CHEMISTRY COMMUNITY

Paul Hage 2G

If rate laws are not given, you can determine the rate-determining step by knowing that it is the slowest step of a reaction. The intermediate reactions will likely have a parentheses by them specifying whether it is a (slow) or (fast) reaction.

Paul Hage 2G

If you are referring to a concentration cell, you would use the equation Ecell = Enaught - RT/nF lnQ. For a concentration cell, it is important to know that Enaught is 0. Therefore all you need to solve for a concentration of either the anode/cathode would be an Ecell value, the temperature, the con...

Paul Hage 2G

Also, for a first-order reaction, ln[A]=-kt+ln[A]₀

Paul Hage 2G

The catalyst wouldn't affect the equilibrium of a reaction because a catalyst speeds up both the rates of the forward and the reverse reaction. The equilibrium would only change if either the forward or the reverse reactions were sped up/favored.

Paul Hage 2G

Also, if you are given a rate law, it is important to know that the slow step is the rate-determining step. This can help you determine which intermediate reaction is the slow one when given the rate law.

Paul Hage 2G

This problems is most conceptual, but some parts of the problem could be approached using the Nernst equation. For example, in part b, you would be diluting the anode concentration; therefore the numerator of the Q expression in the Nernst equation would decrease, causing the overall Q expression to...

Paul Hage 2G

The second page of the thermodynamics outline will be covered by test 2.

Paul Hage 2G

It may also be important to know that, for a concentration cell, the side with the lower concentration is the anode. This makes sense because, for the anode, the metal will be oxidized to enter solution as aqueous ions, increasing the concentration. The opposite will occur for the cathode and this w...

Paul Hage 2G

Typically for cell notation problems, you only need to include the phases of the species that are being oxidized or reduced. However, for problems involving concentration cells, it may be helpful to keep track of the concentrations of each species (because they are usually the same element in the sa...

Paul Hage 2G

Delta G represents the maximum work that can be done by a voltaic cell. This is why it is possible to equate the two equations w(max)=-nFE(max) and deltaG(cell)=-nFE(cell).

Paul Hage 2G

I don't think we will need to know how a rechargeable battery works because it is not part of the section outline for electrochemistry.

Paul Hage 2G

Galvanic cells have a positive Ecell value. However, the E value of the half reactions could be negative, as long as the Ecell value is positive. The magnitude of E for the half reactions must be worked out so that the overall Ecell value is positive.

Paul Hage 2G

A salt bridge can be used in a concentration cell. It allows for the transfer of ions to neutralize both solutions.

Paul Hage 2G

A galvanic cell creates electrical energy from the chemical energy that comes from redox reactions occurring in the electrochemical cell. The parts of the galvanic cell include the anode (typically on the left side) and the cathode (typically on the right side). The anode is the site of oxidation an...

Paul Hage 2G

Platinum is introduced into the cell diagram when there are no solid metals as part of the anode or cathode of the cell diagram. It serves as an inert electrode (provides a surface for the transfer of electrons, but does not affect or participate in the redox reaction).

Paul Hage 2G

Mariah wrote:When is test 2?

Dr. Lavelle just announced that the test will be moved to week 9 in discussion sections.

Paul Hage 2G

Splitting the overall reaction into half reactions is also helpful when balancing redox reactions. This makes it easier to balance the H and the O atoms separately using H20 and H+ ions. It also helps to determine the total electrons needed to balance charges in the reaction.

Paul Hage 2G

Mariah wrote:When is test 2?

Test 2 is next week in discussion sections.

Paul Hage 2G

I believe the material for Test 2 will cover everything from what we began learning after the midterm (Focus 5G Gibbs Free Energy and Equilibrium Constant) up until the electrochemistry topics we learn in lecture this week. It is likely that we will learn the final topics for Test 2 on Friday in lec...

Paul Hage 2G

Electron flow creates electrical energy and this is caused by several chemical reactions. A galvanic cell is an example where chemical energy is transformed into electrical energy. In this case, chemical reactions occur, causing electrons to flow through a wire (part of the exterior circuit of the c...

