarquiz 2F

I had this question too, and my TA said that if you put H+ on one side and 02 on the other, then you cannot balance the equations for the molecules. You can balance the charges for the half reaction, but you cannot have equal number of hydrogen and oxygen molecules on both sides.

Justin Sarquiz 2F

We did an example in class with actual numbers to show how decreasing the volume, increased concentration (n/V) and made equilibrium shift to the side with fewer gas moles. You can use the same numbers to show how increasing the volume makes equilibrium shift to the side with more gas moles.

Justin Sarquiz 2F

Splitting the reaction into half reactions will be useful when we get to galvanic cells. The half reactions help determine which reaction is occurring in each container and help see what is occurring in the galvanic cell.

Justin Sarquiz 2F

Delta G only determines whether a reaction is spontaneous or not. It says nothing about the speed of the reaction.

Justin Sarquiz 2F

You should memorize the oxidation numbers of common polyatomic ions like phosphate and sulfate because these are definitely fair game.

Justin Sarquiz 2F

When delta G is negative, the reaction is spontaneous, so it forms products because the reaction proceeds normally. When delta G is positive, the reaction is not spontaneous, so it forms reactants since the reaction does not proceed without an input of energy.

Justin Sarquiz 2F

Work done at equilibrium is w=-nRT ln V2/V1. According to the graph that Lavelle showed us, the area under the graph for reversible reactions is greater than irreversible reactions. Therefore, reversible reactions do more work than irreversible reactions.

Justin Sarquiz 2F

I believe that insulated systems are isolated because heat outside the insulated system cannot enter or leave.

Justin Sarquiz 2F

Delta U for an ideal gas is 3/2 nRdelta T. Since temperature is constant, delta U equals zero. I am not really sure how to derive U=3/2nRT because I do not think that we covered it in class. I can recall maybe just one (if even that) problem using this equation on the homework, so it probably is not...

Justin Sarquiz 2F

The homework had a lot of problems about LeChatelier's Principle, and Lavelle's survey about the video modules has a lot of examples too. Hope those help

Justin Sarquiz 2F

Spontaneity is determine using the Gibbs Free energy equation G=H-TS. Since -G is spontaneous, we want a large and positive delta S. However, you would need to plug in the values for temperature and enthalpy to really determine spontaneity because enthalpy could be greater than TS.

Justin Sarquiz 2F

Work done on a system means that work is positive. Work done by the system on the surroundings means that work is negative because some energy is lost to the surroundings through work.

Justin Sarquiz 2F

The negative sign in the formula takes into account that the system is doing work, so it is losing energy. For example in w=-P x delta V, if the volume increases (expansion), the system is doing work and is losing energy.

Justin Sarquiz 2F

If a reaction is reversible, then the problem will state that it is reversible or that the reaction is at equilibrium. I think a reaction must be at a constant temperature because if temperature is not constant, then this changes the equilibrium constant. Therefore, we have to use the work equation ...

Justin Sarquiz 2F

You have to use two gas constants in order to cancel out the atm(L). You take that answer and multiply it by (8.314 J x K x mol^-1/ 0.08206 L x atm x K^-1 x mol^-1). The units will cancel until you are left with just Joules.

Justin Sarquiz 2F

For this problem, I used Hess's Law to cancel the NO2. For the final equation, I got 2 NO + 3/2 O2 ---> N205. You know the enthalpy of reaction and the enthalpy of formation for NO and the enthalpy of 02 is zero, so you can solve for the enthalpy for N205. I do not know why the answer key put that d...

Justin Sarquiz 2F

We do not take into account the moles of the strong acid and strong base because they are at the same molarity. Since they are a strong acid and a strong base, we assume that they combine to water, so we can just ignore that information in the problem.

Justin Sarquiz 2F

An inert gas is on both sides of the chemical equation because it is not involved in the reaction. An inert gas can change the pressure of the system though.

Justin Sarquiz 2F

The units of J or kJ does not really matter unless the problem specifies. Make sure to pay attention to the units of the constants though because some constants are given as kJ and some are given in J.

Justin Sarquiz 2F

In addition, enthalpy is a state function because you can add and subtract enthalpy values.

Justin Sarquiz 2F

q is not always heat transfer from a hot object to a cold object. q can be either positive or negative depending on the situation.

Justin Sarquiz 2F

Lavelle has not gone over this topic yet because he is starting on thermochemistry first and then going to thermodynamics. I am guessing that you would use the equation according to the information given in the problem.

Justin Sarquiz 2F

The formula sheet should say how to convert from atm to bar or bar to torr, so you should not worry too much about that.

Justin Sarquiz 2F

Using the same values as stated above, we can also say how equilibrium lies to the right or shifts right or favors products. All three of the terminologies mean about the same thing.

