CHEMISTRY COMMUNITY

Daniel Honeychurch1C

Increasing temperature does not affect activation energy. At a higher temperature, faster moving molecules can collide to overcome the activation energy more frequently. For an exothermic reaction, a higher temperature would favor the reactants, so the forward rate constant would decrease. For an en...

Daniel Honeychurch1C

The initial rate for a reactant is the rate at which it is used up in the reaction (which is stoichiometric). Its unique rate is equal to its rate divided by its stoichiometric coefficient. All reactants and products in a reaction have the same (or opposite sign- depending on if its being used or ma...

Daniel Honeychurch1C

A catalyst is neither formed or used up, so it would appear first on the reactant side and then the product side of the chemical equations. This is different than an intermediate, which is created and then used up. You would need to know which reactions are fast or slow, or analyze the reactions at ...

Daniel Honeychurch1C

Average rate is measured over a time period while instantaneous rate is at a certain time point. If the average rate is measured while the system is constantly at equilibrium, then the average rate will be 0. However, if the average rate is measured when the reaction is not at equilibrium and then r...

Daniel Honeychurch1C

The activated complex is the structure with the maximum energy for a reaction path (highest point on the reaction profile). This depends on temperature because a high temp. has a lower activation energy.

Daniel Honeychurch1C

Ea is usually given, so we use the Arrhenius equation to calculate k.

Daniel Honeychurch1C

Probably not, Lavelle always gives the molecule name along with the formula.

Daniel Honeychurch1C

When NO is doubled, the reaction rate increases by a factor of 4: the reaction order with respect to NO must be 2nd order (2^2 = 4) When both NO and O2 are doubled the reaction rate increases by a factor of 8: we already know that NO is 2nd order, so its doubling alone results in the reaction order ...

Daniel Honeychurch1C

E* is the standard reduction potential which is calculated using the standard hydrogen electrode. E is the cell potential for that specific reaction which depends on concentrations.

Daniel Honeychurch1C

When the cell potential is negative, the reaction is not spontaneous. For the reaction to proceed, there has to be electrical energy imputed into the system.

Daniel Honeychurch1C

All the standard reduction potentials are in the appendix of the textbook. You have to use the cell diagram to write the half reactions and look up the standard reduction potentials.

Daniel Honeychurch1C

Yes you leave out H20 (I think the reason is that its the solvent and in large excess). I believe that H20 is the only one that you leave out.

Daniel Honeychurch1C

When K > 1, the reaction is spontaneous because this makes delta G negative. So when the question asks to find if K > 1, it wants you to find the sign of delta G and thus if the reaction is spontaneous.

Daniel Honeychurch1C

From the table in the textbook:
AgI(s)+e- → Ag(s)+I^-(aq) E°= -0.15 V (anode)
AgCl(s)+e- →Ag(s)+Cl^-(aq) E°= +0.22V (cathode - more positive E°)

The equation, E°= E°cath - E°anode, uses the values from the table. So the answer will be E°= 0.22 - (-0.15) = 0.37 V

Daniel Honeychurch1C

You have to include Pt(s) if there is not a conductor (like a solid) in either the anode or cathode. I noticed that in the textbook, when there's Hg(l), Pt (s) is not included. Does anyone know why?

Daniel Honeychurch1C

n is the number of moles of electrons being transferred from the anode to the cathode. You can find n by writing the balanced half reactions, and the coefficient of e- is n.

Daniel Honeychurch1C

The reducing agent is oxidized because it causes reduction by losing its electrons. The oxidizing agent is reduced because it causes oxidation by gaining electrons.

Daniel Honeychurch1C

In 6 L.3 of the textbook, it says that a positive cell potential indicates that the right side of the cell diagram is the cathode. Conversely, a negative cell potential indicates that the left side of the cell diagram in the cathode. In this question, the cell potential is negative (-0.689 V), so th...

Daniel Honeychurch1C

To use the equation, ln(K2/K1) = Hf / R (1/T1 - 1/T2), you need to know K for a reaction at another temp to calculate K for a different temp. You should instead use the equation, delta G = -RTlnK, which you can use without needing to know K at a different temp.

Daniel Honeychurch1C

I believe that it is a typo, because the reaction in 5G.13 (which is what you graph in 5G.17) is I2 (aq) --> 2I (g).

Daniel Honeychurch1C

Delta G is for the whole reaction, which can be calculated using the standard gibbs free energy of formation for the products and reactants.

Daniel Honeychurch1C

I don't think so because electrons are going to flow from the higher positive potential to the lower potential (this is spontaneous because the higher potential has a greater electron pulling ability). The higher potential side is thus the cathode because the electron current flows out.

Daniel Honeychurch1C

In the example in class, the oxidation states of the atoms were given in the complete reaction. All we had to do was balance the half reactions using the atoms and their ions given in the complete equation.