Paul Hage 2G

In my book, 5J.15 shows 150 degrees celsius, which would match up with the 423 kelvins that you referenced.

Paul Hage 2G

Yes, that is correct. The only difference is that delta G naught represents the delta G at standard conditions (1 atm and 298 K). Therefore, as a reaction approaches equilibrium, delta G naught would not change because it still represents standard conditions, whereas delta G will change because it i...

Paul Hage 2G

For part a), we know that the enthalpy of vaporization can be written in units of kJ/mol. And because CH4 is at its boiling point, we can divide the heat value given in the problem by the amount in moles given in the problem. This will give the enthalpy of vaporization. For part b), you would do the...

Paul Hage 2G

From the values given in the problem, we can calculate the heat absorbed in 30 minutes. Therefore, assuming the heat absorption rate is constant, we can find the heat absorbed in 10 hours. 10/0.5=20, so to find the heat absorbed in 10 hours, you would multiply the heat absorbed in 30 minutes by 20.

Paul Hage 2G

Unless the problem specifies whether we should have our answer in Joules or kilojoules, I believe we could write our answer using either unit. The important thing is to make sure the units for different variables in a specific problem match.

Paul Hage 2G

Some equations that we may need to know are not on the equation sheet. For example, it may be important to know that, if we need to solve for PΔV but we do not know the change in volume or the pressure, we could instead use ΔnRT because they are equivalent when derived from the ideal gas law.

Paul Hage 2G

If the value of x is less than 5% of the initial concentration of reactant, then we can approximate. Typically, we can approximate when Ka is less than 10^-3, but the 5% rule is the best way to be sure that approximating will be acceptable.

Paul Hage 2G

We arrive at the "-w=q" relationship by knowing that for an ideal gas, U(total)=3/2(nRT). Therefore ΔU=3/2(nRΔT). When ΔT=0, ΔU=0. Since ΔU=q+w=0, you can rearrange the expression to arrive at -w=q.

Paul Hage 2G

System with Constant External Pressure: w=-P(external)ΔV
System with Changing Pressure (Reversible Expansion): w=-nRT(ln(V2/V1))
Change in internal energy with ΔP=0 (constant external pressure): ΔU=ΔH-PΔV because q=ΔH and w=-PΔV at constant pressure.

Paul Hage 2G

Cooling a closed system would affect the internal energy of the system because heat will be released by the system, resulting in a negative q value. Unless work is being done on or by the system, the value of w will not change.

Paul Hage 2G

In an isolated system, neither matter nor energy can be exchanged between the system and the surroundings; therefore, the change in internal energy will be 0 over time because the internal energy can only be changed when heat energy enters/leaves the system or when work is done on/by the system.

Paul Hage 2G

A bomb calorimeter is a calorimeter in which the volume remains constant. A bomb calorimeter is used as an example of an isolated system because neither matter nor energy can be exchanged between the system/surroundings.

Paul Hage 2G

This is a Hess's Law application problem. You would first need to set up 3 combustion reactions for each of the 3 compounds listed in the problem. Next, you need to balance those combustion reactions. Finally, you would adjust these reactions (reverse reaction, multiply by a factor of 2, etc) so tha...

Paul Hage 2G

The standard enthalpy of formation of N2 is 0 because it is an element in its standard state.

Paul Hage 2G

So for Hess's Law, the sign of the resulting net enthalpy will tell if the sum of the reactions is endothermic or exothermic right? That is correct. If delta H of the sum of the enthalpies of the multiple reactions is negative, the overall reaction is exothermic. If delta H of the sum of the enthal...

Paul Hage 2G

Delta H may be written as the enthalpy of reaction or an enthalpy of formation that represents kJ/mol. For example, in question 4D.1 in the book, the reaction shows a Delta H of +358.8kJ; to answer part a), which asks "How much heat is absorbed in the reaction of 1.25 mol S8 at constant pressur...

