CHEMISTRY COMMUNITY

Sandhya Rajkumar 1C

By plotting the experimental data vs time, you would be able to identify the order of the reaction depending on whether [A] vs time, ln[A] vs time , or 1/[A] vs time gave you a straight line plot, where A is the reactant. Also, the slope would tell you what k is.

Sandhya Rajkumar 1C

An open system would be a beaker of water, where matter and energy can exchange with the surroundings. A closed system would be a sealed beaker of water, where only energy can exchange with the surroundings. In an isolated system, such as a bomb calorimeter, neither matter nor energy can exchange wi...

Sandhya Rajkumar 1C

The bond strength between a given pair of atoms varies slightly from one compound to another because of the different electronegativities of the other atoms in the compound. When we use bond enthalpy values to calculate standard enthalpy, we are actually using the mean bond enthalpy which is the ave...

Sandhya Rajkumar 1C

Also, the section in the textbook on collision theory discusses reaction profiles, which were also mentioned in class, so I think knowing/understanding those would be useful.

Sandhya Rajkumar 1C

Also, once you've solved for k2, if you plug it in to ln(k1/k2) you get -0.59. But if you plug k1 and k2 into ln(k2/k1) you get 0.59. So to answer your question, ln(k1/k2) doesn't equal 0.59, it equals -0.59.

Sandhya Rajkumar 1C

The higher the activation energy, the stronger the temperature dependance of the rate constant. In this reaction, the forward reaction has a higher activation than the reverse reaction, which is why k of the forward reaction will increase more than k of the reverse reaction.

Sandhya Rajkumar 1C

For part b, the balanced equation shows that for every 2 moles of NH3 formed, 3 moles of H2 are consumed. This means that the rate of formation of NH3 would be 2/3 of the rate of consumption of H2. We do the same thing for part c. For every 2 moles of NH3 formed, 1 mole of N2 is consumed. This means...

Sandhya Rajkumar 1C

The reaction is second order with respect to B because when the amount of B is doubled (increased by a factor of 2), the rate increases by a factor of 4, so instead of it being a 1:1 ratio, which would make it first order, it's a 1:2 ratio, which makes it second order.

Sandhya Rajkumar 1C

The rate of the reaction has units mol*(L^-1)*(s^-1) For a zero order reaction, the rate = k*[A]^0= k, where k is the rate constant. In order for the rate to have units mol*(L^-1)*(s^-1) the units for k have to be mol*(L^-1)*(s^-1) For a first order reaction, the rate = k *[A]^1 and the units for [A...

Sandhya Rajkumar 1C

From my understanding of it, if your stoichiometric coefficient is now 1/a instead of a, you plug in 1/a to where you would plug in a. This leaves you with
(-1/(1/a))*(d[A]/dt) which simplifies to -a*(d[A]/dt)

Sandhya Rajkumar 1C

The concentration of the reactants will decrease over the reaction, so without the negative sign, the reaction rate for the reactants would be a negative number, as it is calculated from the change in concentration over the change in time. However, we want to work with positive reaction rates, so by...

Sandhya Rajkumar 1C

Also, don't forget, you need to use a negative sign when referring to the reactants. For example, if the reaction was aA + bB --> cC + dD,
rate = -(1/a)*(d[A]/dt) = -(1/b)*(d[B]/dt) = (1/c)*(d[C]/dt) = (1/d)*(d[D]/dt)

Sandhya Rajkumar 1C

I believe it will be on electrochemistry and everything that we covered in chapter 14. The assigned homework problems from the electrochemistry section should cover all of the topics we need to know.

Sandhya Rajkumar 1C

After obtaining the half-reactions, you have to multiply the oxidation half-reaction (part b) by 3 so that both half-reactions have the same number of electrons involved. Afterwards, you combine the two equations.

Sandhya Rajkumar 1C

Also, because the problem specifies that the reaction occurs in an acidic solution, you know to add H+ and not OH-.

Sandhya Rajkumar 1C

I was told that we won't be using ICE tables in this class.

Sandhya Rajkumar 1C

I don't think it would hurt to know the derivations, especially because knowing/understanding them could potentially help with conceptual questions.

Sandhya Rajkumar 1C

If you are given ΔG°, you can calculate K by first isolating it. Once you have lnK = ΔG°/-RT you then get rid of the natural log in front of the K by using e. You end up with e^(lnK) = e^(ΔG°/-RT) and the e^ln cancels out, so K = e^(ΔG°/-RT).

Sandhya Rajkumar 1C

The problem asks us to calculate entropy and gives us the number of molecules and the number of arrangements. Using these 2 things, we an calculate the degeneracy(W). To calculate the degeneracy: (# of arrangements)^(# of molecules). For part (a), we know there are 64 molecules aligned in the same d...

Sandhya Rajkumar 1C

When delta H is positive, the reaction is endothermic and is only spontaneous at high temperatures (unless delta S is negative, in which case the reaction is not spontaneous at any temperature, but the reverse reaction is spontaneous). Because this reaction requires heat, it is not favorable, and th...

Sandhya Rajkumar 1C

Yes, they are similar because they are both state functions. To calculate the standard reaction entropy, you subtract the sum of the standard formation entropies of all of the reactants from the sum of the standard formation entropies of all of the products.

Sandhya Rajkumar 1C

Internal energy (U) us a state property because changes in internal energy are a function of initial and final states. It doesn't depend on the path taken, it just depends on the initial and final state. (delta U = Ufinal - Uinitial)

Sandhya Rajkumar 1C

Reaction enthalpies are usually in kJ because we have to multiply either the formation enthalpies or the bond enthalpies (usually given in kJ/mol) by the number of moles, so moles would cancel out and you'd be left with kJ.

