CHEMISTRY COMMUNITY

RyanKopeikin_2I

To my understanding you just add the orders of the reactants in the rate law together to get the overall rxn order..

RyanKopeikin_2I

To be honest I just look at the right side of the equation and figure out how to make it match the left side of the equation, which is usually in units of M/s. So if it's a second order rxn, you have M/s = k (M)(M) so you can kinda figure out that the units of k need to be M^-1•s^-1

RyanKopeikin_2I

In order to determine the rate law from concentration graphs, just examine the graphs and look for the most linear plot. The most linear plot is most indicative that the reaction is of that order.

RyanKopeikin_2I

Water would only be included in the rate law equation if it was part of the rate-determining (slowest) step, and an organic solvent is being used (So water is not the solvent). As long as water IS the solvent, it won't be included in the rate law.

RyanKopeikin_2I

The rate constant is affected by temperature and activation energy. Decreasing the activation energy by using a catalyst would increase the rate constant. Increasing temperature would also increase the rate constant.

RyanKopeikin_2I

Intermediates are produced in one step and consumed in another, and they do not show up in the overall reaction.

RyanKopeikin_2I

You have to properly write out the redox half-reactions and combine them to get the overall balanced redox reaction. Then you plug in values for Q like you would a normal chemical equilibrium question.

RyanKopeikin_2I

I usually do this: example 5mmol = 5x10^-3mole

RyanKopeikin_2I

USUALLY the anode is on the left side of the diagram and the cathode is on the right side, but do not automatically 100% assume this, as it is actually not a given. If you are given standard reduction potentials, always check and compare them to make sure that the left side really does represent the...

RyanKopeikin_2I

The least positive standard reduction potential is usually the anode.

RyanKopeikin_2I

Thank you so much for this explanation, I have also been struggling with this concept.

RyanKopeikin_2I

I haven't looked at this problem specifically, but usually it depends on what other elements are involved in the reaction. For example, the reaction of iron with oxygen (Fe and O2) is going to influence which Fe standard reduction equation you use.

RyanKopeikin_2I

Balance the oxidation and reduction half-reactions, and the coefficient of e- is the value of "n" (moles of electrons).

RyanKopeikin_2I

Anode goes on the left, cathode goes on the right.

RyanKopeikin_2I

A salt bridge prevents the buildup of charges in the two half-cells, which would stop electron flow.

RyanKopeikin_2I

A salt bridge prevents the buildup of charge in the two half-cells so that the flow of electrons doesn't completely stop, and so the salt bridge completes the electric circuit. The cathode starts to become more negative as it receives electrons, so cations from the salt bridge flow to the cathode to...

RyanKopeikin_2I

I like to think of it this way:

strong reducing agent - <---------------- 0 --------------> + strong oxidizing agent

RyanKopeikin_2I

Typically when looking at a cell diagram, the anode (oxidation) will be on the left and the cathode (reduction) will be on the right. The cathode will have a higher reduction potential as well.

RyanKopeikin_2I

There are 2 reactions as a part of this problem, but in general don't include H20 or electrons in cell diagrams! Also do not include stoichiometric coefficients in cell diagrams.

RyanKopeikin_2I

You only use Pt(s) when an inert metal is not provided for you to use as an electrode. I'm assuming the textbook sometimes has Pt(s) on one side of the cell diagram only because that half-reaction does not have a metal conductor given (ex: reduction of hydrogen: 2H+ + 2e- --> H2(g))

RyanKopeikin_2I

I was taught OIL RIG (oxidation is loss, reduction is gain) and I remember that the agents are backwards. The oxidizing agent oxidizes the other species, and the reducing agent reduces the other species.

RyanKopeikin_2I

The purpose of calibrating the calorimeter is to obtain the heat capacity of the calorimeter, which you can later use to find qrxn.

RyanKopeikin_2I

From what I understand, if a gas is being compressed by a piston, work is being done ON THE GAS, and therefore work is positive. If a gas is expanding in a piston, then work is negative, because work is being done BY THE GAS.

