CHEMISTRY COMMUNITY

Jainam Shah 4I

An intermediate is formed in one step and then used up in the next. A catalyst isn't formed as it is present in the beginning and end. The catalyst is not consumed as the reaction progresses.

Jainam Shah 4I

The concentration of one substance has to be constant for you to determine the rate of the other substance whose concentration is changing at different times in the reaction.

Jainam Shah 4I

Its energy so joules or kilojoules are acceptable. However, when using the Arrhenius equation or something of that sort ensure units cancel properly.

Jainam Shah 4I

Ssystem + Ssurroundings = Stotal(universe)

Jainam Shah 4I

K' is the rate constant for the reverse reaction which is why k/k' is the equilibrium constant K

Jainam Shah 4I

The reactants are on top of the enzyme. The enzyme is catalyzing the substrates and that region is also known as the active site.

Jainam Shah 4I

Also, when a catalyst is present the reaction can be zero order, because the catalyst controls the reaction rather than the concentration or amount of reactant present making the reactants a negligible part of the reaction.

Jainam Shah 4I

Yes you can it just changes the equation, but both equations from the sheet when used correctly would give the same answer. Make sure you check units while using either and use the correct given temperature

Jainam Shah 4I

Ka is the disassociation constant for a weak acid and for Kb its for a weak base. You typically use them in ICE tables to find how much an acid or base disassociates. If you are dealing with H+ or H30+ use Ka and for OH- use Kb

Jainam Shah 4I

If the graph of the concentration of the reactants versus time is a negative sloping linear line then it's a zero order reaction. The slope would be just k.

Jainam Shah 4I

The order would either be given or you would given a plot of the concentration versus time in some manner from which you can figure out the order.

Jainam Shah 4I

The water doesn't affect the charge of any substance in the half reactions. It isn't participating in neither the oxidation or reduction so it is left out.

Jainam Shah 4I

The derivation is for a more clear understanding and I believe the equation sheet gives us majority of the equations needed.

Jainam Shah 4I

Yes galvanic cells are always spontaneous and they convert chemical energy into electrical energy. They are essentially batteries which has a positive cell potential hence being favorable.

Jainam Shah 4I

If you are using the standard reduction potential table from the appendix then just use E(not) of the cell = E(not)cathode - Not(anode). That equation accounts for the anode potential to be the standard oxidation potential.

Jainam Shah 4I

Yes you would include water if it is in the gaseous phase.

Jainam Shah 4I

Q is used when the reaction is not at equilibrium

Jainam Shah 4I

Usually the metal electrode that is being oxidized is losing mass in terms of the metal and is dissolving into the aqueous solution.

Jainam Shah 4I

Kinetics is related with the speed of a reaction not the spontaneity. Thermodynamics is the better indicator about whether a reaction is favorable or not.

Jainam Shah 4I

For this class I believe so. But always look at the half reactions to be completely sure

Jainam Shah 4I

The oxidation agent causes another substance to be oxidized and in the reaction the oxidizing agent is being reduced.

Jainam Shah 4I

Use the cell diagrams to come up with the reduction and oxidization half reactions. From that go to Appendix 2B and find the standard potential differences for each reaction. Use Ecathode-Eanode to find the Ecell and don't reverse the potentials from the appendix, because the equation already accoun...

Jainam Shah 4I

It would be the number of electrons that are being transferred. Both the half reactions should be gaining and losing the same amount of electrons depending on which is the reduction half reaction and which is the oxidization half reaction.

Jainam Shah 4I

Reducing agent is the agent that causes some other substance to be reduced. It basically causes the other substance to gain electrons. The reducing agent thus itself is being oxidized and losing its electrons to another substance.

Jainam Shah 4I

We will be accountable for all of the electrochemical outline and then the second page of the thermodynamics outline

Jainam Shah 4I

Appendix 2B has them and remember they are all given as reduction potentials. Keep that in mind while calculating the standard potentials

Jainam Shah 4I

When you have no conducting solids that can serve as an electrode use solid platinum on either the cathode or anode or both in some cases.

Jainam Shah 4I

The single line represents different phases while the comma simply separates molecules in the same phase.

Jainam Shah 4I

G(not) is at standard conditions whereas G itself doesn't have to be at standard conditions.

Jainam Shah 4I

A voltmeter only measures the difference in electric potential. It cannot measure the individual potential of the cathode or the anode half-reactions. If we use a hydrogen electrode whose potential is zero we can use it as a reference to the half reactions in the galvanic cell.

Jainam Shah 4I

The second page of outline 4 is what you will be held accountable for.