Sandhya Rajkumar 1C

In order to increase the energy of an open system, you can add substance to the system, heat the system, or do work on the system. In order to increase the energy of a closed system, you can heat the system or do work on the system. The energy of an isolated system is constant because no matter or h...

Sandhya Rajkumar 1C

I think that atm is best for pressure and L is best for volume because then you can just convert from L.atm to J (1 L.atm = 101.325 J).

Sandhya Rajkumar 1C

Conversely, you could also remove substance from the system, cool the system, or let the system do work on the surroundings.

Sandhya Rajkumar 1C

In the diagram, I believe that the volume is area x displacement of the piston because we need to calculate the change in volume (delta V) not the volume of the entire system. And so in order to calculate the change in volume, we would multiply the area and the distance that the piston has been disp...

Sandhya Rajkumar 1C

I remember that endothermic reactions require/absorb energy because of the root (endo) and that they have a positive delta H because if they absorb energy, then the final product of the reaction will have more energy than the initial reactants (change = final - initial). Exothermic reactions release...

Sandhya Rajkumar 1C

You would make it negative, because for fusion, the process is endothermic and delta H is positive, which means the reverse reaction, freezing, would be exothermic and have a negative delta H.

Sandhya Rajkumar 1C

The reactants have positive enthalpies because the process of breaking the bonds in the reactants requires/absorbs heat/energy, which means it would be an endothermic reaction, which is why delta H is positive. The products have negative enthalpies because the process of forming the bonds for the pr...

Sandhya Rajkumar 1C

In that problem I don't think the concentrations are halved, it says that the stoichiometric coefficients are halved. And because the stoichiometric coefficients become the exponents, taking the square root would be the equivalent of multiplying them by 1/2.

Sandhya Rajkumar 1C

For Lewis acids and bases, I remember that the Lewis acid is the lone pair electron acceptor because acid and acceptor both start with the letter A.

Sandhya Rajkumar 1C

I saw a chart in the textbook that specified some of the polydentate ligands, it's in chapter 17, table 17.4.

Sandhya Rajkumar 1C

In class, we learned that if a reaction requires heat (is endothermic) while forming product, then heating would favor product formation. I read something that explained this using equations: heat+6CO2(g)+6H2O(l)⇌C6H12O6(aq)+6O2(g) In endothermic reactions, heat is on the side of the reactants, so a...

Sandhya Rajkumar 1C

I first find out what the central atom is, and then I draw a single bond between the central atom and each of the other atoms. I then fill in all of the remaining electrons with either bonds or lone pairs and try to keep the formal charge as close to 0 as possible.

Sandhya Rajkumar 1C

As the reply above said, the constant (R) usually used in this class is 0.08206. This constant is used when pressure (P) is measured in atm, volume (V) is measured in liters (L), n is the moles of gas, and temperature (T) is measured in Kelvin (K).

Sandhya Rajkumar 1C

An electron cloud of a lone pair can take up more space than an electron cloud of a bonding pair because it's less restrained because it's held in place by only 1 atom, not 2. This leads to greater repulsion.

Sandhya Rajkumar 1C

Cis- isomers have 2 of the same functional groups on the same side of the double bond, while in trans- isomers they are on opposite sides of the double bond.

Sandhya Rajkumar 1C

Dipole moments can occur between ions in an ionic bond or atoms in a covalent bond. They occur when atoms in a molecule share electrons unequally, which occurs because one atom is more electronegative than the other. The more electronegative atom pulls more tightly on the shared electrons, creating ...

Sandhya Rajkumar 1C

Dipole moments occur when atoms in a molecule share electrons unequally, which occurs when one atom is more electronegative than the other. The more electronegative atom will have a stronger pull on the electrons, creating a separation of charge. The greater the difference in electronegativity, the ...

Sandhya Rajkumar 1C

To add on to the previous reply, Pauli Exclusion Principle states that there can be no more than 2 electrons per orbital and that if 2 electrons are in the same orbital then they spin paired.

Sandhya Rajkumar 1C

I feel like the most important thing for me in terms of studying is doing all of the assigned homework problems, and making sure that I understand the concepts behind them. In terms of online resources, I really only use the video modules that Dr. Lavelle posted on the website.

Sandhya Rajkumar 1C

The atomic radius is the distance between the atomic nucleus and the outermost stable electron of a neutral atom (which is found by measuring the distance between the nuclei of 2 atoms that are touching, and halving that distance.) The ionic radius of an element is its share of the distance between ...

Sandhya Rajkumar 1C

As test 2 covered the assigned homework problems 3-41 from chapter 1, for test 3 it would probably be a good idea to complete the rest of the assigned problems for chapter 1 as well as the assigned problems for chapter 2. And as mentioned above, the tests are in our discussion sections beginning Oct...

Sandhya Rajkumar 1C

Also, he mentioned in class that for electron configurations, we need to know the s-block, the p-block, and the first row d-block.

Sandhya Rajkumar 1C

To add on to the last reply about matter, relatively large objects primarily exhibit particle properties, but to explain the behavior of very small objects, like electrons, both wave and particle properties must be considered.

Sandhya Rajkumar 1C

If you are trying to compare to other wavelengths, converting to nanometers would probably make things easier.

Sandhya Rajkumar 1C

You can use c = λ ν to find the wavelength when you are given the frequency, or vice versa. You can combine this formula with E = h ν to get E = hc/λ to solve for either the photon energy or the wavelength, as long as you know one of the 2 variables.

Sandhya Rajkumar 1C

You could just use M = n/V, you just have to convert the grams of KCl to moles, and then use .125L as your final volume.

Sandhya Rajkumar 1C

In ionic compounds, cations and anions are packed tightly in a crystal lattice pattern and form a giant structure, which is why we refer to them formula units as opposed to molecules.

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