RyanKopeikin_2I

There are no units for degeneracy. You can see this because in the equation S = KblnW, the units of S are J.K^-1, which comes from the Boltzmann Constant

RyanKopeikin_2I

I believe you can interchange qrev and deltaH when the system is at constant pressure. Because at constant pressure, deltaH = q, we can interchange these two values in the expression deltaS = qrev/T

RyanKopeikin_2I

For this formula you need to use Kelvin.

RyanKopeikin_2I

It's definitely J.K^-1 by default, but if you are using the equation delta G = delta H - T*delta S, you will most likely need to convert the units of delta S to kJ.mol^-1 so you can subtract it from the change in enthalpy.

RyanKopeikin_2I

Positive change in entropy means that there is an increase in disorder, meaning that it is possible for the substance to occupy more micro states than previously. Negative entropy means the reverse, that there is a decrease in disorder, and the substance will occupy less micro states because it is m...

RyanKopeikin_2I

Cp is used for conditions of constant pressure, and Cv is used for conditions of constant volume.

RyanKopeikin_2I

Cp is used for conditions of constant pressure, and Cv is used for conditions of constant volume.

RyanKopeikin_2I

It really just depends on the units used in the problem. Usually, when calculating delta G or delta S, you will use 8.314 J.K^-1.mol^-1, and when using the ideal gas law, or any other time when pressure is given, use 8.206x10^-2 L.atm.K^-1.mol^-1

RyanKopeikin_2I

I would assume K stands for the equilibrium constant, because k (or kB) is Boltzmann's Constant, and if it is temperature in Kelvin the "K" will be accompanied by a number, like in 300K.

RyanKopeikin_2I

isothermal reversible expansions have the property q=-w

RyanKopeikin_2I

This is a very helpful summary, thank you for putting this together!

RyanKopeikin_2I

Thank you for posting these steps! Very helpful.

RyanKopeikin_2I

I think the textbook defines internal energy as "the average energy of the molecules in the system." So delta U would be the change in the average energy of the molecules in the system.

RyanKopeikin_2I

Summing up infinitesimal changes in q results in infinitesimal changes in S, if you are wanting to know the relationship between q and S in this equation.

RyanKopeikin_2I

It really just depends on the units you are looking for in the answer, and the units of the variables you are given. It seems like in thermodynamics, we use R = 8.314 J·K-1·mol-1 often, but obviously if you are given pressure in atm you are going to need the R value that has atm in it.

RyanKopeikin_2I

Work is positive when there is work being done on the system, and work is negative when work is being done by the system (on the surroundings).

RyanKopeikin_2I

What I like to do is I read the problem, and any formulas that float to my mind relating to the problem, I write down on paper. Often your brain subconsciously makes connections, and seeing a couple formulas on your page might help you discern which one is the appropriate one to use.

RyanKopeikin_2I

I think it's fair to assume temp is 298K because this is STP, but I think for midterms/final Lavelle will probably specify the temperature.

RyanKopeikin_2I

download it from CCLE but make sure to test it before every midterm, I know respondus has refused to open on my laptop because I had some settings turned on that the program wasn't compatible with. I had to edit my computer's settings in order to get respondus to work properly.

RyanKopeikin_2I

Phase change is when something transforms into another form. For example, a phase change occurs when liquid water becomes water vapor (gas).

RyanKopeikin_2I

Exothermic reactions will always have negative delta H because heat is being released, which is denoted with a negative sign. Endothermic reactions will always have positive delta H because heat is being absorbed, which is denoted with a plus sign.

RyanKopeikin_2I

Just to add onto what has been said, during a phase change, the heating curve experiences a flat section because the added energy is being used to break bonds, so temperature does not change.

RyanKopeikin_2I

When water vapor condenses on a surface, bonds are re-forming between water molecules, and bond formation is an exothermic process (releases heat).

RyanKopeikin_2I

Boiling water breaks the hydrogen bonds in water, creating water vapor.

RyanKopeikin_2I

I would say if K<10^-3 then you can assume X is insignificant, and if K<10^-4 then you can DEFINITELY assume X is insignificant.

