CHEMISTRY COMMUNITY

kaushalrao2H

even if the steps of the reaction mechanism aren't explicitly labeled as slow and fast, you should just try writing the rate law for each of the steps and seeing which one matches the experimentally determined rate law for the overall reaction. if the rate law for a step in the mechanism matches the...

kaushalrao2H

catalysts are not used up in the reaction, so you will see the catalyst in BOTH the product and reactant sides of the chemical equations. intermediates, on the other hand, so not appear in the chemical reaction equation.

kaushalrao2H

in general, the only way to determine the reaction order is experimentally or via the reaction mechanism.

kaushalrao2H

yeah i think for this class we do not need to consider the products' effects on the overall rate law for a reaction. however, if a product DOES have an effect on the overall reaction, then the reaction order for the product concentration would be a negative value in the overall rate law equation.

kaushalrao2H

you should be able to add all of the reaction mechanism (step) reactions together in a Hess's law type manner in order to get the final chemical equation.

kaushalrao2H

since the unit for rate is mol/(liters*seconds), you'd want to ideally use those units for the calculations.

kaushalrao2H

use mols. in the book, some problems give amounts of a substance in terms of mmol (millimol), but you should try to convert this amount to mols by multiplying this value by 10^3.

kaushalrao2H

for the purposes of chemistry 14b, only worry about zero order, first order, and second order reactions.

kaushalrao2H

the rate of reduction/formation of a reactant/product divided by its respective stoichiometric coefficient. every single species in a chemical reaction should have the same unique rate.

kaushalrao2H

in general, remember that the rate constant k is dependent on temperature and the activation energy of the reaction.

kaushalrao2H

electrodes have to be solid, and if none of the species in a reduction or oxidation reaction are solid (such as when two Fe species have different oxidation numbers but are both in aqueous form), then there must be an inert conductor to relay electrons.

kaushalrao2H

only assume that the reaction is not at standard conditions when the prompt itself gives some conditions (such as varying concentrations of reactant/product).

kaushalrao2H

yes, over time, if a reactant is decreasing, then the sign on the rate expression would be negative. however, if a reactant is increasing over time, then the sign on the rate expression would be positive.

kaushalrao2H

if we calculate a cell potential and it comes out negative, then we know that the forward process is nonspontaneous while the reverse process IS spontaneous. therefore, the cell can be a galvanic cell going in the reverse (spontaneous) direction. a cell can be considered galvanic if there is some sp...

kaushalrao2H

in a nonreversible process, S(system) is most likely not equal to S(surroundings), so S(total) would not be equal to zero.

kaushalrao2H

Oxidizing agents oxidize the species that it is reacting with, and reducing agents reduce the species it is reacting with

kaushalrao2H

I think it's fine to leave it either way.

kaushalrao2H

A common material for salt bridges is KCl. The K+ ions dissociate into the cation tub, and the Cl- ions dissociate into the anion tub to counter charge buildup and allow ion transfer.

kaushalrao2H

Yes, the salt bridge essentially counters charge buildup as it allows ion transfer. A common salt bridge is a KCl bridge.

kaushalrao2H

Yes, you are correct. Enthalpy of reaction can also be the enthalpy of formation if the reaction is simply the compound being formed from its constituent elements.

kaushalrao2H

Cp for ideal gases is 4R. Cv for ideal gases is 3R. When in doubt, remember that Cp = Cv + R

kaushalrao2H

Heat and work are not state functions; they are considered "path-dependent" functions

kaushalrao2H

residual entropy is the entropy of a sample at T=0 (absolute zero) arising from positional disorder surviving at that temperature. Gases tend to have higher residual entropy than solids and liquids because they have smaller dipole moments (leading to less orderly arrangements).

kaushalrao2H

Hm refers to the molar enthalpy of the solid or liquid. and yes, i think we need to know to use these delta H for phase change problems.

kaushalrao2H

one Calorie is just 1000 calories!

kaushalrao2H

condensation, freezing, deposition!

kaushalrao2H

The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions.

kaushalrao2H

in reversible expansion, the maximum amount of work is being done by the system because external pressure remains equal to the internal pressure as the gas expands. in irreversible expansion, the external pressure is constant and work is not maximized.

kaushalrao2H

if substance 1 has a smaller specific heat capacity than substance 2, then one gram of substance 1 needs less heat than substance 2 to increase its temperature by 1 degree Celsius.

kaushalrao2H

Yes the above reply is correct. Recall that in a closed system, energy can be transferred by the system to the surroundings, but matter cannot be transferred by the system to the surroundings. Therefore, the internal energy of the system can be changed by work and heat applied to or by the system to...

kaushalrao2H

the qreaction is equal to the qcalorimeter, as related by the equation qreaction = -qcalorimeter. all the heat released from the reaction would be absorbed by the calorimeter.

kaushalrao2H

isothermal reversible expansion describes gas expansion whose direction can be reversed by inducing infinitesimal changes in the internal or external pressure. the pressure on the inside has to be a LITTLE bit more than the pressure on the outside, because the gas would not expand if the pressures w...

kaushalrao2H

Delta U represents a system's internal energy, or total store of energy in a system. If delta U is zero, then there has been no energy transfer (work or heat) between the system and surroundings.

kaushalrao2H

The book quotes, "If the external pressure were to be increased even infinitesimally at any stage of the (ideal gas) expansion, the piston would move in instead of out. Therefore, the work done during a reversible expansion of a gas is the maximum expansion work possible."

kaushalrao2H

a PDF is fine for discussion!

