CHEMISTRY COMMUNITY

Brian_Ho_2B

Do we have to know anything about the equation: Rate of Product formation = k2[E]o[S]/(Km + [S]) and do we need to know anything about Km = (k1' + k2)/k1 ?

Brian_Ho_2B

Are we expected to be able to solve problems like Example 7C.1 in the textbook? If so, can someone walk me through the general procedure of how to do so? There was a problem similar to it in the homework (I believe 7.21 or 7.23) and I couldn't figure out how the final answer included three different...

Brian_Ho_2B

In a reaction mechanism for a catalyzed reaction, how do we identify which intermediate is a catalyst?

Brian_Ho_2B

On the final, if we are given an overall reaction for an unknown mechanism and we are given what the intermediates are but not where they go in the mechanism, are we expected to be able to figure out how to set up each step and where to place the intermediates?

Brian_Ho_2B

The tangent line, or more importantly, its slope, is the instantaneous rate of change of the dependent variable (concentration) with respect to the independent variable (time). It's basically the concept of a derivative. The derivative is the function with respect to time that we use to calculate th...

Brian_Ho_2B

Molecularity accounts for both the number of different types of species as well as their stoichiometric coefficients. For instance, if an elementary step had something like: 2A + B --> something, the molecularity is termolecular, because in total, there are three molecules involved.

Brian_Ho_2B

Although unlikely that something like that would appear on an exam, it is in fact a learning objective that you do know how to derive the integrated rate law equations and the half life equations. If you do need help with the math part of it, as you know there are math assistance resources on Prof. ...

Brian_Ho_2B

The unique rate of change of any reaction is -(1/r)(d[R]/dt) or +(1/p)(d[P]/dt). The instantaneous rates of change of reactants and products are based on their stoichiometric coefficients, but the unique rate of change for the full reaction can be calculated using any of the reactants or products, g...

Brian_Ho_2B

Solubility rules are not covered in any of the outlines nor do I think they were required for Test 2. However, if you do want to learn about them, I believe it is section 6I of the textbook that covers it.

Brian_Ho_2B

In a multistep reaction with at least two elementary steps, the rate law is based on the slowest step of the mechanism, which determines the rate. When you do write the rate law, it must be consistent with the overall observed reaction (the overall reaction of the mechanism) and it must be consisten...

Brian_Ho_2B

Based on logarithmic properties, ln([Ao]/[A])/t is equal to ln([A]/[Ao])/-t. The negative sign applied to a logarithm takes the inverse (^-1) of whatever is in the logarithm.

Brian_Ho_2B

Time is generally written as seconds (as the standard SI unit); however, it can be represented as minutes or hours depending on the duration of the reaction. If you do happen to change time from seconds to minutes or from hours to minutes and so on, remember to change the value and units of your &qu...

Brian_Ho_2B

From what I remember from last quarter, his final is roughly the same length, if not, very slightly longer than the midterm. You will have ample time to finish everything and double check (I recommend triple check) your answers. If you work diligently, you can expect to have about an hour to spare.

Brian_Ho_2B

One thing I noticed during lectures recently is that when we derive the integrated rate laws (whether it be 0th, 1st, or 2nd order), we set the value of the stoichiometric coefficient of [A] as 1 so that -(1/a)*d[A]/dt = -d[A]/dt and therefore -d[A]/dt = k[A]^n. How would the rate laws be affected i...

Brian_Ho_2B

The species on the cathode side are all the relevant species that are in the reduction reaction. Usually when you see a reduction reaction like this one that you're unsure about, I highly recommend looking at the appendix to see how the reduction reaction with those species are written out so that y...

Brian_Ho_2B

I've been getting the same answer that you're getting, so I'm 99% sure that it is an error in the solutions manual.

Brian_Ho_2B

For this problem, you'll have to refer to Appendix 2 in the back of the textbook and look for the oxidation and reduction reaction that involves O2, H+, and OH-. Hint: If your redox reaction involves H+ and OH-, chances are that it involves water as well. Use the cell diagram to figure out which equ...

Brian_Ho_2B

According to the reading in section 6O.1, if there is more than one species that is reducible, the species with a higher potential for reduction is preferably reduced.

