CHEMISTRY COMMUNITY

Michelle Nguyen 2L

The reaction rate increasing by a factor of 1000 is like saying the rate constant is increasing by a factor of 1000. Therefore, using k= Ae^-Ea/RT, you can write (Ae^(-Ea/R(298 K))/(Ae^(-98000 J/R(298 K)))= 1000. That is, the ratio between the two k values is equal to 1000.

Michelle Nguyen 2L

Janet is right! Elementary reactions are the only kind where you can derive the rate law directly from the chemical reaction and it’s stoichiometric coefficients. Normally, reactions are composed of many elementary steps, so we cannot get the rate law just from looking at the overall chemical reaction

Michelle Nguyen 2L

Yup, pretty sure Vivian is correct! Plus, we would need to be told which step is the slow step to use the pre-equilibrium approach in the first place

Michelle Nguyen 2L

Question 15.27 asks you to use the half life for a first order reaction, t(1/2)= ln2/k, which does not depend on the concentration of reactants at all. That means the half life will always be the same length of time no matter how far the reaction proceeds or what the initial concentration is, so you...

Michelle Nguyen 2L

You are correct that the pre-equilibrium assumption depends on information given on which step is the slow step. However, you shouldn't think about the two assumptions as mutually exclusive. That is, rather than thinking you should use either one or the other, it is more like you would often need to...

Michelle Nguyen 2L

I think tests will mostly focus on uni or bimolecular reactions. However, I think we are still expected to know the definition of a termolecular reaction, why they are more rare, and how to write the rate law for a given termolecular elementary reaction.

Michelle Nguyen 2L

I'm not sure if the two are related in some convoluted way, but as of right now at least, we cannot determine the order of a reaction simply by looking at the chemical formula and stoichiometric coefficients of the reaction. For example A->B might be a first order reaction, but so might 3A->2B+C

Michelle Nguyen 2L

The problem should always tell you the order of the reaction unless they want you to verify if it is first order or whatnot, in which case you would likely have to plot ln[A] by time and determine if the graph is a straight line. The decomposition of N2O5 to yield NO2 and O2 gas is a first order rea...

Michelle Nguyen 2L

Yup; knowing that the rate has units (mol/L*s) tells you what units you should use for k, the rate constant. For example, the rate constant for a first order reaction has units (1/s) while the rate constant for a third order reaction Rate= k[A]^3 has units L^2/(mol^2 *s). ( [A] has units mol/L)

Michelle Nguyen 2L

Hi all, You can tell which half reaction to use for water by the fact that water must always be the reactant! Therefore, to find which substance gets reduced at the cathode, compare the substance in question to the cell potential of water getting reduced, and at the cathode, compare to the cell pote...

Michelle Nguyen 2L

Ions will form at the electrode where the ion concentration is lower in order to reach equilibrium. In a concentration cell, the reactants and products are the same ions except their concentrations; hence, the product ions form at the side with the lower ion concentration.

Michelle Nguyen 2L

Think of it just like how we looked at Q and K in Chem 14A. The only addition in electrochem is that we look at cell potential as well, which is related to Q. As the reaction proceeds, concentrations of reactants decrease and that of products increases. This makes a larger Q and when Q is large enou...

Michelle Nguyen 2L

Standard conditions only affect pressure and concentration (1 bar/atm and 1 M for solutions)

Michelle Nguyen 2L

The substance at the anode is oxidized from a solid to an aqueous cation while at the cathode the ion is reduced to the solid form. Therefore, for a current to run we want the substance at the anode to be oxided into the cation, which would happen if the anode presently has a lower concentration of ...

Michelle Nguyen 2L

More strongly reducing means a substance is more easily oxidized and less easily reduced, corresponding to a negative cell potential for the reduction of the substance. Therefore, to find a strongly reducing agent, look for a substance with a very negative cell potential (in a cell where the substan...

Michelle Nguyen 2L

We set deltaG equal to 0 when we want to find the temperatures at which a reaction is spontaneous. This is because at equilibrium deltaG is 0; neither process, forward or reverse, products or reactants is favored. If we have T for which deltaG is 0, then we know that any slight change in T from that...

