CHEMISTRY COMMUNITY

Aarushi Verma 2I

Thank you so much Dr. Lavelle!

Aarushi Verma 2I

LeanneBagood_2F wrote:in terms of equations/laws, which ones can/cannot have catalysts and/or intermediates in them again?
i saw someone mention that intermediates can't be in rate laws but is there anything else?

I believe you cannot have intermediates in the rate law, but you can have catalysts.

Aarushi Verma 2I

Yes the anode will always be oxidized and the cathode will always be reduced.

Aarushi Verma 2I

It can be either kj/mol or J/mol, but you should be careful when using the activation energy in the Arrhenius Equation because the R constant is in J/mol K so you should convert your activation energy from kj/mol to J/mol by multiplying by 1000.

Aarushi Verma 2I

In Matthew's Workshop yesterday, he used the pre-equilibrium approach and for the overall rate, we just left the answer as rate = k[a]... where k = (k1 * k2)/k'1. He also posted his answers for the worksheet if you want to see it under Matthew's Workshop Week 10. This is the only way I've seen this ...

Aarushi Verma 2I

Intermediates are species that are in the mechanism, but not in the overall rate law. The way to distinguish intermediates and catalysts is that intermediates are always produced by a reaction, and then consumed immediately after in the next reaction. On the other hand, catalysts are consumed first,...

Aarushi Verma 2I

When you use the table to find the order, you essentially have to divide two experiments where one reactant is constant and the other is changing. You can look at the table and formulate the equation in your head to get it without having to write down everything. I find this helpful when the factor ...

Aarushi Verma 2I

K is affected by a change in temperature!

Aarushi Verma 2I

You can use experimental data or graphs to figure out the order. To the order of the reaction using the experimental data, you should find two trials where all of the reactants except for one is being held constant and then divide those two to find the order. You should do this for all of the other ...

Aarushi Verma 2I

You can change the rate constant by changing the temperature the reaction takes place in or by changing the activation energy ( you can add or remove a catalyst to do this).

Aarushi Verma 2I

The standard cell potential will be zero because in a concentration cell, the same reaction is occurring on both sides, there's just a concentration difference between the anode and cathode and that's what drives the electron flow. When calculating E naught of the cell, if you subtract the E anode f...

Aarushi Verma 2I

The states of matter are important and Redox reactions and you should include them in your answer for sapling!

Aarushi Verma 2I

In this equation, the n value will the number of electrons transferred between the reactions. The best way to find this out is to separate the oxidation and reduction half reactions and then after balancing them, look at the number of electrons that are transferred to get the n value.

Aarushi Verma 2I

For acidic solutions, you want to use H+ and H2O to balance out the equations, but for basic solutions, you want to use OH- and H20 instead of H+ because you don't want to have H+ in a basic solution.

Aarushi Verma 2I

Once you balance your oxidation and reduction half reactions, n is going to the number of electrons that transferred in the equation. Since its balanced, it should be the same in both reactions.

Aarushi Verma 2I

If the E naught value is positive, then the reaction is spontaneous, but if the E naught value is negative, then the reaction is not spontaneous.

Aarushi Verma 2I

If the E naught value is positive, then the reaction is spontaneous, but if the E naught value is negative, then the reaction is not spontaneous.

Aarushi Verma 2I

STP is at 1 atm and 273.15 K or 0 C.

Aarushi Verma 2I

A thermos or any insulated sealed container would be as close to a real isolated system because no energy can be transferred from the system to the surroundings and no matter can be exchanged either.

Aarushi Verma 2I

W = (number of equal energy states) ^ (the number molecules). You would then plug this W value into the S = Kb ln W equation to get positional/residual entropy.

Aarushi Verma 2I

Delta H = q when there is constant pressure. In problems, they might say isobaric too, but that also means constant pressure.

Aarushi Verma 2I

After calculating work with the P*deltaV equation, you should multiply your answer by 101.325 J/(L atm) because this will convert the value into J. This is super important because q usually ends up in J or kj and you want to be able to add/subtract them for delta U.

Aarushi Verma 2I

Yes I believe we are talking about the entropy of the world because that would technically be delta S total. We know that a process is spontaneous only if delta S total is positive, so here the delta S total is negative which makes it not spontaneous.

Aarushi Verma 2I

Keeping track of the units, like everyone said, is super helpful! That's saved me so many times. You should use the 0.08206 L atm/ K mol for PV=nRT because the units of the constant will cancel out and then you want to use the 8.3145 J/mol K for any equation where you're trying to find energy or ent...

Aarushi Verma 2I

A state function is where the value doesn't depend on the path taken to get there, but only the final and initial values. Free energy is a state function because we only need the final and initial values to find the change and determine whether the reaction is spontaneous.

Aarushi Verma 2I

You should add 273 or 273.15 to the C temperature value to convert to Kelvin and then subtract 273 or 273.15 from the K value to get the C value.

Aarushi Verma 2I

I don't remember going over finding the enthalpy of vaporization in our lectures, but I used the units to figure out how to find it. I realized that I needed to convert the heater value with units W to kj (convert from W to J/sec to J to kj), and then I used the masses given to calculate the moles o...

Aarushi Verma 2I

The R constant you should use depends on the units that the question is asking for. Usually if you're using the R to find energy, then I would recommend using 8.3145 J/mol*K, but if you're using PV=nRT to convert values, then I would use the 0.08206 L*atm/mol because that will give you the right uni...