Jainam Shah 4I

Whichever atom has a lower charge or the oxidation number has decreased has went to reduction as reduction is the gaining of electrons

Jainam Shah 4I

Yes q will always be zero as in an adiabatic system there cannot be exchange of heat between the walls of the system

Jainam Shah 4I

Heat capacity is the amount of heat required to heat a certain amount of substance. That amount is flexible hence it is an extensive property. The heat needed will change dependent on the mass. Specific heat capacity is the heat capacity per gram. That value doesn't change. Regardless if you have 50...

Jainam Shah 4I

Delta G or the change in Gibbs free energy simply states whether a reaction will be spontaneous or not. A spontaneous reaction occurs naturally without input of energy. When Gibbs Free Energy is negative the reaction will be spontaneous

Jainam Shah 4I

If the question is asking for temperature and doesn't specify units you can use either celsius or kelvin. If you are doing calculations then be careful to see the units given in the constant and use kelvin or celsius accordingly

Jainam Shah 4I

The salt bridge helps the electrolyte solutions stay neutral so that the anode doesn't become too positively charged and an imbalance develops.

Jainam Shah 4I

When q=-w that means that internal energy is zero which typically is seen in isothermal reversible expansion.

Jainam Shah 4I

Delta U is basically the internal energy of the system. When a system is isolated the delta U cannot change. Delta U changes by an input of energy via heat transfer or work due to compression/expansion.

Jainam Shah 4I

Yes, that equation assumes that temperature is constant.

Jainam Shah 4I

If you reverse the reaction simply take the reciprocal of the the original K value so 1/K.

Jainam Shah 4I

A state function doesn't depend on what steps you take or how the final state is reached. Only the initial state and final state matter for a state function regardless of path taken. A path function depends on the path or steps taken. Work and Heat are common examples of path functions.

Jainam Shah 4I

Yes exclude the solids and liquids. They usually are in excess as they barely react and the concentration of liquids change very little making the change negligible.

Jainam Shah 4I

W^N or in other words the number of micro states ^ number of particles. If you are given the number of moles then use W^N(a) or number of micro states ^ moles*avogrado's number

Jainam Shah 4I

Negative work indicates that the system is doing work on the surroundings so expansion work is done and the system itself is losing energy. Positive work means that the system is being compressed and the surroundings is doing work upon the system so that means its gaining energy.

Jainam Shah 4I

To break bond you always need an input of energy so it would be endothermic

Jainam Shah 4I

You can have their ratios be equal, but the expressions do not have to be equal to each other.

Jainam Shah 4I

Heat transfer can occur in a closed system or work can be done via compression/expansion. We cannot add any substance in the system, but heat transfer can occur between the system and surroundings. Think of a cold ice pack as a closed system.

Jainam Shah 4I

Specific heat capacity is the heat capacity per one gram of a substance. To use this you would need to know how much of a substance you have in terms of its mass. Molar heat capacity is similar but its the heat capacity per one mole of the substance. When using your q=mcdelta(t) equation pay attenti...

Jainam Shah 4I

That expression holds true for isothermal reversible expansion. As Dr. Lavelle explain today in lecture the net change in temperature is zero. The work does means energy does exit the system, but at the same time energy released by the system renters the system as heat. Since these changes are minut...

Jainam Shah 4I

DHavo_1E wrote:Hello,

What would be an example of an intensive property? Thank you!

Specific Heat Capacity can be considered intensive. No matter how much of a substance you have, its ability to absorb heat energy at a particular temperature is the same.

Jainam Shah 4I

The O2 in its diatomic form is the most stable it can be which is why the standard enthalpy of formation is zero.

Jainam Shah 4I

The energy needed to from boiling water to steam is quite high. While steam and boiling water may not have a considerable difference in temperature when the steam cools down it will release a lot of energy as it condenses, because when the steam was formed it needed a lot of energy. When that steam ...

Jainam Shah 4I

I believe so delta h would be the most reliable way unless you are given initial, final, or a change in temperature. You can also see whether bonds are being broken or formed in a reaction.

Jainam Shah 4I

It is negative, because bonds are being formed and the rxn is exothermic. Forming bonds is more favorable, because it doesn't require an input of energy breaking bonds does.

Jainam Shah 4I

Enthalpy is an extensive property. If you are given the specific heat of a substance per gram than the number of grams affects the enthalpy, for example.

Jainam Shah 4I

K is basically raised to whatever factor you multiplied your reaction by. Lets say you have 2SO2(g)+O2(g)⇌2SO3(g) and you multiply iy by a factor of 2 obtaining 4SO2(g)+2O2(g)⇌4SO3(g). The K expression is K=[SO3]^4[SO2]^4/[02]^2. For the original equation assume we are given K to be 2.8*10^2, then t...

Jainam Shah 4I

Group 1 and group 2 cations are typically cations from a strong acid or base. For example the Na+ of NaOH and K+ from KOH. If you look at the reaction for both of these in the presence of water you would see that since these are strong bases they full disassociate. The cation thus has no point or ab...