Daniel Honeychurch1C

If the temperature changes beyond the vaporization/melting point (ex. from 70 C to 110 C for water) then a phase change occurred.

Daniel Honeychurch1C

I don’t think so because work is at a maximum for isothermal reversible expansion (constant pressure and temp).

Daniel Honeychurch1C

You can calculate using q = mc(delta T). Or if there’s constant pressure, q(p) = delta H.

Daniel Honeychurch1C

Usually the problem will tell you. If not, in reversible expansion, the pressure of the system equals the external pressure (which decreases during the expansion). In irreversible expansion, the external pressure is constant.

Daniel Honeychurch1C

When deciding which work equation to use, you need to decide if the process is reversible (pressure of system = pressure outside) or irreversible expansion.

Daniel Honeychurch1C

To heat the water you also need to heat the kettle so heat can be transferred to the water. When you heat water in a kettle, the kettle doesn't stay cold but requires heat to increase its temperature.

Daniel Honeychurch1C

We are just looking at the change in entropy between reactants and products. So you use entropy (products) - entropy (reactants).

Daniel Honeychurch1C

The change in entropy is calculated by dividing the amount of energy transferred as heat (reversibly) with the temperature
(when the temperature is constant). This equation tells us that entropy increases more at lower temperatures than higher temperatures for a given amount of heat transferred.

Daniel Honeychurch1C

Reversible reactions do more work because they expand against a larger external pressure (which is at equilibrium with the system). In an irreversible expansion, the external pressure of the system is less than the internal pressure.

Daniel Honeychurch1C

In a reversible expansion, temperature is constant because the work is the sum of the infinitesimal small changes. Since these changes are really small, the energy lost from the system through work is instantly supplied with heat from the surroundings. Therefore, the temperature does not change for ...

Daniel Honeychurch1C

Energy is always given in J or kJ so I assume that you should convert it to those units.

Daniel Honeychurch1C

In an isolated system, matter and energy cannot be exchanged between the system and the environment. Therefore, the total energy in an isolated system cannot change.

Daniel Honeychurch1C

A closed system could be a flask with a cap on it (energy can be exchanged with the system but matter cannot).

Daniel Honeychurch1C

When pressure is constant, q(p) = change in enthalpy. Therefore, the change in enthalpy = 765 J.

Daniel Honeychurch1C

A reversible reaction can be undone; for example, a compressed gas can expand to its original state if the piston is free to move. An irreversible reaction cannot be undone; for example, a gas is compressed with a piston and then the piston is locked with a pin preventing the gas from expanding.

Daniel Honeychurch1C

You don't need to do any calculations, just look at the equilibrium constants for 600 K and 700 K. The Keq is smaller at the higher temperature, so increasing the temperature will increase the concentration of the reactants. Therefore, the concentration of NH3 will decrease.

Daniel Honeychurch1C

Flask 3 and 4 have the same concentrations of X2 and X. Therefore, the reaction reaches an equilibrium (no net change of the reactant or product).

Daniel Honeychurch1C

I think the answer is wrong for i) pH. When you do pH = -log([H+]) = -log(1.50) = -0.176 but the answer in the solution manual is 0.176

Daniel Honeychurch1C

Hi, I usually look over all my lecture notes and do as many practice problems from the textbook that I can.

Daniel Honeychurch1C

I believe that pKa and pKb for weak acids/bases are typically between 3-7 but can be outside this range. This all depends on the acid/base.

Daniel Honeychurch1C

If K (equilibrium constant) is very small, < 10^-3, you can simplify your calculations by assuming that x is too small compared to the initial concentration to affect (concentration - x) in the equation. After calculating x this way, you should check using the 5% rule (if x is less than 5% the initi...

Daniel Honeychurch1C

You would only need to include gases and aqueous compounds in your final Kc equation (Products/reactants), not pure substances (liquid and solids).

Daniel Honeychurch1C

According to the textbook for question 5I.29, which has a squared function, you can assume x is too small to affect the calculations when K is really small.

Daniel Honeychurch1C

Since K is so small (3.2 x 10^-34), there is very little of the products at equilibrium. Therefore, we know that x is going to be very low. Using this assumption, we can simplify the expression 3.2 x 10^-34 = (x^2) / (0.22 -2x)^2 to 3.2 x 10^-34 = x^2 / (0.22)^2, since x is so small that it will hav...

Daniel Honeychurch1C

K (equilibrium constant) is a constant because it does not change when concentrations of reactants or products are modified in the same environment (constant pressure and temperature). When concentrations are changed, the system will make more reactants or products to reach equilibrium again and K w...

Daniel Honeychurch1C

The concentration of H2 would increase according to Le Chatelier's principle.