kaushalrao2H

imagine the ice cube as the system, and its immediate surroundings as the surroundings. in order for the ice cube to melt, it needs to absorb heat from its surroundings in order to reach its melting point. thus, the phase change is endothermic because the system takes heat from its surroundings (del...

kaushalrao2H

for K (acid/base) constants, we only use concentrations of aqueous or gaseous substances, and leave out pure substances (solid and liquid).

kaushalrao2H

i think either way is fine, but if you write the formula as HCOO-, you emphasize the point that the carboxyl group in HCOOH has lost a H+ ion.

kaushalrao2H

the reaction is O2 + H20 --> 2(OH-), so the book is just saying that OH- is the conjugate acid of O2 without specifying that there are two of them per reaction.

kaushalrao2H

All "central" atoms (atoms that are bonded to more than one atom) have hybridized orbitals. All "non-central" atoms (only bonded to one atom) do not have hybridized orbitals.

kaushalrao2H

You wrote the question wrong. I think you combined portions of 11.37 and 11.38.

kaushalrao2H

I think that was a book typo. The ligands should always be in alphabetical order.

kaushalrao2H

We'd just need to label the bonds using the atoms' respective orbitals. For example, for the four single bonds in CCl4, the bond notation would be (C2sp3, Cl3p)

kaushalrao2H

ligands definitely impact a change in magnetic properties and color of the transition metal in an aqueous solution. CFT qualitatively describes the strength of metal-ligand bonds and describes how the presence of ligands contributes to the breaking of orbital degeneracy in transition metal complexes

kaushalrao2H

a coordination compound is a neutrally charged compound where one of its ions is a coordination complex, which in turn consists of a central transition metal and a set of attached ligands (bonded by coordinate covalent bonds). these do not relate to hybridization, but we would sometimes need to know...

kaushalrao2H

A ligand is a Lewis base (donates electrons) that forms a coordinate covalent bond with the central transition metal of a coordination complex. They can be either ions or molecules. The number of ligands that can be bonded to a particular transition (d-block) metal is determined by the metal's coord...

kaushalrao2H

A coordination complex is a potentially charged complex consisting of a central transition (d-block) metal and its attached ligands, which are Lewis bases that can be either ions or molecules. A coordination compound is a neutral compound in which one of the ions is the coordination complex. A coord...

kaushalrao2H

i think the coordination number just depends on the specific transition metal used, but know that the number of coordinate metal-ligand bonds generally increases as the positive charge on the metal ion becomes larger

kaushalrao2H

yes we should!

kaushalrao2H

Only diatomic molecules (ex. H2) have a truly nonpolar bond. If the two elements involved in the bond are different, there's going to be at least a little bit of electronegative difference, which results in a slightly polar covalent bond. Polarity of an entire molecule is different than when we are ...

kaushalrao2H

Adding on to the previous reply:

Truly nonpolar covalent bonds arise only when the two elements bonded are identical (ex. diatomic molecules). If the two elements are different, then there's always going to be a little bit of electronegativity difference that results in a polar covalent bond.

kaushalrao2H

Electron affinity generally follows the same trend as ionization energy (with some exceptions, such as how carbon has a higher electron affinity than nitrogen).

kaushalrao2H

think of the quantum numbers as a progression of specificity with regards to identifying a particular electron of an atom. n -> l -> ml -> ms energy level of electron -> subshell within the energy level (n) where the electron is located -> the specific orbital within the subshell (l) in which the el...

kaushalrao2H

Hi, I think there's a typo in the book! Or, the book meant the 5th group OF transition metals (group 7 on the periodic table). If we are talking about group 5 on the periodic table however, you are right. Transition metals in group 5 (such as Vanadium) should have the electron configuration (n-1)d^3...

kaushalrao2H

In lecture, Lavelle covered only the Copper (Cu) and Chromium (Cr) exceptions to the standard electron configuration rules. I don't think you'd need to know any of the other exceptions.

kaushalrao2H

Lets say that we are describing a valence electron in Aluminum. The principle quantum number (n) would be 3, the angular momentum quantum number (l) would be 1 (because 1 signifies the p orbital), the magnetic quantum number (m sub l) would be -1 (because the ml numbers -1 0 and 1 represent the px, ...

kaushalrao2H

I'm pretty sure Test 3 will include only 1.6, 1.7, and Chapter 2.

kaushalrao2H

Nope! For this week's test, you just need to know the equations introduced in 1.1-1.5 in the textbook. You do not need to know the Heisenberg uncertainty or Schrodinger equations for the test, but you do need to know the De Broglie equation, atomic spectra equations (such as [v=R((1/nf^2)-(1/ni^2))]...

kaushalrao2H

For part B you would need to convert the 5.00 mg of sodium atoms into mols of sodium, convert that mol value into the number of atoms (using Avogradro's number), and then multiply the number of atoms by the energy (J) value you found in part A. For part C you basically do the same thing except you k...

kaushalrao2H

I too was a bit confused with that section, but I think the main idea is the idea of Heisenberg uncertainty and that both the position and momentum of a small object cannot be known at the same time. You're right about everything you mentioned in your post also (the stuff about probability density, ...

kaushalrao2H

λν=c

Just think of this formula when answering questions related to wavelength (λ) and frequency (v). The product of the two must always equal c (3.00 x 10^8), so if either the wavelength or frequency decreases, then the other must increase.

kaushalrao2H

Sometimes, when the molar ratios aren't as clean as 1:1:1.33, you're going to have to guess and check and see what coefficient you need to multiply the molar values by.

kaushalrao2H

Q: What's a show that cesium and iodine love watching together?

A: CSI

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