CHEMISTRY COMMUNITY

sahiltelang-Discussion 1J

You can use the pre-equilibria step and use the previous elementary step. You treat the previous step as an equilibrium reaction as there is a buildup of product as the proceeding step is slow. Therefore, we can calculate the equilibrium K in terms of the reactants and products of the elementary ste...

sahiltelang-Discussion 1J

The reducing agent is O2 as seen in the half reaction: 2OH- +O2 --> 2e- + H2O +O3. One of the oxygens from the hydroxide goes from an oxidation state of 2+ to a neutral charge when forming ozone with the oxygen gas. thus it gives off 2e-s.

sahiltelang-Discussion 1J

You would have to separate them by a line.

sahiltelang-Discussion 1J

The pre-equilibria technique is regarding the balanced reaction equation and does not deal with the various elementary steps. Therefore, one does not need to worry about the slow step.

sahiltelang-Discussion 1J

A reversible isothermal expansion is thought to provide the maximum work as there is constant pressure. This allows for all the available energy to be used for pushing the piston rather than being used up by the gas to adjust the pressure to the changing volume.

sahiltelang-Discussion 1J

lnQ and logQ are conceptually the same yet matematically not. ln is the natural log or log base e whiles log is just log base 10. Thermo equations use log sometimes when dealing with ph's as it makes it easier to solve for pH or pOH rather than needing to convert between ln and log. You can calculat...

sahiltelang-Discussion 1J

We are able to determine this if we see that the reaction rate does not change if we change the concentration of this reactant significantly.

sahiltelang-Discussion 1J

For reactions where we see multiple reactants, we design an experiment such that all the reagents' concentrations are constant so that we can change one significantly(which would be at a low concentration) and see how it changes the rate law. Thus by doing this for reactant we can determine the orde...

sahiltelang-Discussion 1J

To find the overall rate of reactions with multiple reactants, one needs to use experimental data to find the rate law (and thus the order of each reactant). To do this, we make the concentration of all except one reactant so that we can assume constant concentration while observing how only one rea...

sahiltelang-Discussion 1J

The main thing you have to remember when you use the Cathode-Anode equation is that both the Enaught values that you use have to be reduction half reaction potential values.

sahiltelang-Discussion 1J

technically, the hydrogen could gain an electron and then produce H2, but most problems that we deal with don't seem to be concerned with this.

sahiltelang-Discussion 1J

No, but it is good to know that liquids and solids don't get put in the rate law and that the rate law is only defined by gasses and aqueous molecules/ atoms. So most likely the reactants expressed in the rate law will be in the same phase.

sahiltelang-Discussion 1J

They derive this equation by using Appendix 2B and coupling the half rxns for the oxidation and reduction of T1+.

sahiltelang-Discussion 1J

Another thing that is important is that if you have different substances that are part of the reaction but do not have an interface but rather together in solution, you group them together with commas separating them. If you have both the products and reactants for the oxidation/reduction that are b...

sahiltelang-Discussion 1J

The Cl is the oxidation step such that Cl- becomes Cl and loses an electron as it forms a covalent bond with another Cl that has also lost an electron.

sahiltelang-Discussion 1J

You can use entropy, enthalpy and temperature to calculate deltaG and then use this value to find the original K using deltaG=-RTln(K). Once K1 is calculated then you can use the Van Hoff equation.

sahiltelang-Discussion 1J

One of the things that we look for if we see a decrease or increase in high entropy substances. For example, if there is a decrease in number of moles of gas from reactants to products that is an example of entropy decrease. Also if we go from many smaller simple compounds to a fewer number of compl...

sahiltelang-Discussion 1J

Since it states assume ideal behavior, we can assume that it is 1 mol of gas within the combustion engine

sahiltelang-Discussion 1J

this is the same equation as deltaS=q/T, except that it shows that this is the maximum amount of entropy. It just mathematically depicts the fact that all the calculations we do are regarding ideal conditions. This equation just shows that it is possible to have less entrophy in a non-ideal scenario.

sahiltelang-Discussion 1J

Pressure is a way of measuring the concentration of a gas in a certain volume. Therefore, B and C forms diatomic molecules and have less atoms and therefore less pressure in this problem in comparison to A.

sahiltelang-Discussion 1J

Another thing that we can look at to determine this is by looking at the deltaG of the products vs the reactants. When both of these two values equal each other then we know that the reaction is at equilibirum. If not then we can calculate the reaction quotient and see how the reaction will proceed

sahiltelang-Discussion 1J

Since it states ideal behavior, we can assume that it is 1 mole of substance.

sahiltelang-Discussion 1J

both components that add up to the change in internal energy are path-dependent, but deltaU is not dependent. Most energy values are state functions, because they can conceptually be thought of as a position. Just like you can have a displacement in your position, the current energy state can be dis...

sahiltelang-Discussion 1J

They are no truly 100% isolated systems in the universe. We have devices though such as bomb calorimeter that are practically isolated so we can just treat them in that manner.

sahiltelang-Discussion 1J

you have to set up the -PV=nRT and for part a make volume constant and make it -deltaP*V=-deltan*R*T. For part b do the same thing, but then make it -P*deltaV=-deltan*R*T. This will give you work done.

sahiltelang-Discussion 1J

We usually use atm and standardized all the values to this as all the constants we have for R have atm. 760 torr =1 atm, 1 atm = 1.01325 bar, and 1 atm =101325 pascals.

