CHEMISTRY COMMUNITY

Stephanie Zhang 2K

hi! yes, catalysts lower the activation energy for both the forward and reverse rates; it changes the pathway of the reaction. I like to visualize it in a reaction profile- catalysts lower the activation energy and thus make the "hump" smaller, which would make both the forward and reverse...

Stephanie Zhang 2K

hi! catalysts are first 'consumed' in one step and then produced in a later step, so technically catalysts are not consumed in the overall reaction. intermediates are the opposite: they are first produced in one step and then consumed in a later step. Hope this helps!

Stephanie Zhang 2K

It would be helpful to write out a reaction profile! the ∆H enthalpy is the change in energy between the reactants and the products; you take the difference between the initial product energy and the initial reactant energy. The activation energy is the energy barrier that the reaction needs to over...

Stephanie Zhang 2K

hi! intermediates are molecules that are not part of the overall reaction; they are first produced in one step and then consumed in a later step. Be sure not to confuse this with a catalyst- catalysts are first consumed in one step and then produced in a later step.

Stephanie Zhang 2K

yes, you usually look at the overall rate law and see which reactants are part of it, and then match that to the elementary step which also has the same reactants. Or, the problem would give you the slow step. Only the reactants in the slow step are included in the rate law. If you see a reactant in...

Stephanie Zhang 2K

Hi! when catalysts are saturated, no matter how much more reactant/substrate are present in the solution, the rate will stay the same since the catalyst can't go any faster than it already is when it's saturated. That means that the reaction rate does not depend on the concentration of the reactant,...

Stephanie Zhang 2K

Yes, you can look at stoichiometric coefficients and use those to determine the rate law of elementary steps, as they depend on molecules "bumping into" each other to form products. And yes, for 2A +B it would be termolecular as A is bimolecular and B is unimolecular; you need 2 molecules ...

Stephanie Zhang 2K

Hi! You can only use the molecularity of each step if you are looking at elementary steps. You use the molecularity of each step in order to write out the rate laws for elementary steps, so you can use the stoichiometric coefficients to write out each rate law. Note that you can not do this with the...

Stephanie Zhang 2K

Hi! To determine any reactant's order, you want to choose to compare between 2 reactions where the only value that changes is the reactant's concentration. Looking at the reactions given, we can see that for Experiment 1 and 4, the values of A and B stay the same, while the value of C changes. Just ...

Stephanie Zhang 2K

Hi! Personally, I like to write out the equations. rate=k[A]^n. The units of rate are always going to M/s. So, then for each order, you can plug in the corresponding order value for n, and then finding the corresponding unit For example, for zero order k[A]^0, so rate=k. That means the rate units of...

Stephanie Zhang 2K

Hi! for this problem, you use the equation ∆G˚=-nFE˚. From our last unit, remember that if ∆G˚ is negative, that means the reaction is spontaneous as written/in the forward direction, and if it's positive, that means the reaction is not spontaneous as written. From ∆G˚=-nFE˚, we can see that ∆G˚ wil...

Stephanie Zhang 2K

Hi! you have the right coefficients for Cl2O7 and H2O but not the rest. Here's my general thought process for doing this problem: So first, you separate the compounds and determine which one is the oxidizing/reducing reagant by seeing how the charges change with Cl. Then, you balance the non Oxygen ...

Stephanie Zhang 2K

Hi! So since the reaction isn't in standard conditions, where [H+]=1 M and PH3=1 atm, we have to use the Nerst equation for nonstandard conditions: E=E˚-(0.0592V)/n logQ. Using the equation 2H+ + 2e- --> H2, we can find the E˚ from the table, and we can determine Q from Q=P/R (make sure to square th...

Stephanie Zhang 2K

Hi! so I generally think of it as you want to get each oxidation state equal to the overall charge of the ion/compound. Since none of these have charges, all their charges are equal to 0. It might be helpful to write each element's oxidation state separately. This is where the rules come in: if an e...

Stephanie Zhang 2K

Hi! for this problem you'd use the Nerst equation, E=E˚-RT/nF lnQ. Since standard condition is [H+]=1 M and PH= 1.0 atm, this indicates that it is not under standard conditions, so you would use the Nerst equation for nonstandard conditions: E=E˚-(0.0592V/n)logQ. From there, you plug in the values. ...

Stephanie Zhang 2K

Hello! Yes, if you're concerned with ∆nRT, you only consider the change in the number of moles of gas! I like to think of it as since we're only concerned with work of expansion right now, the only phase that would be able to "expand" in that way is the gas phase, and so work is only done ...

Stephanie Zhang 2K

For that specific formula, I would definitely use Kelvin! This is because you're looking at the ratio of T2/T1, and so the ratio would be different if you're using ˚C versus Kelvin. For things that are asking for the change in temperature, such as q=mC∆T, that's when you can use either ˚C or Kelvin,...

