CHEMISTRY COMMUNITY

Johann Park 2B

Because we are using the unique rate of E, we put 1/(coefficient) in front of the rate. The coefficient in this case is 2, so we have 1/2 d[E]/dt.

In a general rate, we would not use the stoichiometric coefficient.

Johann Park 2B

diangelosoriano wrote:Is there a way to tell by looking at these equations if the catalyst was heterogeneous or homogeneous?

A catalyst is homogeneous if it is in the same phase as the reactant R.
It is heterogeneous when it is in a different phase as R; in this case it is usually a solid.

Johann Park 2B

If I am understanding your question correctly: I believe that if the early step is the slow step, yes, the overall reaction rate is the rate of that first slow step. If the early step is fast, however, and the second step is the slow step, there is a "bottleneck" effect and a build-up of i...

Johann Park 2B

diangelosoriano wrote:If a catalyst were added, how would this also affect k or activation energy?

When you add a catalyst, the reaction is sped up - k is increased and E_a is lowered. The new pathway has a lower activation energy.

Johann Park 2B

I believe you are talking about the Half-life for a first order rxn: 1) Start with the integrated rate law: ln [A] = -kt + ln [A] o 2) At t = t 1/2 , [A] = 1/2[A] o 3) Substitute in 1/2[A] o : ln 1/2[A] o = -kt 1/2 + ln [A] o 4) Separate variables: ln 1/2[A] o - ln [A] o = -kt 1/2 5) Simplify: ln 1/...

Johann Park 2B

1) Start with: d[A]/dt = k[A]
2) Separate the variables: d[A]/[A] = -k dt
3) Integrate both sides: ln [A] = -kt + C
4) Because at t=0, ln[A]_o = C: ln [A]_t = -kt + ln [A]_o

Johann Park 2B

So when we have a reaction with multiple reactants changing, such as A + B + C -> P, we make one [R] small and the others very large. In practice, we would thus have: k' = [A] N , just like in a normal first-order rate problem (except the fact that we denote k as k'). Then you would do the same by m...

Johann Park 2B

Convert grams to mols using molar masses of H₂ and I₂. Then use .750 L to convert it to Molarity. You should have roughly 0.35 mol/L H₂ and 0.001 mol/L I₂.

Rate = k[H₂][I₂]

Rate = 2.2 x 10^-5 mol/(L x S)

Johann Park 2B

Did you mean the reaction: 2NO₂ -> 2NO + O₂?

Then yes, the decomposition of NO₂ is used to find the unique rate of the reaction which is 1/2(Δ[NO₂]/Δt)

Johann Park 2B

With respect to [B], the reaction is second-order. You use the different values for the rates and concentrations to solve for the exponents (x, y, z) on the concentrations.

Johann Park 2B

The concentration increased by a factor of 1.2 means that the rate of the reaction also increases by a factor of 1.2:

1.2Rate = k [1.2CH₃Br]^x[OH^-]^y

Johann Park 2B

The balanced reduction half reaction should be Cr₂O₇^2- (aq) + 14H⁺ (aq) + 6e^- -> 2Cr³⁺ (aq) + 7H₂O (l)

Johann Park 2B

The elements in the anode are to the left of the salt bridge and the cathode on the right. From there, I believe each side is written in order of reactants to products.

Johann Park 2B

When dealing with statistical/residual entropy and the make-up of molecules and crystals, we will use S = K + ln W to calculate is as a value. When dealing with reactions and changes, we have delta S = q/T and other derivations that reflect delta S as a state function.

Johann Park 2B

Internal energy is a state function dependent on temperature, thus, when temperature does not change (definition of isothermal), there is no change in internal energy. Additional note: Dealing with ideal gases, contributions to U are kinetic energy and intermolecular interactions. The latter is 0, b...

Johann Park 2B

Standard Gibbs Free Energy entails 1 bar of pressure and a specified temperature (in this case being 25^oC (298 K)). Otherwise, it should be noted as just a "Gibbs Free Energy" value.

Johann Park 2B

For "free expansion" we know that ∆U = 0 and w = 0. Since ∆U = q + w, q = ∆U - w = 0 (q_SYS = 0),

q_sys = -q_surr, thus q_surr = 0.

Calculating ∆S_surr = q_surr/T = 0/323 = 0; ∆S_surr = 0.