sahiltelang-Discussion 1J

The idea behind a calorimeter is that it isolates the reaction such that the reaction occurs in the first insulated chamber and whatever change in energy (heat) occurs in the first 'chamber' causes for the same change in the second layer of insulation that is measured. Thus the change in temperature...

sahiltelang-Discussion 1J

you add on the sublimation of graphite as part of the reactants in the equation for calculating the estimated enthalpy of formation. This value is provided in the problem.

sahiltelang-Discussion 1J

In most problems they usually provide you the phase they are in (usually in the subscript of the chemical equation). If not, then you can look at the temperature of the reaction to determine the state. If temperature is not given, assume it is at 25 degrees C (standard conditions).

sahiltelang-Discussion 1J

Based on the problem we can find out what they are asking. If it asks how much energy for a certain amount of reactants in a given equation then provide kJ, but on the other hand when they ask for the heat of formation then provide it as kJ/mol. For the problem you mentioned it asks for a standardiz...

sahiltelang-Discussion 1J

Yes, we get the value 14, by taking the negative log of Kw. Thus Kw describes the autoprotolysis of H2O at 25C. At different temperature values there are different Kw values and thus the pH and pOH add to a different value, albeit still being the negative log of Kw.

sahiltelang-Discussion 1J

here we know that it is a base, and therefore we know that we are going to add an additional proton to the complex. Thus we will add to the ammine group which is more likely than oxygen, which is more electronegative, to have a proton being added to it. It would not be correct to use the NH2OH2, bec...

sahiltelang-Discussion 1J

I think examples wise we should know the ones mentioned in the lecture - the semimetal hydroxides- for the amphoteric compounds and then water for the amphiprotic compound. The concepts behind it is that amphoteric compounds can act as acids and bases and amphiprotic compounds can both accept and gi...

sahiltelang-Discussion 1J

Take the concentration of hydronium or hydroxide ions and divide it by the original concentration of the weak acid or base. Then multiply it by 100 to get the percentage.

sahiltelang-Discussion 1J

You have to mathematically derive Q, the reaction quotient to determine the direction of the reaction. By comparing it to K, one can then use a ICE table to determine the equilibrium concentrations or partial pressures.

sahiltelang-Discussion 1J

solvents are not included in equilibria. Take a dissociation equation of KCl in H2o. The reaction equilibrium constant K is determined by the products, K+ and Cl- ions, and is not dictated by H2O or the KCl solid. It is important to note though that the H2O is necessary to break up the ionic bond to...

sahiltelang-Discussion 1J

For 41, the compound is a chain of 3 carbons such that the first two have a double bond the second and third have a single bond. The first carbon has two bonds with H, the second with one bond with H and the last C with a triple bond with N. The N has a lone pair. The first carbon has three regions ...

sahiltelang-Discussion 1J

Hemoglobin is comprised of four myoglobin. Each myoglobin has a imidazole ligand compound that can bind with an electron rich compound such as carbon monoxide or more preferably O2.

sahiltelang-Discussion 1J

It does matter as different atoms are more electronegative than others and have different needs to become stable. In your example with CH3, H already has a bond thus giving it a full shell of electrons, making it improbable that it would have that radical. Carbon on the other hand is lacking a bond,...

sahiltelang-Discussion 1J

It is due to the fact that the jump from the nitrogen family to the oxygen family introduces a new electron in the p orbital such that it is the first to be paired. Having a complete half sub shell is much more stable than that addition of the electron pair.

sahiltelang-Discussion 1J

Phosphorous can expand to the d orbital allowing it to bond more, beyond its octet.

sahiltelang-Discussion 1J

The d orbital is a wave function that is derived from Schrodinger's equation. Based off of this periodic function, we are able to model the d orbital spatially.

sahiltelang-Discussion 1J

This is because Oxygen is more electronegative and thus would have the lone pair rather than Cl. Thus since Cl does not have as strong of a grip on e-, O will have the lone pair while Cl will have the lone electron.

sahiltelang-Discussion 1J

Yes, this is the way to complete these problems but you have to keep in mind about the exceptions in the column starting with chromium and the column starting with copper as they only have one in the s, thus you would be taking two off the d.

sahiltelang-Discussion 1J

I believe that the test will cover everything from last Wednesday and Friday's lecture till material from October 27th.

sahiltelang-Discussion 1J

Atoms always want to be in the lowest energy state possible. Thus it requires less energy for an atom to have electrons in a 4s orbital rather than a 3d orbital. This can get confusing bc 4 is considered a higher energy level.

sahiltelang-Discussion 1J

Our TA remarked about this in our discussion section to use the constants provided by Professor Lavelle on his website (subsequently the same for the test).

sahiltelang-Discussion 1J

It it possible for an electron to move to a lower energy level. It doesn't effect the equation but rather you just need to make sure you do FINAL - INITIAL for change in energy.

sahiltelang-Discussion 1J

for combustion chem equations- the hydrocarbon mostly reacts with oxygen to form water and carbon dioxide. if the original reactant contains nitrogen it will still react in the same manner but it will also give off nitrogen gas as a product.

sahiltelang-Discussion 1J

What are the differences between the Molecular and Empirical formulas? More specifically, how do you identify one from the other? The empirical formula is the ration of atoms of a compound while the molecular is the actual number of atoms. For example, in a room of children there is 20 boys and 25 ...

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