Stephanie Zhang 2K

Hello! Since ∆U is the change in internal energy, ∆U= q + w. At a constant pressure, q= ∆H, and that just comes from the general definition that ∆H is the change in heat transferred at a constant pressure. Also, at a constant pressure, w=-p∆V. Then, you can just plug that into the equation, and you ...

Stephanie Zhang 2K

Hi! an open system is where both matter and energy can move in or out. A closed system is where only energy can move in/out, but matter can't move. An isolated system is where both energy and matter can't move in/out of the system. I would consider whether you need to keep in mind the movement of ma...

Stephanie Zhang 2K

Hello! Usually when you use the w=-p∆V formula, you get units of L atm. So, to convert that to Joules, you have to use the conversion that 1 L atm= 101.325 Joules. The conversion is towards to top of the formula sheet, under the specific heat capacities. Then from there, you can convert the Joules t...

Stephanie Zhang 2K

Hello! This is similar to a Hess's Law problem. In order to find the equation needed, you have to rearrange the two equations given. If a substance appears with the same number of moles on either side of the "arrow", you can cancel it out in the reaction when you add the two equations toge...

Stephanie Zhang 2K

Hi ! I think this has to do with one of the lectures from last week, where since entropy is a state function you can essentially treat the irreversible process as 2 steps, since all that matters is the initial and final states. The first step: find the change in entropy from a change in volume using...

Stephanie Zhang 2K

Hi! For the first step, you find the moles of gas using the ideal gas law PV=nRT. Don't forget 100kPa=1 bar, and use the correct R. Then, you can calculate change in entropy from a change in temperature using the ∆S=nClnT2/T1 equation (it should be on your equation sheet). Make sure that the tempera...

Stephanie Zhang 2K

Hi! the units for seconds should be cancelled: you multiply W by J/s, and then multiply that by 60s/1min, by the number of minutes provided to get the total heat that the heater contributes over that time (in Joules). Make sure to convert that to kJ for calculating the enthalpy of each substance! Fo...

Stephanie Zhang 2K

Hydrogen burns in an atmosphere of bromine gas to give hydrogen bromide gas. (a) What is the standard Gibbs free energy of the reaction H2 (g) + Br2 (g) --> 2HBr(g) at 298 K? (b) if 120.mL of H2 gas at SATP combines with a stoichiometric amount of bromine and the resulting hydrogen bromide dissolves...

Stephanie Zhang 2K

Hello! For this problem, you saw from part B what the limiting reactant was. To find the number of moles consumed, the limiting reactant will have all its moles consumed. For the excess reactant, you multiply the initial moles of the limiting reactant by the ratio of the moles in the chemical equati...

Stephanie Zhang 2K

Hi! For part A, first you find the amount of heat released by the compound by multiplying the mass of it by the molar mass and then by the combustion (since that's the amount of heat released per mole of the substance). This gives you the amount of heat released by the substance, or q. Keep in mind ...

Stephanie Zhang 2K

I find that using the ice tables helps a lot in order to keep all the numbers in order (although it isn't completely necessary). Since most weak acids and bases that we typically use only gain/lose one H+, you can just use the Ka/Kb equation and set it up that way, with each concentration having a c...

Stephanie Zhang 2K

Hi! In the Week 4 Wednesday lecture around 30 minutes in, Dr. Lavelle gives a really good example. If you're pumping up a bicycle tire, you get tired really quickly because you're losing energy and putting work into the system. The work goes into the system in order to compress the air inside of the...

Stephanie Zhang 2K

Hi! So what I did was set the q equations equal to each other. Since q=mc∆t, that means m1C∆T=-m2C∆T . The one that starts at a higher temperature has a negative q value because it's losing heat that is gained by the other water. Since both of them are water in liquid form, you can assume that their...

Stephanie Zhang 2K

For acids, it's known as percent deprotonation because acids give off a H+ proton, so the acid itself is loses a proton. It's also known as the percent ionization the acid goes from a neutral charge to having a negative charge, so the compound becomes ionized. Bases have a percent protonation becaus...

Stephanie Zhang 2K

Hi! Usually when the K value is less than 1E-4, that indicates that the reactants are more stable at equilibrium. So, that means that the change in the initial reactant concentration will not be very different from the equilibrium reactant concentration. Additionally, you should check afterwards tha...

Stephanie Zhang 2K

Hi, so if the reaction is endothermic, you treat heat as a reactant, and so a temperature increase would result in more products. Then, from K=P/R, that means K will increase. Conversely, if the reaction is exothermic and you add heat, that means that the reverse reaction will be favored. So, more r...