Brian_Ho_2B

Good catch! It appears that this is another error in the solutions manual. You are correct in the fact that a pH of 1 means that [H+] = 0.1M. Other than that, you should be able to solve the rest of the problem. Remember to use the Kw value of water to find [OH-] values to calculate your Q in the Ne...

Brian_Ho_2B

The units of "k" may vary depending on what the order of the reaction is; a half-life is always measured with a unit of time, usually seconds (or minutes/hours if it is a very slow reaction).

Brian_Ho_2B

I'm almost certain that this is an error in the textbook; I've done that same problem twice and I've gotten your answer twice. You may want to see Prof. Lavelle's updated solutions manual errors on his website to see if this is one of the errors. But I can assure you that your work is correct.

Brian_Ho_2B

"Calculate the potential of a cell constructed with two nickel electrodes. The electrolyte in one compartment is 1.0 m Ni(NO3)2(aq). In the other compartment, NaOH has been added to a Ni(NO3)2 solution until the pH 5 11.0 at 298 K. See Table 6I.1." Can someone explain how to approach this...

Brian_Ho_2B

How do chemistry students come up with chemistry jokes?

Google

Brian_Ho_2B

How do chemistry students come up with chemistry jokes?

Google

Brian_Ho_2B

I'm a bit confused on the conceptual basis of problem 6O.3. Four metal cations are placed in an electrolyzed solution of pH = 7 and the question asks us which would be likely reduced at the cathode. According to the answers in the book, the ones that are to be reduced at the cathode are the species ...

Brian_Ho_2B

"Understand why reaction free energy depends on pressure and the equilibrium constant." I understand the relationship between free energy and the equilibrium constant, but I'm not sure what pressure has to do with gibbs free energy. Is it referring to changing pressure by changing volume a...

Brian_Ho_2B

The book says that in an electrolytic cell, the substance with higher reduction potential (if there's more than one reducable substance) is preferrably reduced. If that is the case, how come the answer to 6O.3 part c is that Ni2+, which has a red. potential of -0.23V preferrably reduced over water, ...

Brian_Ho_2B

Let's say we have a typical cell diagram of a galvanic cell with the format x(s) | x'(aq) || y'(aq)|y(s), and because it is a galvanic cell, we find that E cell not is a positive value which we will call "+V". If the cell diagram was reversed as y|y'|x'|x, would the value of E cell not sti...

Brian_Ho_2B

I cannot wrap my head around this one. It asks us to calculate E (M2+/M) for the cell diagram M|M2+||Cu2+|Cu and we are given that Cu|Cu2+ is +0.34 and that the E cell is -0.689. However, calculating E(M/M2+) using E cell = E cathode - E anode gives me positive 1.029. The answer is -0.349. This answ...

Brian_Ho_2B

I am confused on this as well, when do we need to introduce Pt into the system? We introduce Platinum as an electrode to the side(s) that have their reducing or oxidizing agents in aqueous phases. The platinum acts as a conductor of electrons for the aqueous agents. However, this question doesn't e...

Brian_Ho_2B

The method that I use for a more a more advanced or complicated redox reaction: What you do is you start off with the completely balanced redox reaction, and then what you essentially do is work backwards one step in the balancing process. You take the full redox reaction and split it up into the tw...

Brian_Ho_2B

Yes, and it covers the second page of thermodynamics and the entirety of electrochemistry. We will be starting kinetics this week but that won't be on the second test.

Brian_Ho_2B

Cell diagrams are giving me a bit of a hard time. I was doing this problem that asked me to write the diagram for a redox reaction with Au+ (aq) --> Au (s) + Au3+ (aq); note that this is unbalanced. I figured that since Au+ is reduced to Au(s) and Au+ is oxidized to Au3+, that the cell diagram shoul...

Brian_Ho_2B

Every time I open the course outlines for the two topics that are being covered in test 2, the pdf page opens up but nothing really shows up. Is anyone else having this problem? If you don't have this problem, could you please post the images of the learning outcomes and the homework problems?

