CHEMISTRY COMMUNITY

Jennifer Su 2L

For the snowflakes, delta H is negative because the snowflakes lose heat, and delta S is negative because going from a liquid to a solid is less chaotic. Negative delta H minus a negative delta S makes a negative delta G. For the sublimation, delta H is positive because heat is going in, and delta ...

Jennifer Su 2L

So first you want to find the concentration of [A] after the time period. Since they tell us that delta[B]=0.19M, we can multiply it by (1 mol A)/(2 mol B) to get delta[A]. 0.19 mol B/L x (1 mol A/2 mol B) = 0.095 M of A To get the final [A], simply subtract delta[A] from [A]not [A]final=0.15M-0.095...

Jennifer Su 2L

For parts c and d, you need to first find the value of k by using the half life formula. t1/2 = 0.693/k =355s, so k= 1.95x10^-3 s^-1 You can then plug this into the first order rate equation to find the time (t). ln[A]= -kt + ln[A]not c) ln(0.15)= -(1.95x10^-3 s^-1)*t + ln(1), so t=9.7x10^2 s d) ln(...

Jennifer Su 2L

Does anyone know the answer to these and why?

1. The precipitation of snowflakes inside a cloud at -10 degrees Celcius and 0.839 atm. Is deltaG <0, =0, or >0?

2. The sublimation of dry ice in a warm room. Is delta G <0, =0, or >0?

Jennifer Su 2L

Rate constant is the term "k" itself, and it depends on the temperature, (T), frequency factor (A), and activation energy (Ea). Rate of reaction depends on both "k" and the concentration of reactants. For example, for a reaction aR --> bP, RATE= (-1/a)*(d[R]/dt)= (1/b)*(d[P]/dt) ...

Jennifer Su 2L

Looking at the equation: k= A * e^(-Ea/RT) = A / e^(Ea/RT) When activation energy (Ea) increases, the term "e^(Ea/RT) increases as well. Since that term is in the denominator, the bigger the denominator, the smaller the overall rate (k). The other factors that affect the rate, k, are the freque...

Jennifer Su 2L

You arrange it so that Enot is the most positive. So you should assign the half reaction with the larger Enot as cathode, and assign the half reaction with the smaller Enot as the anode. This way, Enot(cathode)-Enot(anode) will be the largest possible.

Jennifer Su 2L

Since deltaG=-nFE, the reaction is spontaneous in the forward direction when deltaG is negative, which happens when E is positive. On the other hand, the reaction is spontaneous in the reverse direction when deltaG is positive, which happens when E is negative.

Jennifer Su 2L

In a concentration cell, Enot is always 0V, E is typically low voltage, and you always make the lower concentration the product ("it is in solution").

Jennifer Su 2L

The order of a rate law is the sum of the exponents of its concentration terms. More specifically, the reaction order is the exponent to which the concentration of that species is raised, and it indicates to what extent the concentration of a species affects the rate of a reaction, as well as which ...

Jennifer Su 2L

Because the balanced half reactions are:
(6Fe^3+) + (6e-) ----> (6Fe^2+)
(2Cr^3+) + (7H2O) -----> (Cr2O7^-2) + (6e-) + (14H+)

So 6 moles of electrons are transferred.

Jennifer Su 2L

You should solve for Temperature "Z" where deltaG=0 and explain that the reaction is at equilibrium at this point. Then to answer the question, you should write it as an inequality (ie: when Temperature> tempZ, the reaction is spontaneous in the forward direction). If you want to be extra,...

Jennifer Su 2L

Follow up question: is n supposed to be (moles of electrons in reactants)-(moles of electrons in products) or is it (moles of electrons in products)-(moles of electrons in reactants)?

Jennifer Su 2L

Since the equilibrium constant is large, we can conclude that at equilibrium there are more products than reactants. Therefore, the products are more stable, while the reactants are more unstable in comparison.

https://lavelle.chem.ucla.edu/wp-conten ... am_ans.pdf

Jennifer Su 2L

You can take a look at problem 9.65 (6th edition). The solution manual states: "To understand what happens to deltaG as temperature is raised, we use the relationship deltaG=deltaH-T*deltaS. From this it is clear that the free energy of the reaction becomes less favorable (more positive) as tem...