Paul Hage 2G

Bond enthalpies of diatomic molecules (Br2, I2, Cl2, N2, O2, H2, F2) are accurate because they are measured for those specific molecules; however, all other bond enthalpies are averages from many different molecules, so they are not completely accurate.

Paul Hage 2G

When the K value is less than 10^-3, it is safe to assume that the value of x will be negligible compared to the initial concentration. However, to make sure that x can be excluded from the denominator in the K expression, you can check that x is less that 5% of the initial concentration.

Paul Hage 2G

A small equilibrium equilibrium constant (K<10^-3) means that there is a large equilibrium concentration of reactants (reactants are in the denominator of the K expression, so when the concentration of reactants is large, K will be small). Because there is a greater concentration of reactants at equ...

Paul Hage 2G

This will always be true when the stoichiometric coefficients in the balanced chemical equation are the same, which is often the case when an acid or base reacts with water. The only way to make sure that concentration of a conjugate acid always equal to the concentration of hydroxide ions or the co...

Paul Hage 2G

Conjugate acids and bases will typically have either one less H+ ion or one more H+ ion than their reacting counterparts. For example, in the reaction HCl + H2O --> Cl- + H3O+, you can immediately notice that on the reactants side, HCl has one more proton than the Cl- on the products side and the Cl...

Paul Hage 2G

When there is an OH group or a carboxyl group (COOH), those groups will act as acids and become deprotonated by donating an H+ ion. The molecule will more readily lose H+ ions from its OH or COOH groups.

Paul Hage 2G

For this problem, you would set up the expression of K for this reaction: [BrCl]^2/[Cl2][Br2]. Next, you would plug in the equilibrium values given in the problem. Because you know that [BrCl]^2/[Cl2][Br2] = K = 0.031, you can solve for the equilibrium concentration of Br2.

Paul Hage 2G

The N2H2 structure does not have resonance because there is only one Lewis structure that satisfies the octet rule for N and also uses the 12 valence electrons in the molecule. If you remove one lone pair from each of the Nitrogen atoms to create a triple bond between them, the resulting Lewis struc...

Paul Hage 2G

Usually, concentration would be used in ICE tables, but the only way to know what to use is by checking whether the Kp or Kc is given in the problem or whether the problem gives you initial concentrations or initial partial pressures. If you use the pressure of a gas, it would typically be in units ...

Paul Hage 2G

The equilibrium constant is not affected by a change in volume. A change in volume causes the reaction to shift more strongly in one direction to minimize the effect of the change, but does not actually change the equilibrium constant. Changing temperature is the only variation we have learned that ...

Paul Hage 2G

When the partial pressure of NO is increased (the concentration of NO increases because P=(conc)(RT)), the partial pressure (and thus concentration) of NH3 will increase because the reaction will shift towards the reactants to minimize the effect of the change. When the partial pressure (and concent...

Paul Hage 2G

You would exclude the solid and liquid reactants/products from your calculations for the reaction quotient. Aqueous and gaseous reactants/products would be the only species you would include in calculations for both the reaction quotient and the equilibrium constant.

Paul Hage 2G

Keq is the generalized equilibrium constant, whereas Kc is the equilibrium constant found using concentration values. You would use Kc when you are specifically calculating the equilibrium constant using the molarity values of each reactant/product.

Paul Hage 2G

We should use variables for change concentration; however, sometimes the variable x is negligible (when the quotient "x/initial concentration" is less than 0.05). When this occurs, the variable x can be eliminated when it is being subtracted from an initial concentration, approximating the...

Paul Hage 2G

When pressure changes due to a change in volume, the concentrations of reactants/products in the system would change; however, the equilibrium constant cannot be changed due to a variation in the concentrations of reactants or products in the system. The position of the equilibrium would change, but...

Paul Hage 2G

Also, Kp can only be used for gas phase reactions (because an aqueous reactant or product would not have a partial pressure value), whereas the Kc constant can be used for reactions where both gaseous and aqueous reactants/products would need to be considered while calculating the equilibrium consta...