Brian_Ho_2B

When comparing temperatures, deltaS and deltaH are assumed to stay constant for both temperatures which could be why deltaS/R seems to be ignored. Why can you assume that delta S is constant when you're using the van't Hoff equation to compare temperatures? We can assume that the delta S (not) is c...

Brian_Ho_2B

Two ways that we know that O3 is being reduced is we know that in a redox reaction, one thing must be reduced while the other must be oxidized. We know that the other element in this problem (I don't remember what it was but I remember it was oxidized), was oxidized, so the other substance O3 must h...

Brian_Ho_2B

In order to add up the reduction and oxidation half reactions to get the balanced reaction, we need to multiply the bottom half reaction by 2 so that the electrons cancel out. Now both sides have 2 electrons, indication that 2 mol of electrons are being transferred per mole of reaction. That is why ...

Brian_Ho_2B

Substances with a higher, positive standard reduction potentials are not able to reduce H+ to H2, while substances with lower, negative reduction potentials can reduce H+ to H2.

Brian_Ho_2B

The nernst equation wouldn't be the best equation to determine if a reaction, especially a redox reaction, is at equilibrium. If you're looking at a general equation, I'd recommend using the equations discussed in the second page of thermodynamics: delta G = delta G not + RTln(Q) at equilibrium, del...

Brian_Ho_2B

Oxidation occurs in the anode and reduction occurs in the cathode because by definition, the cathode is what attracts electrons because of its positive charge.

Brian_Ho_2B

Does the system have to be at equilibrium in order to use this equation? The system does not have to be at equilibrium or standard conditions to use this equation. The purpose of the "+ RTlnQ" is to account for differences in temperature and concentrations of reactants and products (not a...

Brian_Ho_2B

In lecutre, Professor Lavelle said, when we use the Van't Hoff equation to calculate K at two different temperatures, the ∆H and ∆S don't have any subscripts because enthalpy and entropy aren't temperature dependent. How do we know they aren't temperature dependent? What is the reasoning behind tha...

Brian_Ho_2B

I probably speak for most people when I say that the midterm was difficult. There were a lot of really tough problem-solving heavy questions that I struggled with even though I felt prepared. During the midterm, it was really hard for me to keep focused and calm when I saw questions I felt like I c...

Brian_Ho_2B

In discussion, we talked about how electrons flow from the anode beaker to the cathode beaker due to the spontaneity of one metal being reduced and the other being oxidized. We briefly discussed how a salt tunnel can connect the beakers so that the charged imbalance (after a long time of redox react...

Brian_Ho_2B

Why is internal energy equal to 0 in an isothermal reaction? To add on to what others have said, in an isothermal system such as the isothermal, reversible expansion, delta U equals zero, which may mean that either the system is isolated (eg a very insulated beaker) OR the work is equal to negative...

Brian_Ho_2B

Does anyone know when we will get our midterms back?

Brian_Ho_2B

Hi, I'm having trouble understanding the difference between the ∆G and ∆Gº terms in the equation ∆G = ∆Gº + RT lnQ. Is ∆Gº the standard Gibb's free energy difference between the reactants and products, and ∆G the Gibb's free energy difference between the initial state and equilibrium (since we deri...

Brian_Ho_2B

Can something be both an oxidizing and reducing agent? If so, how is that possible and would we make two separate half reactions? It ultimately does depend on what it’s reacting with. If it is in a galvanic cell with a metal with a higher reduction potential, then the higher potential becomes the o...

Brian_Ho_2B

Yes, fusion is melting.

Brian_Ho_2B

Carbon tries to befriend his two new roommates Nitrogen and Oxygen.
Unfortunately they didn't bond well together.

Brian_Ho_2B

I know that the units of entropy are joules (or kJ) over degrees Celcius or Kelvin. Is using kJ/C interchangeable with kJ/K? I believe that's what Prof. Lavelle did when we were doing practice problems in lecture today, but I wanted to confirm whether or not those two units for entropy are interchan...

Brian_Ho_2B

Are we expected to include the change in entropy due to change in pressure when we are adding up the entropies to find the total change in entropy? I was doing some practice problems and there was a change in volume, temperature, and pressure (I found that out by using PV = nRT and saw that the valu...