RyanKopeikin_2I

For part C: Calculate the amount of CS2 produced when 399 kJ of heat is absorbed.

399kJ x 4 mol CS2/358.8kJ x (molecular weight of CS2)/1mol CS2

RyanKopeikin_2I

I definitely don't think you will need to memorize these, however if you wan't to get familiar with ones you see a lot (like a C-C bond) that wouldn't be a bad thing.

RyanKopeikin_2I

Exothermic reactions have weaker reactant bonds and stronger product bonds. On the other hand, endothermic reactions have stronger reactant bonds and weaker product bonds.

RyanKopeikin_2I

Endothermic reactions have stronger reactant bonds (large energy requirement) and weaker product bonds (small energy release).
Exothermic reactions have weaker reactant bonds (small energy requirement) and stronger product bonds (large energy release).

RyanKopeikin_2I

We will definitely be given these values, don't worry too much! :)

RyanKopeikin_2I

I was told by my TA that we are only allowed to have a periodic table & the equation sheet in front of us when taking the midterm. Are we also allowed to have some paper for scratch work? I can't imagine doing ICE tables without writing them out.

RyanKopeikin_2I

When the textbook asks you to calculate the reaction enthalpy, half the time the solution manual lists the correct answer as ΔH = 35.5 kJ and half the time it lists it as ΔH=35.5 kJ.mol^-1 . I was under the impression that the latter was the correct units, but now I'm just confused. Anyone know if t...

RyanKopeikin_2I

This is a great demonstration of the Conjugate Seesaw:
> The stronger the acid, the weaker it's conjugate base.
>The stronger the base, the weaker its conjugate acid.

So if you have a weak acid, you must have a strong conjugate base!

RyanKopeikin_2I

K can help you determine a lot of things about a reaction! for example:
1) Ratio of products to reactants present at equilibrium
2) How soluble the reactants are
3) the strength of an acid or base

RyanKopeikin_2I

Lavelle has gone back and forth between saying k<10^-4 and k<10^-3 so honestly I would say as long as it's less than 10^-3 you can assume the value of X is negligible, and if it's less than 10^-4 then It's VERY reasonable to make the assumption.

RyanKopeikin_2I

pKa is the NEGATIVE log of Ka : -log(Ka) = pKa . To convert from pKa to Ka, you would do: Ka = 10^-pKa. Just a little trick in general, the presence of that "p" (as in pH, pOH, pKa, etc) means taking the -log.

RyanKopeikin_2I

As someone who had Dr. Scerri for chem 14A, I'm not really sure what to expect from Dr. Lavelle's midterms. Are there any tips for studying you guys would recommend? How to best prepare? Thanks so much :)

RyanKopeikin_2I

I would strongly recommend putting concentrations in the ICE box, even though I know Sappling often uses moles. Using concentrations reduces the amount of steps you will need to do later, and decreases the likelihood you will make an error like forget to convert moles to molarity.

RyanKopeikin_2I

To my understanding, the log sig fig rule is: # of sig figs in molar concentration = # of DECIMAL PLACES in pH value for example: pH = -log( 1.8 x10^-5) = -(-4.74) = 4. 74 However, I just started doing the textbook problems, and it doesn't seem like the textbook uses this sig fig rule at all! for ex...

RyanKopeikin_2I

Kp = equilibrium constant using the equilibrium partial pressures for the reactants & products.
Kc = equilibrium constant using the equilibrium molar concentrations for the reactants & products.

RyanKopeikin_2I

When you add an inert gas to a closed system, you aren't decreasing the concentrations of the gasses already in the system, you are just adding that inert gas in as well. Because of this, K is not affected and therefore equilibrium remains unchanged.

RyanKopeikin_2I

For some of the textbook questions, when you are required to look at Table 5G.2 to get the value of K, you need to make sure you are looking at the row with the correct temperature that was given in the problem, or you will end up with the wrong K.

RyanKopeikin_2I

I just want to clarify something while I caught it --- when K<10^-3, you do NOT ASSUME THE VALUE OF X=0!! This is a mistake that Lavelle pointed out specifically in lecture. Instead, the value of X is considered to be NEGLIGIBLE, so it won't change the concentrations of the reactants a significant a...