Jainam Shah 4I

Le Chatelier's principle is basically a principle that explains how a system after a sudden change or perturbation works to restore equilibrium and it highlights how things such as pressure, volume, and concentration do not affect the K constant itself, but simply it will affect how the reaction ini...

Jainam Shah 4I

Leave out pure liquids and solids. Water as a liquid isn't included, because it remains in excess. Its acting as a solute and the changes in concentration of water are so minute we don't have to account for them.

Jainam Shah 4I

When K is less than 10^-3 it essentially means that the acid or base dependent on the given problem is so weak that it barely disassociates. So any protonation or deprotonation can be ignored and we can approximate x to be a small value and ignore it in either the Ka or Kb expression calculation.

Jainam Shah 4I

They are inversely related. When volume decreases the number of moles remain the same, but the frequency of the collisions increase. The particles collide more often and this increases the pressure. When volume increases the particles have more space. They collide less and the pressure decreases.

Jainam Shah 4I

Pressure change only affects a reaction if it is changing the concentration. Which is why when we change the volume the moles stay constant, but we effectively change the concentration or molarity. The concentration in turn affects the reaction. When we introduce an inert gas it is extremely stable ...

Jainam Shah 4I

Q is the reaction quotient and tells us where the reaction is at currently relative to equilibrium. Note that for Q you do not need to use concentrations or partial pressures at equilibrium. However, for K we do. If Q<K product formation is favored and if Q>K reactant formation is favored. When Q=K ...

Jainam Shah 4I

K does not have units. The brackets signify its molarity or concentration of a reactant or product. As Dr. Lavelle discussed in class the true way to explain is via the ratio of the activities of the reactants and product. K is simply giving us the relative information of the rates and which reactio...

Jainam Shah 4I

The forward and reverse reaction rates are equal at equilibrium hence there is no change in concentration.

Jainam Shah 4I

If you are given concentrations and asked to find Kp you can rearrange PV=nRT to solve for the partial pressures of the reactants and products. Conversely, if you were given partial pressures and asked to find Kc you could use the equation to convert the partial pressures into concentrations.

Jainam Shah 4I

If you remember the manipulation of the ideal gas equation we can go from the partial pressure to the concentration of a reactant or product and vice-versa. This is why even though all the products and reactants are in the gas phase the book provides the Kc expression. If you were doing calculations...

Jainam Shah 4I

Kc is the equilibrium constant that uses concentration and you typically use this expression when you deal with aqueous solutions or are given molarities. Kp is the equilibrium constant that uses partial pressure and you use it when one or more of your reactants or products is in the gas phase.

Jainam Shah 4I

Yes, any change in the concentration of either the reactants and product will shift the equilibrium towards either the left or the right for a temporary period of time. Since both the forward and reverse reactions are happening simultaneously the equilibrium will shift back and after some time equil...

Jainam Shah 4I

The solvent will be in large excess in the reaction and is present on the left and right side of the chemical reaction numerically. Any change in the concentration will be very minimal so it nearly stays at a constant concentration.

Jainam Shah 4I

Only pure solids and pure liquids are omitted from the equilibrium constant expression, because the concentration of a solid doesn't not make sense and for pure liquids the change in concentrations is very insignificant because it acts as a solvent.

Jainam Shah 4I

Polyprotic acids are specific acids that are capable of losing more than a single proton in acid-base reactions. From the molecular formula if you have more than one hydrogen the acid can be Polyprotic. To determine the strengths between Polyprotic acids with the same number of hydrogen look at the ...

Jainam Shah 4I

Compared to AP Chem, we are encouraged to think more deeply and combine concepts when solving problems. Our midterm questions for example utilized many different concepts in just one question. This class is certainly fair and not demanding in the course load, but you do have to practice many problem...

Jainam Shah 4I

Water can act as a bronsted acid by donating an H+ proton to become OH-. While it can act as a bronsted base by accepting the proton becoming H3O+.

Jainam Shah 4I

The anion of a weak acid tends to be unstable, because a weak acid's reaction is reversible. Something like HCl would completely disassociate so the chlorine would be stable and it would make no effort to snag a H+ because there is no HCL left in the water. The conjugate base or Cl- is stable. In so...

Jainam Shah 4I

The course outline for that structure mentions cisplatin, hemoglobin, and myoglobin. I would also make sure to know how the chlorines in cisplatin bind to DNA and how exactly it functions as a chemo drug.

Jainam Shah 4I

Take 3-4 minutes to glance over the test first. Figure out what questions will take more time and which ones will not. Try to attempt the ones that you have a good idea of and make sure you do them completely correct. So when you check your work you barely have to check those. Then do the problems t...