Brian_Ho_2B

In the isothermal, reversible expansion, delta S(surr) is zero because there is no net heat being transferred to/from the system from/to the surroundings, hence the name "isothermal". Because delta S = q/T and "q" is zero, it's reasonable that there's no change in entropy of the ...

Brian_Ho_2B

The equation for change in entropy in general is q/T or if under constant pressure, deltaH/T. At lower values of T, values of heat or enthalpy will have a greater impact on entropy because T is in the denominator in the equation for delta S. That is why adding heat to a system of high temperature wo...

Brian_Ho_2B

The change in moles of gas due to the reaction is a surplus of two more gas moles IF there are four moles of CO2 created. However, we are given that there is only one mole of CO2. Using quick stoichiometry, 1 mol CO2 * (+2 mol gas/4 mol CO2) = +1/2 mol gas.

Brian_Ho_2B

Molecules/atoms with higher mass have more entropy because they have more subatomic particles, which can occupy more microstates within the molecule/atom's electron clouds and nuclei. There are definitely more formal and correct explanations as to why molecules that are larger and more complex have ...

Brian_Ho_2B

For the purposes of this class, expansions that are isothermal will most likely always be reversible, since we only learned two equations for work for the two possible expansions we learned about (isothermal, reversible vs. irreversible against constant pressure). As for your second question, an iso...

Brian_Ho_2B

Microstates are all the possible configurations of atoms/molecules in a system. The value of W in the equation S = k*ln(W) is equal to degeneracy, which is the number of all microstates. W = a^n, where a = possible positions/states and n = number of particles/molecules that can occupy the positions....

Brian_Ho_2B

-w = q for isothermal reversible expansions because temperature is constant. This is because when we apply heat to such a system, all of that heat is converted into the work used for the expansion. This relation also allows us to derive the second law of thermodynamics: delta S = q/T, where temperat...

Brian_Ho_2B

Part b asks for the heat absorbed due to the oxidation. Heat absorbed is endothermic and is generally positive values, because heat is gained.

Brian_Ho_2B

That equation is the internal energy of an ideal, monoatomic gas. Work done by an isothermal reversible expansion is w = -nRTln(V2/V1).

Brian_Ho_2B

Entropy can be expressed as both a statistical property and a thermodynamic property. The statistical definition of entropy is S = k*ln(W), where W = degeneracy = a^n, where a = number of states in a system and b = number of molecules/particles. The thermodynamic definition of entropy is delta S = q...

Brian_Ho_2B

For constant pressure, qp = enthalpy, and qp doesn't necessarily equal delta U. For constant volume, qv = delta U = delta H IF there is no nonexpansion work involved. For isothermal, reversible reactions where delta T is zero, q = -w, delta S = q/T, and
w = -nRTln(V2/V1).

Brian_Ho_2B

Standard enthalpy of formation is the amount of energy required or released when forming one mole of a molecule. Standard reaction enthalpy is the net amount of energy used or released per mole of reactions occurring. Both have the unit kJ/mole. You use the sum of all the enthalpies of formation of ...

Brian_Ho_2B

W = -PdeltaV is another way to calculate work using the format of "force times a distance", except we are using pressure and volume. That is why we have the conversion factor of 101.325 Joules per liter per atm so that we convert the units of -PdeltaV into the same unit that we get from w=...

Brian_Ho_2B

The integral is the general process of calculating the work done over a distance. The general equation for work is the integral from point A to B of the force function with respect to the distance that the force is applied upon. The -PdeltaV formula is the same idea; but we used the integral of pres...

Brian_Ho_2B

A reaction where it starts off with more energy and releases energy is exergonic, but there's no way to determine the sign of delta H without further information. We need to know how much energy was added as heat, how much work was done on/by the system, and the conditions it is under (eg. constant ...

Brian_Ho_2B

HCl oxidizes zinc to produce hydrogen gas and chloride ions. A piece of zine of 8.5g is placed in a vat of 0.800L and 0.500M HCl. If the initial temperature is 25 degrees C, what is the final temperature? Assume the heat capacity and density of the HCl solution is the same as those of water. I'm not...