Johann Park 2B

I believe you are asking about Problem 9.19? If so, we need to find the standard entropy of the vaporization of water at 85 o C because that is what the problem is asking of us. We have ∆S TOT = ∆S 1 + ∆S 2 + ∆S 3 . ∆S 1 = Heat Capacity of liquid water x ln T 2 /T 1 or 75.3 x ln 373/358 ∆S 2 = Stand...

Johann Park 2B

Volume is inversely proportional to the pressure, and with a decrease in volume (increase of pressure), the random movement of gaseous molecules decreases. Thus, the molar entropy decreases. I'm pretty sure that in this case we are regarding T as a constant.

Johann Park 2B

∆S_TOT = ∆S_SYS + ∆S_SURR. When heat is leaving the system, the value is negative. As heat leaves the system, it enters the surroundings.

Johann Park 2B

Setting T = 333K is not the final answer; it's the "boiling point." But it is necessary to find that value to come up with the final answer: T > 333K favors the forward reaction and makes delta G negative.

Johann Park 2B

You have the correct initial equation with (1.38 x 10^-23) x ln(6^{6.02 x 10^23}). Using properties of natural logs, move the exponent NA in front of the natural log:

(1.38 x 10^-23) x (6.02 x 10²³) x (ln 6) and this will get you 14.9 J/K

Johann Park 2B

Using mean bond enthalpies will result in an answer close to that of the answer from enthalpy of formation, but will be different because the bond enthalpy method is less accurate (uses averages).

Johann Park 2B

The problem states that this is a combustion reaction at 25°C. The two products that come from combustion are CO₂(g) and H₂O - the H₂O is liquid water because the combustion happens at room temperature.

Johann Park 2B

To balance the second equation, you must add the coefficient 3/2 to O₂. This does not change the enthalpy of the reaction because the enthalpy of an element in its natural elemental state, such as O₂ gas, is zero.

Johann Park 2B

An irreversible expansion happens instantaneously while a reversible expansion maximizes work. For the reversible expansion, you solve with dw = -PdV, and this calculus accounts for work in continuously through intervals.

Johann Park 2B

Yes, in thermodynamics we must remember that energy is conserved and that q_sys + q_surr = 0, and thus q_sys = -q_surr; the system being the ice and the surroundings being the water.

Johann Park 2B

Remember that the enthalpy of a naturally occurring element (such as O₂ gas) is always 0. So I believe changing the coefficient of O₂ would not affect the enthalpy of the reaction.

Johann Park 2B

That's correct! ΔH RXN is equal to Energy required to break bonds (positive values) minus energy required to form bonds (or using addition and making sure to include that the values are negative.) The example in class was C=C (+612 kJ/mol), H-Br (+366 kJ/mol) -> C-C (-348 kJ/mol), C-H (-412 kJ/mol),...

Johann Park 2B

In order to get the desired reaction, you have to choose and manipulate the two given reactions. In doing so, you would have to first choose Reaction 1, reverse it, and multiply the coefficients by 3/2 (make sure to change the value for enthalpy too). Then with Reaction 2, I believe we need it to ma...

Johann Park 2B

Choosing what to include in heat capacity depends on what is being asked/given. The heat capacity of just the calorimeter is different from the heat capacity of the entire calorimeter system (calorimeter and water). The heat capacity of the entire system (C) is the sum of the heat capacities of the ...

Johann Park 2B

I think it's a good thing to know how to identify polyprotic acids and maybe give the section a quick read, but I don't believe it'll be necessary to spend too much time focusing on the pH of polyprotic acids, which is why he omitted the homework problems on that section.

Johann Park 2B

Identifying resonance in a structure is a way to find delocalized pi bonds in a molecule. In resonance structures, electrons are able to move to stabilize the molecule -> this movement is delocalization.

Johann Park 2B

Using what we know about Lewis Structures and VSEPR models, you first have to draw out the molecule with its atoms and lone pairs. If there is no symmetry and atoms' dipoles (as a result of electronegativity values of atoms) do not cancel out, then the molecule is most likely polar (like H 2 O). If ...

Johann Park 2B

When asked to just name the hybridization orbital, you would, for example, write sp³. However, when discussing the hybridization bonding model, you would refer to an atom's specific hybrid orbital (for example: 2sp³) overlapping with another orbital.