Jainam Shah 4I

Yes, if a ligand has two lone pairs that means there are two potential binding sites for the central atom to attach too. That is the transition metal can bind to that ligand at two places hence with two bonds. The ring structure has more to do with a chelate complex.

Jainam Shah 4I

The roman numeral refers to the oxidation state of the transition metal. Many transition metals can have different charges. Further with the Roman numeral you can figure out the charge of a ligand by comparing the charge on the transition metal and the overall charge on the entire complex. For examp...

Jainam Shah 4I

Larger molar mass for the most part implies that the molecule has more electrons and the element is further down the periodic table hence more atomic shells. This allows the molecules to induce a greater dipole spontaneously as more electrons can spontaneously rearrange themselves and attract to a p...

Jainam Shah 4I

Draw the lewis structure will allow you to see the bonding regions around the central atom. If the complex simply has 4 ligands attached to the central transition metal its a tetrahedral complex. If there are two lone pairs and 4 ligands then its a symmetrical square planar shape.

Jainam Shah 4I

Any molecule that donates hydrogen ions or protons to another molecule can act as a Bronsted acid.

Jainam Shah 4I

I second what the comment above says. Further, you should also determine if the question is asking to determine the IMF's the molecule can have with another molecule of itself or another molecule that the problem specifies. If you have a OH, HN, or HF bond you know there is possibility for H-bonding...

Jainam Shah 4I

You are correct. Lone pairs on the central atom repel against the electrons in the bonding pair. This causes the atoms attached via the bonding pair to come closer together due to the repulsion which is why the expected angle is less than normal

Jainam Shah 4I

Sigma bonds lie along the internuclear axis meaning the internuclear axis passes through two overlapping orbitals. This means they are no nodal planes and the atoms can rotate freely. For pi bonds the orbital lie perpendicular to the internuclear axis and there is a nodal plane so they cannot rotate.

Jainam Shah 4I

Cisplatin has the ability to form a coordination compound with DNA molecules. The 2 Cl on cisplatin bond with bases on the DNA which stop cell division which is why it makes for a good chemotherapy drug.

Jainam Shah 4I

Electron arrangement takes account for lone pairs to be a region of electron density while molecular shape is only determined by the atoms bonded to the central atom and their bonding regions. Only time electron arrangement and molecular shape are the same is when there are no lone pairs present.

Jainam Shah 4I

As all the previous posts mentioned, he doesn't curve individual grades. However, if at the end the final average out of 500 points is lower than what he expects he may change the grading scale to match an acceptable class average. So a 93 isn't always an A and a 90 isn't always an A minus. However,...

Jainam Shah 4I

Ionic bonds always supersede Hydrogen Bonding in terms of strength of their attraction. That is why NaCl has a much higher melting point when compared to HCl.

Jainam Shah 4I

They all are identical and simply depend on the electron distribution and how it gets distorted. The more polarizable an atom or molecule is the stronger the IMF's or the induced dipoles.

Jainam Shah 4I

You can do the homework from Section 3F if you wish, because that is what we covered right after the midterm.

Jainam Shah 4I

For SO2, the lone pair on Sulfur and the two lone pairs on oxygen create a dipole moment. You can draw a dipole moment from Sulfur to Oxygen for each bond. You have two dipole moments going towards the same direction hence they do not cancel. So from that we know we know that the molecule is polar a...

Jainam Shah 4I

CHI3 would have the higher boiling point. Since Iodine is a larger atom than Fluorine it has greater polarizability and can induce stronger dipoles.

Jainam Shah 4I

The arrangement in which H and Cl are closest together will have the strongest intermolecular attractions. The closer they are the stronger the dipole is, thus stronger intermolecular attraction and forces are present. I believe structure 2(II) has H and Cl to be the closest together.

Jainam Shah 4I

The electrons that are shared are delocalized meaning a specific single or double bond doesn't have full control of them. They are delocalized and the effect is spread among the entire structure. So a single bond has some double bond character while a double bond has single bond character. This is w...

Jainam Shah 4I

The hybrid I believe is just one structure with a dotted line for example showing how the electron can either act as a lone pair or be used to form a double bond. I would rather focus on resonance structure and focus where double bonds can be made and favorable formal charges.

Jainam Shah 4I

Atoms whose radii are larger tend to have longer bond lengths. Consider HF, HCL, HBr, and HI. The iodine will have the largest atomic radius thus when the H bonds with the Iodine the bond will be longer.

Jainam Shah 4I

There are more electrons involved in the bond hence there is a stronger attraction between the atoms. They are pulled closely so the length of the bond shortens compared to a single bond where only 2 electrons are being shared.

Jainam Shah 4I

The equation you stated is a derivation of -hr/nfinal^2 - (-hr/ninitial^2). If you factor out the -hr you should get an equation similar to what you posted. As the others stated the equation we learned in class is easier to see conceptually.

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