Brian_Ho_2B

You'll want to use the standard enthalpies of formation and the coefficients of the balanced equation to find the enthalpy of the reaction. \Delta H = 2(-395.72kJ\cdot mol^-1)-2(-296.83kJ\cdot mol^-1)=-197.78kJ\cdot mol^-1 Since the question says that 0.030mol of SO2 are consumed, y...

Brian_Ho_2B

I'm looking at problem 4.17, where in part e, it asks for the change in enthalpy. The information given is that we have 2SO2 + O2 produces 2SO3 (all gasses) at 25 degrees C and 1.00 atm. I know that the work done is positive 40J, but I have no idea how to find change in enthalpy for this problem. Ca...

Brian_Ho_2B

There is a chart in the book that lists the standard reaction enthalpies of different combustion reactions. On a side note, how did you do part A of 4.7? I've been stuck on why the answer is 3.72 kJ

Brian_Ho_2B

Test 2 covers material only between the midterm and up to test 2.

Brian_Ho_2B

We found that the mass of water was 0.100g based on the fact that 0.050M HCl reacted with 0.050M NaOH in 1.0L. However, that neutralization reaction would've produced 0.050 mol H20 and 0.050mol NaCl. Why did we make the bold approximation of having water equal to 0.100 L when based on the stoichiome...

Brian_Ho_2B

Steam causes severe burns at 100C vs water at 100C because not only was there a lot of heat to supply H2O to that temperature, but there was a LOT more heat supplied to get it from a liquid to a gas phase. That combined amount of heat is much more heat that just water at 100C. When steam touches ski...

Brian_Ho_2B

Why doesn't the calorimetry formula q=-C∆T include mass like q=mC∆T. And why is the q value negative if the calorimeter is gaining the energy lost by the reaction? The equation is q=-C∆T because it is the heat capacity of a certain change in temperature per mole or per gram of mass. The q value is ...

Brian_Ho_2B

q is the transfer of heat (thermal energy specifically) and delta H is change in enthalpy, which is equal to change in the system's internal energy added to internal pressure times change in volume. The full equation for delta H = w + q + P*delta Volume. If pressure is the same on the outside and in...

Brian_Ho_2B

Reversible expansions allow the gas to return to its original volume naturally on its own. Irreversible expansions change to a certain volume and stay there. There are different equations for the work done by expansion for each of those cases. For reversible expansions, w = -nRT*ln(V2/V1) and for ir...

Brian_Ho_2B

When you are calculating the difference of energy used to raise the temperature, why does the expansion of the volume change the amount of heat change? Calculating the difference in energy used to raise temperature depends on which variables are constant. If we are talking about a gas, then a gas's...

Brian_Ho_2B

You want to use either value depending on the information present. The problem would typically provide you with one or the other, and you may have to convert a value in moles to grams or the other way around. These are two different values; specific heat is joules per gram per degree Kelvin and Mola...

Brian_Ho_2B

It states in the book that breaking bonds is always endothermic and requires energy. However in biology, breaking a phosphate group from ATP to form ADP releases energy. Can someone help me understand this?

Brian_Ho_2B

The standard enthalpies of formation of diatomic gases like O2 are zero because these are very stable gases that require very little energy or input to form. They naturally form and don't require or release much enthalpy. This is why when you calculate a reaction's standard enthalpy using the standa...

Brian_Ho_2B

The heat lost by the copper is heat gained by the water, meaning that the equation should be q(copper) = -q(water). I made a similar mistake forgetting that negative sign. Also remember to set the temperature in terms of Kelvin.

Brian_Ho_2B

This unit includes how to derive an equation for the work of an isothermal, reversible expansion and it includes integration. Should we be reviewing our calculus skills and techniques for these units and future chem 14B units? Or is the derivation trivial and we just need to understand the derived f...

Brian_Ho_2B

The general rule is that the stronger an acid/base is, the weaker its conjugate. The weaker an acid/base is, the stronger its conjugate. Group 1/2 cations (with the exception to Magnesium, but that's a topic for a more advanced chemistry class) do not affect pH because they are mostly stable and tho...