Johann Park 2B

Knowing that the volume of the containers are the same, we only have to look at the amount of O 2 in each container. Our products and reactants are the same in both containers, and only the amount (moles) of O 3 changes from .10 to .50 -> therefore, O 2 will also have more moles in container 2 than ...

Johann Park 2B

We know that ethylenediamine is bidentate because it has 2 atoms that would coordinate to Pt in this case. ethylenediamine, C₂H₄(NH₂)₂, uses both its NH₂ atoms to bond with the metal atom

Johann Park 2B

I think this is a matter of notation but when asked the hybridization of the molecule, do we have to write the coefficient, notating quantum # n, with it or does it matter?

E.g. Ammonia, NH₃: "sp³" or "2sp³"?

Johann Park 2B

Hybridization of orbitals requires a "higher state of energy" of an atom (like the Carbon example Prof. Lavelle used in lecture for CH 4 ) and thus only occurs when it is involved with bonding. Really I believe it is to be considered as a model to describe bonding and explain observed stru...

Johann Park 2B

If I am understanding your question correctly, in looking at BrO + , we're split between choosing a double bond, Br and O with two lone pairs vs. a triple bond where Br has two lone pairs and O has one? Because we can draw the structure of :Br: = :O:, with a double bond and two lone pairs on each at...

Johann Park 2B

The midterm covers up to Chapter 3, which is Lewis structures; this means that you won't need to apply the specific molecular shapes. I also believe we won't need to apply certain placement of lone pair electrons because the VSEPR model (where we minimize e- repulsion) is in Chapter 4 and Prof. Lave...

Johann Park 2B

Prof. Lavelle said the correct answer (of Octet O and lone Cl) is based off molecular orbital arguments just outside the scope of this course (everything plus what was said above^). A Lewis structure connecting Cl and O with a single bond, giving Cl the octet and O the single lone electron would res...

Johann Park 2B

Yes, you will basically write the electron configuration the same way as a neutral ground-state atom. You must still consider Aufbau's Principle, Hund's Rule, and the Pauli Exclusion Principle. You will add to the outermost orbital, then to the next higher orbital.

Johann Park 2B

In many-electron atoms, electron-electron repulsion causes increasing orbitals to be increasing in energy (because electrons in "greater" orbitals are shielded from the full attraction of the nucleus & we know that when electrons are more strongly attracted to the nucleus, they are low...

Johann Park 2B

P x , P y , and P z just specify the different locations/orientation of electrons in a P-orbital (Whether an electron lies on the x-/y-/z-axis on the 3D Cartesian graph). I believe that Prof. Lavelle said this concept is only to help our understanding of these different orbitals, so when writing the...

Johann Park 2B

If I am interpreting your question correctly: EMR is able to oscillate through empty space because EM waves don't need a medium to propagate itself. If we were to imagine a separated electromagnetic field - the electric field generates a magnetic field at a different angle and vice versa. In short, ...

Johann Park 2B

Looking at my notes from one of the lectures, I remember Prof. Lavelle saying that anything ≤ 10 -15 m is too small to detect. In the examples he used, a classical object had something of a wavelength of 1.64x10 -34 m, which is way too small. While for example, an electron at 5.3x10 6 m/s would resu...

Johann Park 2B

When talking about Electromagnetic radiation, you are correct that as frequency decreases, the wavelength must increase due to their inverse relationship. When we look at how the electric and magnetic fields oscillate as light goes through space with a decreased frequency, the field has less energy ...

Johann Park 2B

I believe that it would be from the wave model to the particle model. The textbook notes that the results of some experiments (Photoelectric effect) compelled scientists to view that electromagnetic radiation is particlelike.

Johann Park 2B

Usually in these cases, the mass of the two products will be given, as well as the mass of the unknown compound. Normally this will be a combustion question. In combustion, you have O 2 reacting with the compound to produce Carbon Dioxide and Water vapor. For example, if you have the oxidation: C x ...

Johann Park 2B

In Problem F21, after you find all of the mmol values of the elements in didemnin-A, the smallest value is 0.0113 mmol of Nitrogen, so you would divide all values by 0.0113, leaving you with 8.18 C : 13.0 H : 1.00 N : 2.00 O. Note that this isn't an incorrect ratio, but you still have to multiply ea...

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