Daniel Honeychurch1C

If K>Q, more products will be formed from the reactants to reach equilibrium. If K<Q, more reactants will be formed from the products to reach equilibrium. If K=Q, the reaction is at equilibrium.

Daniel Honeychurch1C

Yes, bar is a unit for pressure (1 bar = 100,000 Pascals).

Daniel Honeychurch1C

Kc is the equilibrium constant in terms of molar concentrations, and Kp is the equilibrium constant in terms of partial pressure of gasses.

Daniel Honeychurch1C

No, the concentration of reactants and products do not have to be the same. However, at equilibrium, the net change of reactants and products is 0. Although reactions are still occurring, the rate of the forward and backward reactions are equal.

Daniel Honeychurch1C

I has a larger ionic radius which makes the H-I bond longer and weaker. Therefore, HI dissociates more than HF and is a stronger acid.

Daniel Honeychurch1C

Yes, if the anion or cation in a salt are conjugates to weak acids or bases, they will affect the pH.

Daniel Honeychurch1C

You use those prefixes when the ligand already has a prefix in its name or if its polydentate.

Daniel Honeychurch1C

I think that you have to draw the Lewis structure of the ligands. A polydentate ligand will have multiple atoms with a lone pair and an orientation that allows for those atoms to bind with a central atom in two places (this means that the atoms with lone pairs are close to each other).

Daniel Honeychurch1C

Lavelle will probably do the same questions for all the lectures to be fair to all of his students.

Daniel Honeychurch1C

For K2[Ni(CN)4],

2 K+ ions (total +2 charge)

1 Ni atom (charge ?)

4 CN- ions (total -4 charge)

The entire compound is neutral.

Therefore, 0 = 2 + oxidation number of Ni + -4
oxidation number of Ni = 2

Daniel Honeychurch1C

When pH = pKa, the amount of conjugate acid and conjugate base are equal.

Daniel Honeychurch1C

Yes, the additional oxygen atom pulls the electron density away from the H-O bond. This weakens the H-O bond which makes the acid stronger.

Daniel Honeychurch1C

Yes! Stronger acids dissociate more which increases [H3O+] and lowers the pH.

Daniel Honeychurch1C

Yes, since water can act as an acid (accept a proton to make H3O+) or a base (donate a proton to make OH-), water is an amphoteric compound.

Daniel Honeychurch1C

No, the hybridized orbitals are 2sp, 3sp^2, 4sp^3.

Daniel Honeychurch1C

I don't know if we will get marked down, but I would make sure to write it showing which atom is bonded to the central atom as the common convention.

Daniel Honeychurch1C

I think that the molecular formula will be given when drawing those molecules.

Daniel Honeychurch1C

You need to draw the lewis structure to see if the central atom has lone pairs. Tetrahedral has 4 bonded atoms and no lone pairs while square planer has 4 bonded atoms and 2 lone pairs.

Daniel Honeychurch1C

Lavelle discussed radicals when going over Lewis structures: a radical is an atom with an unpaired electron.

Daniel Honeychurch1C

Intermolecular forces (london, dipole-dipole, hydrogen bonding) are between different molecules while intramolecular forces are within one molecule (ionic bonds, covalent bonds).

Daniel Honeychurch1C

I believe that Lavelle only said that hydrogen bonding requires a H bonded to N, O, or F and the H is close to a lone pair of an electronegative atom.

Daniel Honeychurch1C

When atoms are replaced by lone pairs to make a T-shape, the atoms are from the equatorial positions because these atoms had a bond angle of 120 degrees. The axial atoms are not replaced because those atoms are only 90 degrees from the equatorial atoms. This maximizes the distance between the lone p...

Daniel Honeychurch1C

SiF4 (-86 C) has a higher boiling point than SiH4 (-112 C). SiF4 is larger and has more electrons, making it more polarizable. This increases the induced dipole-induced dipole forces.

Daniel Honeychurch1C

I believe that the atom with the lower electronegativity with be more polarizable. Therefore, polarizability decreases across a row.

Daniel Honeychurch1C

HI has a higher boiling point than HBr because HI has more electrons and thus stronger induced dipole-induced dipole forces.

Daniel Honeychurch1C

The strongest intermolecular interaction is ion-ion (-250 kJ/mol) because the charge of an ion is a lot greater than the charge of a dipole.

Daniel Honeychurch1C

Yeah you can probably do any problems from material not covered in the midterm, for example 3F.

Daniel Honeychurch1C

There are always induced dipole-induced dipole (london) forces between two molecules. If there is a dipole in the molecule, there will be dipole-dipole forces between those molecules. If a molecule has a H bonded to a N, O, or F and there is at least one lone pair of electrons, there will be hydroge...