Justin Sarquiz 2F

When the concentration of either the reactant or product is changed, the system temporarily goes out of equilibrium. If the concentration of reactants is increased, then the concentration of products will increase until the ration of products over reactants is the same as usual. This means that the ...

Justin Sarquiz 2F

Making a reaction go faster is adding a catalyst. Adding the catalyst only increases the rate of the reaction and does not increase the amount of product at equilibrium.

Justin Sarquiz 2F

K is the concentration of the products divided by the concentration of reactants. Therefore, K tells us the relative concentrations between products and reactants. If K >1, then the concentration of products is greater than the concentration of reactants at equilibrium and vice versa.

Justin Sarquiz 2F

All you need to know about enthalpy so far is that negative enthalpy means exothermic and a positive enthalpy means endothermic. Therefore, heating up an endothermic reaction causes equilibrium to favor products, and cooling an exothermic reaction causes equilibrium to favor products too.

Justin Sarquiz 2F

The equilibrium constant is actually the activity of the product divided by the activity of the reactant. Calculating the activity is its own separate equation, but in that equation the units actually cancel. Therefore, the activity of the products and reactants have no units making K have no units ...

Justin Sarquiz 2F

You only look at the moles of gas, not the moles of aqueous or liquids or solids. If we look only at the moles of gas, we see that there are only 6 moles of gas on each side, so compressing the system does not shift equilibrium.

Justin Sarquiz 2F

concentration= n/V K=[HCl]^2/ [Cl2][HI]^2 If the volume decreases, then the concentration increases. If you plug in values for the concentration, you will see that the Q is actually less than K. Since Q<K, the reaction proceeds to the right. He did a similar example in class where he plugged in valu...

Justin Sarquiz 2F

This reaction is breaking the N2 bond to form 2N atoms. Breaking bonds require heat making this reaction endothermic.

Justin Sarquiz 2F

You would use partial pressure for equation with gases and concentration for equations in aqueous solution. Technically, writing the answer with concentration instead of partial pressure is the same thing I believe. But in problems where you have to solve for K, the problem would give you either con...

Justin Sarquiz 2F

The concentration of the products and reactants do not have to be the same at equilibrium. The concentration of products does not change at equilibrium, and the concentration of reactants does not change at equilibrium too.

Justin Sarquiz 2F

This question is asking which reactant is more stable (Cl2 or F2). Since the question is asking for the stability of the reactants, we must see which K value is smaller because that reaction would favor the reactants over the products more. If the question asked for the stability of the products, th...

Justin Sarquiz 2F

Changing temperature in a system in equilibrium will change the equilibrium constant. Changing the concentration or volume or pressure in a system in equilibrium would not change the equilibrium constant because the ratio between products and reactants would remain the same.

Justin Sarquiz 2F

The equilibrium constant is the same for concentration or partial pressure. If you put the molecule in brackets, then that means you are using concentration. If you put the "P" then that means you are using partial pressure. It just depends on what information the problem gives you.

Justin Sarquiz 2F

I do not know specifically when light acts as a wave, but you should know that light has both wave and particle qualities. Light acts as a particle according to the photoelectric effect. Light acted as a wave when shined through a hole in the wall or through a crystal because light was diffracted in...

Justin Sarquiz 2F

Group 1 elements do not affect the pH, so we can leave out potassium in the equation.

F- + H20 ------> HF + OH-

Since F-, gained a proton, KF would be considered basic. The reason this cannot form a hydronium ion is because KF has no hydrogen to donate to water.

Justin Sarquiz 2F

I do not think you have to worry about sp2d orbitals because this is pretty uncommon. For the final, you should just be familiar with sp, sp2, sp3, sp3d, and sp3d2 hybridization.

Justin Sarquiz 2F

Delocalized electrons are associated with resonance structures. If a molecule has resonance, the molecule is like a "blend" of all the resonance structures combined. Therefore, the bond lengths that participate in the resonance would all be a "blend" of each other and be the same...

Justin Sarquiz 2F

Bond length can be determined by looking at the size of the atom. The larger the atom, the longer the bond length and vice versa.

Justin Sarquiz 2F

Carbon does not always have sp3. The trick I use to calculate hybridization is that I count the number of electron densities around the atom. If there is two electron densities, then hybridzation is sp. If there is three, then sp2. If there is four, then sp3. If there is five then, dsp3. If there is...

Justin Sarquiz 2F

There were different versions of the test, but essentially each lone pair on a N, O, or F counts as one hydrogen bonding site. Each hydrogen attached to a N, O, or F also counts as a hydrogen bonding site.

Justin Sarquiz 2F

I am not sure about bases, but for acids the longer the bond length, the stronger the acid because it is easier to lose an H+. Therefore, the larger the atom, the longer the bond length, the stronger the acid.