Jennifer Su 2L

When deltaG is positive, it means that the free energy of the products is MORE than the free energy of the reactants. Therefore, an input of free energy is needed for the reaction to occur, so the reaction is NOT spontaneous. When deltaG is negative, it means that the free energy of the products is ...

Jennifer Su 2L

Gibb's Free Energy can be calculated through using enthalpy and entropy. deltaG=deltaH - T*deltaS When a reaction has a positive enthalpy (+deltaH): deltaG will be positive if entropy is negative (-deltaS) deltaG will be negative if entropy is positive (+delta S) and T*deltaS is greater than deltaH ...

Jennifer Su 2L

if K>>>10^3, then products are favored over reactants, and therefore products are more stable. if K<<<10^-3, then reactants are favored over products, and therefore reactants are more stable. if K=1, then products and reactants are exactly equally favored, and therefore are equally stable. for value...

Jennifer Su 2L

Can someone please explain why 3E is false? Because heat is required during melting or boiling (phase change transition), temperature of a sample can remain constant even though heat is being supplied. For example, if you have boiling water, it will remain at 100 C even though heat is being added, ...

Jennifer Su 2L

q(ice) = -q(water) FOR ICE: Melting: q=deltaH(fusion)= 6.01 kJ/mol x (25.0 g / 18 g*mol-1) = 8347.2 J Increasing Temp of Melted Ice: q=mCdeltaT= (25.0 g) x (4.184 J/K*g) x (Tf-273K) = 104.6(J/K)*Tf - 28555.8J FOR WATER: Cooling the Water: q= mCdeltaT= (265 g) x (4.18 J/K*g) x (Tf-298K) = 1107.7(J/K)...

Jennifer Su 2L

You have to break this problem up into two steps: 1. Find m * Heat of fusion 2. Find m * C * delta T When you find those two values, you add them up, and that is the total heat needed. Is it n*Heat of fusion or m* Heat of fusion? For the first part we use moles and the second part we use mass right?

Jennifer Su 2L

I had a different test, but generally speaking, by DECREASING the volume, you are INCREASING the pressure. With an increase in pressure, the reaction will tend towards the side with LESS moles. So you should count the coefficients (which tells you the number of moles) of the balanced equation on the...

Jennifer Su 2L

Outline 3 is covered completely, Outline 4 is covered until problem 9.47 (after that the problems are on Gibb's Free Energy).

Jennifer Su 2L

I would assume anything covered in lecture is fair game, but it's probably smarter to focus on the application of formulas since the formulas themselves will be given on the constants and equations sheet.

Jennifer Su 2L

-Heat Capacity- heat required to raise the temperature of an object by 1C. -Specific Heat Capacity- divide heat capacity by the amount of substance present (g); heat required to raise the temperature of 1 gram of substance by 1C -Molar Heat Capacity- divide heat capacity by moles of substance (mol);...

Jennifer Su 2L

Because you can't change the volume of solids and liquids, so delta V will be 0. Therefore, PdeltaV will equal 0, which makes it insignificant.

Jennifer Su 2L

Towards the end of lecture on Friday Week 4, Lavelle talked about how just knowing delta U (or delta H) isn't enough to know if a reaction will occur and that we also need to know the entropy (S). Can someone explain how knowing delta U and S will tell us how the system will change? In other words, ...

Jennifer Su 2L

I believe he said we don't have to worry about deriving it! He didn't explain how he got 3/2 nRT.

Jennifer Su 2L

Steam burns causes more severe burns than water burns. This is because heat is required during melting or boiling (phase transition), and therefore the temperature of the sample REMAINS CONSTANT even though heat is being supplied. When you get burned by water, since it is not in a phase transition, ...

Jennifer Su 2L

On the website it says under announcements that "Midterm covers: Chemical Equilibrium, Acid and Base Equilibria, Thermochemistry, Thermodynamics to the end of entropy."