Aarushi Verma 2I

The external pressure is the pressure pushing onto the piston from the surroundings.

Aarushi Verma 2I

The First Law of Thermo essentially says that energy cannot be created or destroyed. So if we're looking at the system vs surroundings, a release of energy from the system (negative change) will be energy absorbed by the surroundings (positive change) and vice versa.

Aarushi Verma 2I

The integral equation is used in a reversible system and the external pressure is changing in really small increments (which is the reason we have to use the integral). You won't actually have to use it on a test though thankfully.

Aarushi Verma 2I

You should use J/g C because that's what the textbook uses, but also just be careful of what units the problem gives temperature in. They might give it in kelvin just to throw you off or something like that.

Aarushi Verma 2I

All isolated systems are not vacuums, rather they are just systems where no matter or energy can be transferred so the internal energy of the system stays the same.

Aarushi Verma 2I

If there is an equal amount of moles on both sides, then any pressure changes won't affect the reaction because pressure only affects gases, not liquids and solids.

Aarushi Verma 2I

Honestly, since the calculation takes less than a minute I would do it anytime you're unsure if you can use the approximation. If the K is 10^-4 then I would definitely do that but if its 10^-5 you probably won't need to, but it can't hurt.

Aarushi Verma 2I

You can take out the x in the expression when its (concentration - x) in denominator when the K value for the equation is less than 10^-4. 10^-4 is a safe value for approximation, so if you're unsure you can always go through with the problem to find x and then use the % protonation or deprotonation...

Aarushi Verma 2I

We won't have to memorize the bond enthalpy values or anything like that, we should be given this value and then we can use the values to then figure out the standard reaction enthalpy.

Aarushi Verma 2I

I'm confused on the difference between the standard reaction enthalpy, enthalpy of formation, and enthalpy of combustion. Can we use the three methods from friday's lecture to find all three or are there specific methods to finding one of them.

Aarushi Verma 2I

You can use the 5% rule for percent protonation and deprotonation and then if the rule doesn't work, then I would go back and use the quadratic formula to get the more accurate answer.

Aarushi Verma 2I

Most of the time, the reactants side will have a negative change and the products side will have a positive change. One way to be sure is to calculate Q using the initial concentrations. If K < Q, then the reaction will move towards the reactants which means that the product change is negative and r...

Aarushi Verma 2I

I mostly use PV=nRT in equilibrium to convert gases to concentration and vice versa, depending on the information given in the question. The equation is also helpful for Le Chatelier's Principle because you can see how pressure and volume are related to see volume changes would affect the reaction.

Aarushi Verma 2I

We leave solids and liquids out because they do not have a concentration so we cannot put them into our K expression. We also don't include H20 in the K expression when it acts as a solvent because its concentration is so high on both the reactants and products side that we would see no difference. ...

Aarushi Verma 2I

It's a good thing to keep track of the temperatures in the question because if the temperature changes, then K will change as well. Otherwise, if the temperature stays constant for the whole question, then you can assume that K stays the same even though the concentration of reactants or products ch...

Aarushi Verma 2I

Hi! Since molarity (M) is defined as the concentration over the volume, we can manipulate our ideal gas equation to get (n/v) on one side and get P/rt on the other side. This is super helpful to go between molarity and pressure especially since we use a lot of gases in chemical equilibrium.

Aarushi Verma 2I

To find pKa, you would need to have the Ka value, which you can get by setting up the equilibrium equation with products at the tops and reactants at the bottom. Once you have Ka, just take the -log(Ka) which will give you the pKa value.

Aarushi Verma 2I

The concentration of water is so high on both sides, that the change in water that happens during the reaction is so small so that we don't include it in the K expression. Even if we were to include it, there wouldn't necessarily be a difference in the water value because the value is so small that ...

Aarushi Verma 2I

Yes since Kw is dependent on temperature, it will change if the temperature changes.

Also for your second question, a neutral solution is when both [H3O] and [OH] are 1.0*10^-7 (both equal) so water would only be neutral at that time too.

Aarushi Verma 2I

In a neutral solution (like water), H30 and OH have to be equal. But if the solution is acidic, that means there is more H3O than OH and if the solution is basic, then that means there is more OH than H3O.

Aarushi Verma 2I

You can use either partial pressure or concentration of the gas, as long as you say Kp and Kc respectively and know which one the problem is asking for.

Aarushi Verma 2I

According to Le Chatelier's Principle, if there is a higher number of reactants, then the system will accommodate for that change by making more products to get to equilibrium. So compared to the original scenario, an increase in the reactants will also cause an increase in the products too.

Aarushi Verma 2I

Solvents are usually going to be water but are liquids and have the symbol (l) and they are not included in the equilibrium equation. You want to include aqueous (aq) solutions in the equilibrium equation.

Aarushi Verma 2I

Hi! What helps me remember the relationship between Q and K is if Q is smaller than K, then the reaction still needs more time to get to the equilibrium value and therefore it needs to make more products (shifts to the right), but if Q is larger than K, then the reaction still needs to go to equilib...

Aarushi Verma 2I

If the volume decreases, then the concentration increases which would shift the equilibrium towards side with less moles of gas. If the volume increases, then the equilibrium will shift towards the side with more moles of gas because the concentrations will decrease. Eventually, the system will adju...

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