Justin Sarquiz 2F

In its simplicity, each lone pair around a N, O, or F each count as one hydrogen bonding site. A hydrogen bonded to a N, O, or F also counts as a hydrogen bonding site.

Justin Sarquiz 2F

Add -ate to the end of the transition metal's name if the charge within the brackets is negative.

Justin Sarquiz 2F

The reason there are 3 Bromium atoms in part b is because Chromium has a +3 charge and Bromium has a -1 charge. Therefore, it requires 3 Bromium atoms to make the molecule neutral.

Justin Sarquiz 2F

Any three points make up a plane. Think about a plane as a sheet of paper; this is how I visualize it. If three atoms are on the same plane, that means that the three atoms can be attached to each other through a "sheet of paper".

Justin Sarquiz 2F

When talking about coordinate covalent bonds, we want to look at the lone pairs in order to determine whether a bond will form, not the charges. We need to check that an atom has a lone pair of electrons that it can donate, so a coordinate covalent bond can form.

Justin Sarquiz 2F

The atom with larger polarizability will have the stronger LDF force. In other words, the larger the atom, the stronger the LDF. We can use this information to determine, which molecules have a higher melting point.

Justin Sarquiz 2F

Since the r is to the power of 6, the distance has the most impact on the potential energy of a molecule. In other words, the potential energy is strongly dependent on the distance between atoms.

Justin Sarquiz 2F

Dipole dipole is present if the molecule is polar. In order to have Hydrogen bonding, a hydrogen must bond with an electronegative oxygen, nitrogen, or fluorine, and these three elements must have an extra lone pair in order to bond.

Justin Sarquiz 2F

The type of bond does not really matter when determining polarity of the molecule. We look at the electronegativities of the atoms to determine whether a molecule is polar or not.

Justin Sarquiz 2F

Water actually has a bent and not a linear shape because the two lone pairs of electrons sit next to each other rather than on opposite sides. If the two lone pairs are on opposite sides, then water would be nonpolar.

Justin Sarquiz 2F

In addition, knowing the VSEPR formula for a molecule can help you determine the shape of the molecule. However, molecules with the same VSEPR formula will have the same shape, but the bond angles can be different if they have different atoms attached.

Justin Sarquiz 2F

SCl4 has a Lewis structure of a central atom S with four Chlorines around it with a lone pair on the central atom. A central atom with four bonded atoms and one lone pair has a seesaw shape. ICl3 has a Lewis structure with I as the central atom and 3 Chlorines surrounding with with two lone pairs. A...

Justin Sarquiz 2F

If you do a lot of practice, you should have the angles memorized without really focusing solely on the angles. Practicing naming molecular shapes and angles will help you memorize angles without even realizing it.

Justin Sarquiz 2F

A tetrahedral shape refers to four electron densities around a central atom. Trigonal pyramidal refers to three atoms around a central atom and then one lone pair. Trigonal pyramidal is technically tetrahedral shape since there are four electron densities around the central atom. But it is more spec...

Justin Sarquiz 2F

The lone pairs only affect the shape if they are on the central atom. Lone pairs on the surrounding atoms have no effect on the shape of the molecule.

Justin Sarquiz 2F

Heisenberg's Uncertainty Equation always uses "kg" as the mass and "m" as the x value. The speed of light equation c=v(lambda) always uses meters for the wavelength. Ephoton- threshold energy= kinetic energy. This equation uses Joules/photon as the unit of energy and the mass in ...

Justin Sarquiz 2F

I am pretty sure that the Bohr Frequency Condition is given on the equation sheet. Even if it is not, the equation is E=hv, so it is not that hard to memorize.

Justin Sarquiz 2F

We should use the expanded octet in order to find the lowest energy Lewis structure. This means that we want the Lewis structure with the lowest formal charges. Therefore, it is crucial to solve for the formal charges whenever drawing Lewis structures because it shows whether the Lewis structure dra...

Justin Sarquiz 2F

The Bohr Frequency Condition says how the frequency of light is directly proportional to the energy of light. This was proved by the photoelectric effect that showed how increasing the frequency of light increased the energy of light. Increasing the intensity of light did not increase the energy of ...

Justin Sarquiz 2F

The exceptions to the octet rule are P, S, and Cl because they have an unfilled d subshell. However, I do not know the maximum amount of electrons they can hold, but I do not think that it is that relevant. Just know that these three elements can have more than 8 electrons.

Justin Sarquiz 2F

First, you want to determine how many valence electrons the element has. Since Ga has 3 valence electrons, Ga wants to lose 3 electrons to fulfill the octet rule giving it a +3 charge. Another example is Fluorine. Fluorine has 7 valence electrons. Therefore, Fluorine wants to gain 1 more electron to...