Michelle Nguyen 2L

For a reaction that is thermodynamically favored (delta G is negative), K is larger than 1, showing that products are favored. If delta G is positive (the reaction is not spontaneous), then K is less than 1 and reactants are favored (the reverse reaction is spontaneous). At equilibrium, neither prod...

Michelle Nguyen 2L

Yup Taryn has the correct reasoning; at higher temperatures the S of products and reactants are different but since both are raised to the same temperature, we assume that the difference in entropy between the two remains relatively the same. Of course delta S does change in reality, but for our pur...

Michelle Nguyen 2L

Hmm that’s a good question; I’m not sure because it seems that the equation for change in entropy when temperature changes is derived from the constant temperature equation which only works for reversible pathways, yet the textbook doesn’t seem to specify that this second equation is only for revers...

Michelle Nguyen 2L

The first equation tells you that adding the same amount of heat at a higher constant temperature results in a smaller increase in temperature, but addition of heat still causes increase in entropy. The second equation is for entropy change when the temperature changes as well; this equation tells u...

Michelle Nguyen 2L

I can’t be 100% sure of my answer but there are cases such as isothermal expansion where we look at a reversible process. Furthermore, knowing the maximum amount of work that can be done by a process can tell us how efficient or inefficient the process is when we do it in reality. Hope that gives at...

Michelle Nguyen 2L

I also remember Dr Lavelle saying that isothermal reversible expansion of a gas will not be on the test; and yes, reversible expansion of a gas causes the gas to do the maximum amount of work in expanding.

Michelle Nguyen 2L

Think of a piston compressing a gas inside a container; the atmosphere is providing the external pressure whereas the pressure of the system is the pressure of the contained gas. In a reversible reaction, the external pressure and pressure of the system are equal such that the slightest change in ei...

Michelle Nguyen 2L

Yeah I'm pretty sure we only need to worry about expansion work; even the textbook adds in a note every time (so far) that we should assume no non-expansion work is done.

Michelle Nguyen 2L

Since we are being asked to calculate the amount of heat that must be added to the kettle (containing water), I would consider it as part of the system, the whole system being kettle+water

Michelle Nguyen 2L

As the textbook assumes no non-expansion work is done, and deltaU= heat(q) + work(w), when the volume is constant throughout the reaction, this means no expansion work is done and so w= 0 and deltaU= q. The definition of deltaH is heat transferred at constant pressure, and takes into account that en...

Michelle Nguyen 2L

Usually in problems where the presence of the calorimeter is taken into consideration, the problem will provide you with the heat capacity for the calorimeter, and not the specific heat capacity. This is actually easier as the equation you have to solve then becomes q= -(heat capacity of calorimeter...

Michelle Nguyen 2L

Phase changes are physical and not chemical changes. They result from forces such as hydrogen bonds and van der Waal interactions getting weaker or stronger, and not from changes in chemical bonds between atoms in the same compound. For example, for ice to melt, the water molecules must obtain enoug...

Michelle Nguyen 2L

Yup; if we’re looking at the energy of the system, compression is work being done on the system, hence a transferring of energy from the outside to the system. Work done by the system, as in expansion, instead uses energy of the system, resulting in a loss of energy in the system.

Michelle Nguyen 2L

I believe compressing a gas would be considered doing work on the gas. Since there are different ways we can do work on gases and different ways gases can do work, I don’t think work done by and work done on a gas are always equal; this is only the case for a net work done/on of 0.

Michelle Nguyen 2L

When distinguishing between a Lewis acid and a Bronsted acid, don’t look to the chemical formula necessarily but rather the chemical equation- try to see how bonds were broken and formed. If a certain element accepted electrons then you can consider it a Lewis acid. If a certain element lost an H+ t...

Michelle Nguyen 2L

You can also see that As2O3 would be amphoteric by looking at the periodic table; it’s located near the diagonal band separating the metals and the non metals and thus is likely to correspond to the category of elements that form amphoteric oxides (our textbook has a chart that shows exactly which s...

Michelle Nguyen 2L

The K constant is the ratio of products to reactants at equilibrium; it therefore does not change for the same reaction under the same conditions (e.g. temperature) no matter what the initial concentrations are. Thus, when we add product or reactant to the system, we will see the concentrations of s...