Stephanie Zhang 2K

Hi! Here's the problem: An experimental automobile burns hydrogen for fuel. At the beginning of a test drive, the rigid 30.0-L tank was filled with hydrogen at 16.0 atm and 298K. At the end of the drive, the temperature of the tank was still 298K, but its pressure was 4.0 atm. (a) How many moles of ...

Stephanie Zhang 2K

Hi! So the problem is: How much heat is needed to convert 80.0 g of ice at 0.0°C into liquid water at 20.0°C (see Tables 4A.2 and 4C.1)? I calculated the enthalpy required to melt the ice: I converted the 80.0 grams of ice to moles, then multiplied that by the enthalpy of fusion: 6.01 kJ/mol x 4.43 ...

Stephanie Zhang 2K

Hi! So AlCl3 from my understanding is made of Al^3+ and 3CL-. Cl- is the conjugate base of a strong acid, which results in a "neutral" solution. Al^3+ is a strong metal cation, and so it should make the solution more acidic. I believe in Table 6D.1 it give you the acidic character and Ka v...

Stephanie Zhang 2K

Hello! so for part a) how I went about this problem was I noticed that the muscles produce lactic acid. Since it's an acid, that indicates that the solution will become more acidic, or the [H3O+] will increase if the lactic acid is added. Since in the original reaction, [H3O+] is one of the reactant...

Stephanie Zhang 2K

Hi! So Q is the reaction quotient, and it can be used when the chemical reaction isn't in equilibrium. K is the equilibrium constant, so it is used only when the chemical reaction IS in equilibrium. Both Q and K have the equation of product concentration divided by reactant concentration, or [P]/[R]...

Stephanie Zhang 2K

Calculate the pH and pOH of each of the following aqueous solutions of strong acid or base: (e) 13.6 mg of NaOH dissolved in 0.350 L of solution I would really appreciate it if someone could go over the process to solving this problem! This is my general thought process so far: 1) Convert mg to g, t...

Stephanie Zhang 2K

Yes, I agree with the response above! As a general rule, always make sure to double check the units of the quantities given. Since you need the concentrations of the species in order to find Qc, you have to make sure each value you plug into the general Qc equation has the units of mol/L, which is a...

Stephanie Zhang 2K

I agree with the above post! the "k" in kPa just stands for 1000 (like kilogram or kg is 1000 grams).
So, when converting from kPa to bars:
1 bar= 100000Pa
1 kPa= 1000Pa
1 bar= 100 pKa

That's why 18kPa= 18/100 or 0.18 bar

Stephanie Zhang 2K

I kind of thought the same thing as you guys. One thing that may help that I learned during my discussion section: So Kw= [H+][OH-]= 1.0E-14 @ 25 degree Celsius This is because apparently at that temperature, the concentration of H+ ions is equal to 1.0E-7, and the concentration of OH- is also equal...

Stephanie Zhang 2K

For this one, try to follow a general pattern of solving the problem: 1) check to see whether the equation is balanced (in this case it is) 2) Since the K provided is Kp, you can just use the partial pressures instead of concentrations for the species. 3) set up your ICE table, with PCL5 initially s...

Stephanie Zhang 2K

The general definition is that the system in equilibria minimizes the effects of any change. If you add a reactant/remove a product, the system will form more product in order to reach the original K. In order to do so, the reactant concentrations will decrease. If you add a product/ remove a reacta...

Stephanie Zhang 2K

Hi,

I'm a bit confused on how exactly to tell when a reaction involves fully dissociated ionic compounds in a solution (for some reason I can't recall this from 14A)? Thanks!

Stephanie Zhang 2K

Hi! I'm not quite sure if this answers your question, but I'll try my best. In this example, since Q<K, that means it proceeds towards the products as you stated. Both Q and K use [P]/[R]. So, in this case if Q<K, that implies that the reaction currently has more [R] than it would in equilibrium (si...

Stephanie Zhang 2K

Yes, I agree with the answer above! I just thought of it this way: you start with reactants, so as the reaction proceeds the amount of reactants left will decrease and so x will be negative. Similarly, as the reaction proceeds the amount of products will increase, so x will be positive for the chang...

Stephanie Zhang 2K

Hi, just a quick clarifying question. Does (aq) or (l) indicate a solvent (which ones aren't included in equilibrium constants)? Thank you!

Stephanie Zhang 2K

Hello! I agree with the post above. Some points that I wanted to clarify (because I was confused on it as well) might be that you have to convert the moles of PCL5 to concentration units, or mol•L-1. That's why you divide it by .5 L (and make sure to convert the 500ml to L). Also, with the +x and -x...

Stephanie Zhang 2K

Hello, I'm not sure whether this is the right place to post it but I just had a general clarifying question. At first, I thought that moles and M were the same thing, so I'm a little confused. Does M represent mol•L^-1?

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