Brian_Ho_2B

Heating curves represent the temperature of a system as a function of the amount of thermal energy/heat added. Heating curves are used to figure out the values of the melting/fusion point and vaporization/condensation points, which are when the heating curves have a slope of zero. When the curves ha...

Brian_Ho_2B

Another thing that you would want to be aware about is that just because you see gases in a reaction, it doesn't necessarily mean you should use partial pressures and Kp. Gases and gas reaction equilibrium constants can use either concentrations or partial pressures.

Brian_Ho_2B

As far as we know, temperature is the only relevant factor that can change K. Pressure cannot change K, whether you change the pressure by adding more gases like He2 or by changing the volume. At most, changing the pressure can throw the reaction off equilibrium, but the K value is the same.

Brian_Ho_2B

A system's internal energy is increased by 982 J when 492 J of heat was supplied. Because change in energy is equal to the sum of work and heat, it makes sense that work must have been involved for the the change in energy to be that high. Using that equation, it makes sense that 982 J = 492 J + w, ...

Brian_Ho_2B

The full equation for gases is similar, but much more complete. (P + an^2/V^2)(V - nb) = nRT, where a = average attraction between particles and b = volume of one mole of particles. The values "a" and "b" are typically very small, and that's why we typically use PV = nRT. In a la...

Brian_Ho_2B

You can find the percent protonation of a basic solution by calculating [HB+]/[B initial] times 100%. The [HB+] is the concentration of the base's conjugate acid.

Brian_Ho_2B

The equilibrium constant tells us the relative stability of the reactants and the products. If K is more than one or exceeds one by a lot (K>1 or K>>1), then the reaction favors products, implying that products are more stable. If K is less than one or is very very small (K<10^-3), that means the re...

Brian_Ho_2B

There are other examples of molecules beside water than can do autoprotolysis due to them being amphiprotic. Examples include ammonia, acetic acid, and H2SO4. These aren't really significant; the autoprotolysis of water is the important one you have to know for acid/base equilibria.

Brian_Ho_2B

A high percent deprotonation means that a weak acid was able to produce a sizeable amount of hydronium ions compared to other weak acids. A high percent protonation means that a weak base was able to produce a sizeable amount of hydroxide ions compared to other weak bases. Reminder: Percent Deproton...

Brian_Ho_2B

It really doesn't matter which value: n, R, or T, that you divide by the volume. Communitive rule of multiplication and division. Just make sure you divide by the volume only once, don't divide each value n, R, and T by volume and then multiply those together; that would be nRT/(v^3). You want P = n...

Brian_Ho_2B

How does increasing or decreasing the partial pressure of a reactant affect the partial pressure of the product, and vice versa for a change in product and its effect on the reactant? Think of equilibrium as an unbalanced see-saw where one side is supposed to have a different "weight" tha...

Brian_Ho_2B

Question 5J.5b in the homework gives the following reaction: H20(aq) + C(s) = H2(g) + CO(g). Will the reactants or products be favored by an increase in total pressure due to compression, and why? Writing the value for K gives us K= P_CO*P_H2/P_H2O. This means partial pressure, I'm not sure of any ...

Brian_Ho_2B

We omit the units because in actuality we are supposed to be calculating equilibrium constants and reaction quotients using the activities of each reactant or product of a reaction. Activities are unitless, only scalar values, so it makes sense that calculating a value K with unitless values produce...

Brian_Ho_2B

It is called partial pressure instead of just pressure because there are several gases in a given volume at once. Each gas has a certain amount of moles that occupies part of the total pressure. The sum of all the partial pressures of the gases present equals the total pressure. If only one gas was ...

Brian_Ho_2B

What did Dr. Lavelle mean in lecture 1 today when he said that 10^-12 is actually and would be correct if it were written as 10^-7 for the concentration of H3O+? Correct me if I misinterpreted your question; this is my best explanation for what you're asking. If Hydronium concentrations were 10^-12...

Brian_Ho_2B

Is it possible for a pH level to go under 0 or above 14? Is it possible for hydronium concentrations to be above 1M (for pH less than 0) or hydroxide concentrations more than 1M (for pH greater than 14)?

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