Paul Hage 2G

BaBr2 has bonds that are primarily ionic because the electronegativity between Ba and Br is very large. This is because Ba is in Group 2 of the periodic table and Br is in Group 17. Following the electronegativity trend (electronegativity increases across the periodic table), we can see that there i...

Paul Hage 2G

Another important equation to know is the one depicting the occurrences in the photoelectric effect experiment: Energy of incoming photon = threshold energy of the metal + kinetic energy of the ejected electron. This equation can be rearranged to isolate any of the 3 "variables". It is imp...

Paul Hage 2G

Oxidation number is important for naming coordination compounds, where we would need to express the oxidation number of the transition metal cation in Roman numerals in parentheses after naming the metal cation. The oxidation number is also important because it contributes to the overall charge of t...

Paul Hage 2G

Acid strength is based on how easily the molecule can give off a proton. Because HF has the shortest and therefore strongest bond, it would be somewhat difficult to break the H+ off the molecule, whereas the relatively long bonds in HCl, HBr, and HI make it easier for the molecules to give off a pro...

Paul Hage 2G

If a molecule is a Bronsted acid, it will donate a proton. If a molecule is a Bronsted base, it will accept a proton. To determine whether a molecule is an acid or a base, you would need to know that typically a molecule with an H is a Bronsted acid, especially if the H appears at the beginning of t...

Paul Hage 2G

I believe that they are just different ways to identify acids/bases. The two definitions just represent different properties of acids/bases.

Paul Hage 2G

Hi guys, I'm having trouble understanding the rule regarding strength of inter-ionic forces. How does the strength of an inter-ionic force vary with changing factors? This is part of the conceptual equation we went over quickly in class: the 1/r^6 as part of the interaction potential energy equatio...

Paul Hage 2G

Do pi bonds also occur in higher level subshells like the f-subshell? Or is there a different type of bond beyond that? Only s orbitals are unable to form pi bonds because of their lack of angular nodes. "f" electrons rarely participate in bonding because they are well-shielded, but pi bo...

Paul Hage 2G

Natalie Benitez 1C wrote:Can we refer to bent as angular or v-shaped as well or are we limited to just saying bent for exams?

I recall hearing that bent or v-shaped would be accepted for exams. However, all three are the same shape. I would refer to it as bent on exams to be on the safe side.

Paul Hage 2G

For me, the method I use to name coordination compounds is: 1) Identify whether the complex ion is a cation or anion in the compound 2) If it is an anion, name the cation first by using its elemental name 2a) Name the ligands with their prefixes before naming the transition metal cation (alphabetica...

Paul Hage 2G

Because CH2Cl2 is tetrahedral-shaped, the bond angles would be around 109.5 degrees. Therefore, no matter how the Lewis structure is drawn, the dipole moments will still be going towards chlorine in different directions, so the dipole moments would not cancel.

Paul Hage 2G

The boiling point of HCl would be lower because only covalent/intermolecular forces must be overcome. For ionic compounds such as NaCl, the attraction between the oppositely charged cations/anions requires much more energy to overcome, so the boiling point of NaCl would be higher.

Paul Hage 2G

Dipole-induced dipole would be a separate interaction that occurs when a polar molecule causes a fluctuation in the electron arrangement of a nonpolar species, which in turn leads to a temporary dipole in the nonpolar species. It would likely fall somewhere close to the dipole-dipole interaction, bu...

Paul Hage 2G

What is the difference between these two bonds and why are they significant? The electrons in a sigma bond are referred to as localized between the internuclear space due to the way in which the orbitals overlap; on the other hand, pi bond electrons can be delocalized and can move around the molecu...

Paul Hage 2G

PriscillaLi_3G wrote:Do we have to know Sigma and Pi bonds for the test?

Yes, sigma and pi bonds are the last concepts we need to know for the test. Hybridization will not be covered on Test 2.

Paul Hage 2G

If the polar bonds in the following shapes are identical, with the same atoms connected to the central atom, then the molecule will be non-polar: linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. This is because the atoms will be arranged symmetrically around the central at...