Michelle Nguyen 2L

I thought this problem was confusing though in that the solution manual seems to assume that the pressure within the flask remains constant throughout the reaction. I am predisposed to believe this cannot actually be the case since PV= nRT, hence with volume and temperature staying the same, pressur...

Michelle Nguyen 2L

you would use aqua when the water molecule is acting as a ligand bound to the central metal atom, in which case the H2O would be found within the brackets in the chemical formula. Hydrate is used when the water molecule is not part of the coordination sphere, in which case it would be listed as H2O ...

Michelle Nguyen 2L

yes, both names will be accepted in Lavelle's course!

Michelle Nguyen 2L

I'm not sure if I can explain this clearly but I think the textbook wants you to be able to differentiate between electronic arrangement, which treats lone pairs and bonding pairs the same in determining shape, and molecular structure, which is a specification of electronic arrangement that takes in...

Michelle Nguyen 2L

A molecular structure of trigonal planar shouldn't have any lone pairs. If you're referring to an electronic arrangement that is trigonal planar, then that can either have a molecular structure with three bonded atoms, making it trigonal planar, or two bonded atoms and one lone pair, making it bent,...

Michelle Nguyen 2L

For some of the structures it's just necessary to memorize the bond angles. This is actually easier than you'd think- the linear shape is a line and a line is 180 degrees, hence a linear molecule has bond angles of 180; a trigonal planar molecular shape has bond angles of 120, just like how the inte...

Michelle Nguyen 2L

Usually noble gases would not form any bonds but since xenon is such a heavy noble gas with valence electrons so far away from the nucleus and experiencing very little attraction that it can be forced to make bonds with very electronegative elements such as fluorine. As stated above, you probably do...

Michelle Nguyen 2L

To look out for exceptions to the octet rule in the form of expanded valence shells, remember that nonmetals in periods 3 and above can hold more than 8 electrons using their d orbitals; to remember this, it helps to recognize that the first d orbital is in the n=3 shell. Thus, elements in periods 1...

Michelle Nguyen 2L

All the possible resonance structures contribute to the form of the resonance hybrid; however, more stable forms (which occur where the formal charges of the bound atoms are closest to zero) contribute more to the resonance hybrid than less stable forms, and the more unstable a structure is, the les...

Michelle Nguyen 2L

Electrons usually enter the 4s orbital before the 3d because the 4s is initially lower in energy. However, from Scandium and onwards, the 3d orbital of transition metals actually becomes lower in energy than the 4s, which is why we write 3d before 4s in the configuration. (In fact it has been determ...

Michelle Nguyen 2L

(Replying to the post about 1.69) The number for the work function 2.93 eV is an experimentally calculated value that differs for different metals. The other number, 1.602x10^-19, is a conversion factor that can be found in the back of your book to convert eV to the SI unit Joules. Hope that helps!

Michelle Nguyen 2L

Dr. Lavelle has mentioned that we should know the visible light region, as stated above, and also know that infrared radiation has the next shortest wavelength before red light (infrared radiation has longer wavelength than red/visible light) and ultraviolet radiation has the next shortest wavelengt...

Michelle Nguyen 2L

No, it's not necessary to know the specific lower energy levels of those line series; don't worry about it :D

Michelle Nguyen 2L

Another way of looking at it is when frequency decreases, wavelength increases and when you draw that out, the slope of the waves look flatter. Since slope is basically the same as "extent of change" (slope is the change in y over the change in x), a flatter or I guess you could say smalle...

Michelle Nguyen 2L

Yup, Avogadro's number is really just a number- it can be used to mean 6.022 x 10^23 units of anything

Michelle Nguyen 2L

Yup! After finding the number of moles of each compound, find the number of moles of K+ in each molecule using the subscripts (eg multiply the number of moles of K2S by 2 to get the number of moles of K+ in K2S). Add them all together and you get the total moles of K+ ion in solution. Divide by lite...

Michelle Nguyen 2L

The limiting reactant for the second reaction is not necessarily the same as for the first since you have different reactants (P4O6 instead of P4 for the second reaction) and different amounts of oxygen available for each reaction (you used up some of the 5.77 g of oxygen for the first reaction alre...

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