Justin Sarquiz 2F

Sorry I meant to say that Chromium and Copper takes 1 electron from the s subshell not the p subshell.

Justin Sarquiz 2F

Yes always write the 3 before the 4 subshell because the 4th subshell has slightly higher energy than the 3 subshell. The Chromium and Copper exception refers to the fact that these elements grab one electron from the p subshell to make Chromium have all half full d5 subshell. Copper grabs one elect...

Justin Sarquiz 2F

The equation for formal charge that I use is the one given during class, which is FC= V - (L + S/2) where V is the number of valence electrons, L is the number of lone pair electrons, and S is the bonds shared. Unfortunately, I do not know, which equation you used in discussion to find the formal ch...

Justin Sarquiz 2F

A radical is an atom, molecule, or ion that has an unshared pair of valence electrons. Therefore, for this problem, you want to draw out the Lewis structure for the following molecules or ions and see whether there are any unshared pair of valence electrons around any of the atoms. If so, then the m...

Justin Sarquiz 2F

You should calculate the formal charges in order to see which structure has the lowest energy. Once you get the hang of it, calculating the formal charge only takes a couple of seconds, so it will not take up that much of your time.

Justin Sarquiz 2F

If they asked for the electron configuration of an anion or a negative charge, you would just write out the electron configuration for the element normally, and then add an extra electrons to the end depending on the size of the negative charge.

Justin Sarquiz 2F

Metals are solid, shiny, good conductors, ductile, and malleable. Metals have like to lose electrons because they form ionic bonds and want an octet. Nonmetals are brittle, bad conductors, not ductile, not malleable, and some are liquid. They tend to gain or share electrons. Metalloids have mixed pr...

Justin Sarquiz 2F

Instead of memorizing the elements, just remember that the exceptions to the octet rule are the the first four elements because they would need to gain too many electrons to have an octet, which is very difficult.

Justin Sarquiz 2F

Uncertainty values do not always have to be integers. We could have 5 +- 1.3. The uncertainty value would be 1.3(2)= 2.6.

Justin Sarquiz 2F

This question is asking about the electron configuration in the valence shell, which is the last shell. Since this problem is asking about the Group 5 transition metals, the valence shell must be (n-1)d^3 because transition metals in Group 5 have an exponent of 3, but it is also (n-1) because the fo...

Justin Sarquiz 2F

(delta p) (delta x) > h/(4pi) delta x > h/(4pi)[(m delta v)] delta x > h/(4pi)(8 x 10) The change in velocity is 2 times the +/- number. So change in velocity is 2 x 5=10. delta x > 6.59 x 10^-37 m The solution manual makes an error and says that the change in velocity is 5 when it should be 10, so ...

Justin Sarquiz 2F

When we use the energy level equation, I think we are under the assumption that the electron is in an energy level with the nucleus. Once the electron leaves the influence of the nucleus, then this equation does not apply. Therefore, as n approaches infinity, we make the assumption that the electron...

Justin Sarquiz 2F

The equation for energy levels is E=-hR/n^2. The negative sign is due to the fact that we call zero energy at the energy level of infinity. This equation tells us that the more energy levels, the higher the energy because the smaller the negative. The less energy levels, then the lower the energy be...

Justin Sarquiz 2F

Yes, the Bohr Frequency Condition and the equations for energy are specific only for Hydrogen only because Hydrogen only has one electron. If we use another element that has multiple electrons, then technically those electrons would be repelling each other, so the Bohr Frequency Condition would not ...

Justin Sarquiz 2F

In the photoelectric effect lab, a light was shined on a metal surface in order to measure what energy was needed to remove an electron from different metals. If a certain wavelength did not remove an electron, then we would expect that increasing the intensity should remove an electron if light act...

Justin Sarquiz 2F

Everyone is correct about saying how the limiting reactant determines the amount of product formed. We can calculate the theoretical amount of product formed from our limiting reactant. Some problems would also give us the actual amount of product produced from the experiment, which would be the act...

Justin Sarquiz 2F

It is also important to note that one must look at the lowest amount OF MOLES when determining the limiting reactant. Make sure to balance the equation first and then compare the ratio of moles of the reactants to see which reactant has less moles according to the ratio.

Justin Sarquiz 2F

Planck's constant, Avogadro's number, and any other constant will be given. I don't think the prefixes will be available including Angstrom.

Justin Sarquiz 2F

We can manipulate frequency by changing the wavelength since frequency and wavelength are inversely proportional [c=lambda (v)]. Increasing the intensity only increases the amplitude of the wave and has no effect on the frequency or wavelength.

Justin Sarquiz 2F

Avogadro's constant tells us that there is 6.022 x 10^23 molecules or atoms in a mole. If we ever have grams or moles of an element, we can use Avogadro's constant to determine the number of molecules or atoms.

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