Paul Hage 2G

So can dipole moments be temporary or all they all permanent? Temporary dipoles can occur in Van Der Waals forces, which happens when two atoms of equal electronegativity experience a temporary dipole due to the shifting of electrons which could create a slightly positive charge with fewer electron...

Paul Hage 2G

All molecules have London dispersion forces due to the constant movement of electrons and the arising variations in the arrangement of electrons around a molecule.

Paul Hage 2G

Hydrogen bonding is one of the strongest intermolecular forces because it involves a permanent dipole and there exists a high electronegativity difference between the H atom and an N,O, or F atom, leading to a greater attraction.

Paul Hage 2G

Dipole-dipole: Occurs when two molecules with atoms of different electronegativities interact to create a slight attraction between the partially positive atom in one of the molecules and the partially negative atom in the other molecule. Example: In HCl, the H is partially positive and the Cl is pa...

Paul Hage 2G

Dissociation energy is always positive because bond-breaking is always an endothermic process, as the system requires energy to break a bond.

Paul Hage 2G

Effective nuclear charge is the net positive charge felt by an electron. This is why the effective nuclear charge is decreased when there is a large amount of electron shielding between the nucleus and the electron in question. Effective nuclear charge plays a role in the atomic radius. As you move ...

Paul Hage 2G

Selenium would have a greater electronegativity because its valence shell is closer to the nucleus. Because the 3 other elements given in the problem are one row under Selenium, the distance between the nucleus and the valence shell of these 3 atoms would be greater, making the atom have less of an ...

Paul Hage 2G

I don't think it would be possible to know that C and S have nearly identical electronegativities without being given the values; however, because they are relatively close on the periodic table (and neither of them are one of the 3 most electronegative elements (F,O,Cl)), you could assume that thei...

Paul Hage 2G

Generally, when the electronegativity difference between two atoms >2, the bond will have ionic character because there will be a less equal sharing of electrons between the two atoms. When the electronegativity difference between the two atoms <1.5, the bond will generally have covalent character. ...

Paul Hage 2G

You would add a double bond instead of a lone pair around an atom when adding the bond would bring the formal charge of the atom closer to 0 or when you want to complete the octet of one of the bonded atoms but cannot add a lone pair because this would lead to a Lewis structure that has more electro...

Paul Hage 2G

In 1E.7a, this would be the electron configuration of an excited state of C because there are 2 electrons placed in a single orbital in the 2p subshell, whereas the ground state electron configuration would show one electron in each of the first two orbitals in 2p. In 1E.7b, this would be the electr...

Paul Hage 2G

From my understanding, the answer to that depends on which definition of valence electrons is being used. For instance, if valence electrons are being defined as the number of electrons in the outermost energy level, then manganese would have 2 valence electrons. However, if valence electrons are de...

Paul Hage 2G

The effective nuclear charge felt by an electron in the outermost shell of an atom would be equal to the atomic number minus the number of shielding electrons (core electrons, not including valence electrons). Hope that helps.

Paul Hage 2G

The metalloids are boron, silicon, germanium, arsenic, antimony, tellurium, and polonium. These elements can be pictured on an imaginary staircase on the periodic table. Anything to the right of these elements is a nonmetal. You are correct about the metals (generally in the s and d blocks, except f...

Paul Hage 2G

For the Cu+ ion, the electron is removed from the 4s subshell, making the electron configuration of Cu+ [Ar] 3d10. The electron is removed from the 4s subshell because a full 3d subshell is stable. Therefore, there would be no 4s1 as part of the electron configuration because the single electron was...

Paul Hage 2G

A question I do have is that when reviewing the notes in Lecture, it states that 4s, 4p has higher energy than 1s, 2s,2p, 3s,3p, and 3d. So I am quite confused and perplexed on how the orbitals work. From what I understand, the 4s and 4p subshells are higher in energy than 1s, 2s,2p, 3s,3p, and 3d....

Paul Hage 2G

In 1B.9, you would combine the two equations because you are looking for the energy of the photon. Since we know the wavelength of violet light is 420x10^-9, we can rearrange the E=hv equation to eliminate the frequency value which we do not know and replace it with a relationship involving two valu...