RyanKopeikin_2I

I agree with what has been said above, be careful to watch out for when the question asks how adding solids/liquids affects equilibrium! The answer is it won't shift the reaction!

RyanKopeikin_2I

During lecture today, Dr. Lavelle said that if K<10^-3 , we can assume X is very small and it's value is negligible. However, in one of the audio-visual lectures, he said very clearly that we want to use K<10^-4 because this was a better value to use....so I just want to confirm FOR SURE which is th...

RyanKopeikin_2I

Here's how I worked through this problem: 15. The photosynthesis reaction, 6 CO2(g) + 6 H2O(l) ⇌ C6H12O6(aq) + 6 O2(g), is endothermic. What effect will the following changes have on the equilibrium composition. a) Water is added. According to Le Chatelier's Principle, adding H20 (reactant) will cau...

RyanKopeikin_2I

Here's how I worked through this question: The hydrolysis of ATP: ATP (aq) + H2O (l) ⇌ ADP + PO42-(aq) delta H° = -30 kJ.mol-1 The delta H for this chemical reaction is negative (-30 kJ.mol-1). This means energy is being released, and the reaction is exothermic. Heating an exothermic reaction result...

RyanKopeikin_2I

Here's how I solved this problem: a) Photosynthesis: 6 CO2 (g) + 6 H2O (l) ⇌ C6H12O6(s) + 6 O2 (g) delta H° = +2802 kJ.mol-1 For this rxn notice how delta H°is positive (+2802 kJ.mol-1). Positive delta H means there is an input of energy, so this rxn is endothermic. For endothermic reactions, heatin...

RyanKopeikin_2I

Water would not be included in the equilibrium constant (K) if it is in solid or liquid form. Liquids (solvents) and solids are so abundant in a chemical reaction that their molar concentrations do not change significantly during the reaction. Hopefully this reasoning helps a bit :)

RyanKopeikin_2I

While Lavelle said in one of the modules that for the purposes of this class, 1 bar=1 atm, I personally haven't come across any practice problems yet where you need to convert from bar to atm. I think you will be given one or the other, so don't worry too much :)

RyanKopeikin_2I

When solid NH4HS and 0.400 mol NH3(g) were placed in a vessel of volume 2.0L at 24 degrees C, the equilibrium NH4HS(s) -->NH3(g) + H2S(g), for which Kc= 1.6 x 10-4, was reached. What are the equilibrium concentrations of NH3 and H2S? My issue with this question is that the solution manual gets x= 8 ...

RyanKopeikin_2I

E15 says the molar mass of the metal hydroxide M(OH2) is 74.10 g/mol. What is the molar mass of the sulfide of this metal?

I have no idea how to solve this problem, could anyone explain it to me please?

RyanKopeikin_2I

My guess would be grams but you really need the entire context of the problem to know for sure.

RyanKopeikin_2I

For sig figs in answers, you always use the LEAST number of sig figs. For example if you are working with 1305, 1.2, and .033333, you would use 2 sig figs for the answer because 1.2 only has 2 sig figs.

RyanKopeikin_2I

I'm pretty sure most of the time when you balance chemical equations, you will not need to use fractions as stoichiometric coefficients. So I'd say for the most part don't worry about it, just make sure the coefficients are ALWAYS whole numbers. You may not run into many problems with fraction coeff...

RyanKopeikin_2I

So wait if you are given units in pm, should you always convert to m? Also how many pm's are in 1 m? Not even sure what a pm is to be honest.

RyanKopeikin_2I

If the number is 5 or greater, you would round up. So in the example you gave, if you only wanted 2 sig figs, 1.65 would then become 1.70, and if the number was 1.35 it would become 1.40 and so on.

RyanKopeikin_2I

Basically MgSO4 X 7H20 is the chemical equation for magnesium sulfate heptahydrate. As you can see, "hepta" is the prefix for 7, and hydrate means H20. So in order to calculate the molar mass of magnesium sulfate heptahydrate, you need to write out the full chemical equation so you can mak...

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