Daniel Honeychurch1C

Yes, hydrogen bonding is a dipole-dipole force but a lot stronger (-20 kJ/mol for hydrogen bonding compared to -2 kJ/mol for regular dipole-dipole)

Daniel Honeychurch1C

There are always induced dipole-induced dipole forces between molecules. There will be dipole-dipole forces if both molecules have dipoles and are oriented with the positive side near the other molecule's negative side. There will be hydrogen bonding if a H is bonded to a O, F, or N and it can inter...

Daniel Honeychurch1C

Lauren Sanchez 3D wrote:So triple bonds pull the nuclei even closer then? Which then makes them even stronger than double bonds?

Yes, in triple bonds, there are 6 shared electrons which pulls the two atoms even closer than in double bonds which makes triple bonds stronger and harder to break (more energy is needed).

Daniel Honeychurch1C

Single bonds are longer because they have the weakest strength. In double and triple bonds, there are more electrons shared between the two atoms which pulls the atoms closer due to the attractive forces between the electrons and protons of the two atoms.

Daniel Honeychurch1C

I think that we can turn in problems from the section of chemical bonds not covered by the midterm.

Daniel Honeychurch1C

The dissociation energy of bonds will always be given because its based on experimental data.

Daniel Honeychurch1C

A dipole can also be induced by two nonpolar molecules causing induced dipole - induced dipole attraction (London dispersion forces). This weak intermolecular force is caused by the distorting of the two atoms' electron clouds.

Daniel Honeychurch1C

I think you remove the electrons from the d sub-shell first because it has a higher energy than the s sub-shell.

Daniel Honeychurch1C

The ground-state electron configuration for Ni is [Ar] 3d8 4s2. When 2 electrons are removed to make Ni+2, the 4s2 electrons are removed because they are in the highest energy sub-shell. Therefore, the ground-state electron configuration for Ni+2 is [Ar] 3d8

Daniel Honeychurch1C

Molecules are more stable when the atoms with a higher electronegativity have a negative formal charge. Since oxygen is more electronegative than sulfur, SO4-2 is more stable when oxygen has a -1 formal charge instead of sulfur.

Daniel Honeychurch1C

The central atom usually has the lowest ionization energy. For NHF2, N is the central atom - N has 3 single electrons in its valence structure so it can form 3 bonds, while F and H only have 1 single electron so they will form 1 bond. Therefore, N can form single bonds with H and 2 F atoms.

Daniel Honeychurch1C

Decolonization means the electrons are not associated within one bond. For example in benzene, electrons aren't localized in 3 carbon double bonds and 3 carbon single bonds. Instead, the electrons are spread throughout all the carbon bonds so that each bond has a length between a single and double b...

Daniel Honeychurch1C

The 4s sub-shell has a higher energy than the 3d sub-shell. The first electron removed is from the sub-shell with the highest energy (furthest from nucleus).

Daniel Honeychurch1C

Yeah, it has to do with shielding. In hydrogen (one electron), 2s and 2p have the same energy when the electron is excited. However, when there are multiple electrons, the 2s sub-shell has a slightly lower energy than 2p.

Daniel Honeychurch1C

The f-sub-shell is the highest sub-shell for ground state electrons. However, when atoms are excited, electrons can go to higher shells.

Daniel Honeychurch1C

I don't think that you will get marked off for not using positive signs but the spin quantum number usually has a positive or negative sign by convention.

Daniel Honeychurch1C

The actual values of the atomic radius are always given. You just need to know that cations are smaller than their parent atoms because they have less electrons which means there is less electron repulsion. Anions are larger than their parent atoms because they have more electrons and thus more elec...

Daniel Honeychurch1C

Since c = λv, λ = c/v = (3.0 x 10^8 m/s) / frequency. For both problems, plug in the given frequencies to get the wavelength. Then convert to manometers ( 1 m = 10^9 nm, 1 m = 10^12 pm).

Daniel Honeychurch1C

Effective nuclear charge is important because it tells us how difficult it is to remove a valence electron. The stronger the effective nuclear charge, the harder it is to remove an electron. The effective nuclear charge is based on the number of protons attracting the electrons and the number of oth...

Daniel Honeychurch1C

When reading, you are exposed to radiation from visible light (600 nm is in the visible spectrum). You are able to read because visible light reflects off the book and goes into your eye.

Daniel Honeychurch1C

The Balmer Series (n = 2) is in the visible spectrum while the Layman Series (n = 1) is in the UV spectrum.

Daniel Honeychurch1C

We just learned about Schrodinger's equation to understand that electron orbitals are math functions based on the probability of where electrons could be, we don't need to use the equation.

Daniel Honeychurch1C

Yes, the energy level refers to the electron shell. The lower the energy level, the closer to the electron is the nucleus.

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