Jennifer Su 2L

It comes from the gas constants (this is provided on the list of constants). Gas constant, R = 8.314 (J)/(K*mol) = 8.206 x 10^-2 (L*atm)/(K*mol) Since the values equal each other, 8.314 (J)/(K*mol) / 8.206 x 10^-2 (L*atm)/(K*mol) equals to 1. Cancel out the units and you get 101.316 (J)/(L*atm), whi...

Jennifer Su 2L

Q is called the "reaction quotient," and can be calculated anytime during the reaction. K is called the "equilibrium constant", and can be calculated only when the reaction has reached equilibrium. For a specific reaction (at a constant temperature and pressure), you can get diff...

Jennifer Su 2L

To add on, Constant P. Calorimetry (I guess this is the "normal" kind) gives you the enthalpy values. Both types are used to measure specific heat capacity.

Jennifer Su 2L

I believe qp is equivalent to deltaH (change in enthalpy), and it represents the heat at constant pressure. qp=n*Cp*T where n is in moles, Cp is the molar heat capacity, and T is the change in temperature. I think he's going to talk continue to talk about this in Wednesday's lecture, but if anyone c...

Jennifer Su 2L

This will allow you to convert the partial pressure of gas to molar concentration.
PV=nRT
P=(nRT)/V
P= (n/V)*RT
P= Concentration *(RT) and Concentration= P/(RT)
R and T are normally constant for these problems.

Jennifer Su 2L

Increasing the pressure by introducing an inert gas has no effect on the equilibrium composition. However, compression of a reaction mixture at equilibrium tends to drive the reaction in the direction that reduces the number of gas-phase molecules.

Jennifer Su 2L

My TA said he will be lenient on the first test in regards to significant figures, but it probably depends on how your TA grades. Lavelle posted two links under "Math Assistance" on his website about significant figures that you can review. https://lavelle.chem.ucla.edu/wp-content/supporti...

Jennifer Su 2L

To add on to the post above, the concept that the stronger the acid, the weaker its conjugate base, and the stronger the base, the weaker the conjugate acid is known as conjugate seesaw. It's also helpful to remember that Ka x Kb = Kw and pKa + pKb = 14.

Jennifer Su 2L

Weak acids have Ka<10^-3 and weak bases have Kb<10^-3. Between two Ka and two Kb, the one with the bigger number is stronger. Since Ka x Kb = Kw, they are inversely related, which is why a strong base is paired with a weak conjugate acid and a strong acid is paired with a weak conjugate base. This i...

Jennifer Su 2L

I think I counted wrong. I forgot that numbers like 20.0 is only 1 sigfig and not 3. Thank you for your help!

Jennifer Su 2L

I did not have Lavelle before and was wondering about how we should round our answers. I noticed that the answers in the answer manual for the 6th edition textbook often don't follow the rules of significant figures. For example, let's say there is a given value of 0.385M in the problem. I would ass...

Jennifer Su 2L

The problems from acids & bases are more relevant to what is currently being covered in lecture, but you can do problems from chemical equilibrium too since we're still doing ICE tables.

Jennifer Su 2L

Just remember that pure water has the neutral pH of 7, so as the comment above states, [H3O+] and [OH-] should always be equal. If they are not equal, the solution will either be more acidic (higher concentration of H3O+) or more basic (higher concentration of OH-).

Jennifer Su 2L

As long as you explain that products are favored since K>1 and add/clarify that it isn't strongly favored since K<10^3, it should be correct.

Jennifer Su 2L

Quoting the 6th edition textbook on page 427, it says "when a reaction involves fully dissociated ionic compounds in solution, the equilibrium constant should be written for the net ionic equation, by using the activity for each type of ion." I was wondering how you would know if there are...

Jennifer Su 2L

To add on, the products of a spontaneous reaction are generally more stable than the reactants. Therefore, a spontaneous reaction can occur without any additional input of energy, and as said in the reply above, it normally releases energy.

Jennifer Su 2L

This will depend on whether you are changing the coefficient of a reactant or product. Let's say the equilibrium equation has aA and bB as the reactants, and cC as the product, with the lowercased letters representing the coefficients. Therefore, K=(partial pressure of C)^c/(partial pressure of A)^a...

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