Paul Hage 2G

Does anyone know why the uncertainty in the equation wasn't 10 because 5+5 is 10? You are correct. The uncertainty in velocity for that problem should be 10 m/s because it represents the spread of possible values. This mistake is mentioned in the Solutions Manual Errors link on the Chem 14A class w...

Paul Hage 2G

I think that, as long as you have the order correct, your answer would be sufficient. I believe the important aspect of the problem is understanding the relative radii differences between atoms. You could find this out with the trend that atomic radius increases down the periodic table and decreases...

Paul Hage 2G

So, can we reorganize this equation to find velocity? Would V=(h/wavelength*m) be correct? Yes, you can reorganize the equation to find any value you are looking for, as long as you have 2 of the 3 unknown variables (velocity, mass, wavelength). In the case you mentioned, that equation for velocity...

Paul Hage 2G

The 350 x 10^-12m/s came from the conversion from picometers to meters. 1pm=10^-12m.

Paul Hage 2G

You would use the Heisenberg Uncertainty Equation: ΔpΔx≥h/4pi. This equation is equivalent to mΔvΔx≥h/4pi. So you would now isolate Δv and sub in the values we know. We know, from the problem, that the uncertainty in position, Δx, is 5.0m. The mass is 8.00kg. h is Planck's constant, 6.626 x 10^-34 J...

Paul Hage 2G

The first step of the problem would be to balance the 3 equations: Fe+Br2-->FeBr2 3FeBr2+Br2-->Fe3Br8 Fe3Br8+4Na2CO3-->8NaBr+4CO2+Fe3O4 To answer the problem, we have to use the molar ratios of the reactants and products in the 3 reactions above to convert 2.5t of NaBr to grams of Fe. You would firs...

Paul Hage 2G

Yes, for that part of the problem, you would just have to divide the work function by Avogadro's number in order to find the energy required to remove an electron from one sodium atom.

Paul Hage 2G

For this problem, you would need the following information: the s subshell can hold up to 2 electrons, the p subshell can hold up to 6 electrons, the d subshell can hold up to 10 electrons, and the f subshell can hold up to 14 electrons. Because each orbital can hold two electrons, you would divide ...

Paul Hage 2G

To solve this problem, you would have to use the mass of carbon dioxide to find the moles of C (convert to moles of CO2 using molar mass and then use molar ratio of C in CO2 to find the moles of C). You would then use the mass of H2O to find the moles of H (0.174g H2O/18.02g x mol^-1 H2O to find the...

Paul Hage 2G

I believe that there is charge conservation in chemical reactions; therefore, electrons and protons would be conserved throughout a chemical reaction because the atoms on each side of the equation are balanced.

Paul Hage 2G

The molar mass of the molecular formula is 194.19 g/mol, so you would divide this value by the molar mass of the empirical formula (about 97), which gives the value 2. So you multiply the empirical formula by 2, giving the molecular formula C8H10N4O2.

Paul Hage 2G

When doing problems, I have found that solving for the number of formula units in a mol of a substance works the same as finding the number of atoms/molecules in a mol (6.022 x 10^23 formula units in one mol). However, I believe the term "formula units" is just a different way of classifyi...

Paul Hage 2G

After reading these posts I am confused about whether or not a zero is considered a sig fig if there is a number like 13.0 Since the zero is trailing the decimal point would it or would it not be a sig fig when doing arithmetic calculations? In the example you gave (13.0), the 0 would be considered...

Paul Hage 2G

I don't think we are supposed to comment on the actual/specific benefits of the solution; rather, we would just have to state whether the solution would be present with a high enough concentration of substance X so that the substance could have any health benefits. However, after the 90 doublings of...

Paul Hage 2G

0 is considered a significant figure if it appears between two non-zero numbers (i.e. 104) 0 is also considered significant if it comes after nonzero numbers after a decimal (i.e. 1.300) However, the zeros that come after a decimal but before the nonzero numbers are NOT significant (i.e. 0.0003) If ...

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