CHEMISTRY COMMUNITY

Kim Tran 1J

The presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 125 kJ per mol to 75 kJ per mol. a) By what factor does the rate of reaction increase at 298K, all other factors being equal? What is the relationship between Ea, catalyzed and...

Kim Tran 1J

They tell you that one of the reactions is F2 (g) + 2H+ (aq) + 2e- -> 2HF (aq) and since you are finding the Ka of HF, HF must be the reactant (the denominator) when you calculate Ka. So you would want to flip the given half-reaction 2HF (aq) -> F2 (g) + 2H+ (aq) + 2e- and find the other half reacti...

Kim Tran 1J

What is a concentration cell? How is this different from a galvanic cell and an electrolytic cell?

Kim Tran 1J

14.97 Use the data in Appendix 2B and the fact that for the half-reaction F2(g) + H+ (aq) + 2e- -> 2 HF (aq), E=+3.03 V, to calculate the value for Ka for HF. The log K equation is the one shown in our notes, but the solution manual uses ln K. I tried to use the log K equation but it resulted in a d...

Kim Tran 1J

Calculate the reaction quotient, Q, for the following cell reactions, given the measured values of the cell potential. Balance the chemical equations by using the smallest whole number coefficients: a) Pt(s) | Sn4+(aq),Sn2+(aq) || Pb4+(aq),Pb2+(aq) | C(gr), E=+1.33 V The answer I got for this proble...

Kim Tran 1J

14.41 Calculate E for each of the following concentration cells:
b) Pt(s) I H2 (g, 1 bar) I H+ (aq, pH = 4.0) II H+ (aq, pH = 3.0) I H2 (g, 1 bar) I Pt(s)

In the solution manual, it says that n=1. Why is that?

Kim Tran 1J

Part C of 14.23 is supposed to be spontaneous. To determine spontaneity you would have to solve for delta G using delta G = -nFE. If delta G is negative, then the reaction is spontaneous as written (favors forward reaction). If delta G is positive, then the reaction is unfavorable and not spontaneou...

Kim Tran 1J

14.25 Arrange the following metals in order of increasing strength as reducing agents for species in aqueous solution: (a) Cu, Zn, Cr, Fe; (b) Li, Na, K, Mg; (c) U, V, Ti, Al; (d) Ni, Sn, Au, Ag I understand how to do (a) because they all share a common reaction in which 2 electrons are lost. But fo...

Kim Tran 1J

The coefficients are fractions because you trying to balance the reactants such that you would only get 1 mole of product. The questions does not state how many moles, but you can assume 1 mole since you are solving for standard entropy of formation

Kim Tran 1J

Assume that the standard change in enthalpy and entropy are independent of temperature and calculate the change in the standard Gibbs Free Energy for each of the following at 80 C. Over what temperature range will each reaction be spontaneous under standard conditions? a) B2O3 + 6HF -> 2BF3 + 3H2O W...

Kim Tran 1J

You can calculate the standard entropies of formation (delta S)
If delta S is negative, then the compound will be less stable at a higher temperature.
If delta S is positive, the compound will be more stable at a higher temperature

Kim Tran 1J

Unless stated otherwise in the problem, I think you are to assume the problem is asking for the change in Gibbs Free Energy for the creation of one mole of product hence why it would be 1/2N2 + 3/2H2 --> NH3 and not N2 + 3H2 --> 2NH3.

Kim Tran 1J

Enthalpy of formation is specifically referring to ∆H during the formation of 1 mole of the substance. In the appendix near the top for each column, it should be labeled with the units and does say kJ/mol for enthalpy of formation

Kim Tran 1J

Are there any situations in which we would need to apply adiabatic and diathermic? or are these just concepts we should know?

Kim Tran 1J

An ideal gas is a gas whose molecules don't have volume and don't interact with each other. It's a theoretical concept used to make calculations easier.

Kim Tran 1J

Suppose that 1.00 kJ of energy is transferred as heat to oxygen in a cylinder fitted with a piston; the external pressure is 2.00 atm. The oxygen expands from 1.00 L to 3.00 L against this constant pressure. Calculate w and U for the entire process by treating the O2 as an ideal gas. I got an answer...

Kim Tran 1J

I think the textbook used the equations: -q=qcal and Ccal= qcal/change in T Ccal is the heat capacity of the calorimeter q is the change in internal energy. So you can determine qcal is 22.5 kJ then you can plug this into Ccal=qcal/change in T change in temp= T2 - T1 = 23.97 C - 22.45 C = 1.52 C Cca...

Kim Tran 1J

A calorimeter was calibrated by mixing two aqueous solutions together, each of volume 0.100L. The heat output of the reaction that took place was known to be 4.16kJ, and the temperature of the calorimeter rose by 3.24 degrees C. Calculate the heat capacity of this calorimeter when it contains 0.200 ...

Kim Tran 1J

Your answer should be right. It should be +490 J or +4.90 x 10^2 J (3 sig figs).

The same question in the 6th edition has the answer listed as +4.90 x 10^2 J.

Kim Tran 1J

Work and energy are different. Work refers to a way that energy can be transferred by a system to its surroundings. Energy is the capacity of a system to do work.
So if a system can do a lot of work, it possesses a lot of energy.

Kim Tran 1J

The units for work is just joules right? Or is there more?

When calculating work, what units should we use for pressure and volume?

Kim Tran 1J

Yes, you are right.

The larger the Ka, the stronger the acid. The smaller the pKa, the stronger the acid
And the same applies for bases, the larger the Kb, the stronger the base. The smaller the pKb, the stronger the base

Kim Tran 1J

The ideal gas law (PV=nRT) is used to convert between molar concentration and partial pressure (Kc and Kp).

Kim Tran 1J

When dilution is involved you would use this equation: M1*V1=M2*V2 where M1 = the initial concentration in molarity, V1 = the initial volume before dilution, M2 = the concentration in molarity of the dilute solution, V2 = the volume after dilution So like in (d) you have 2.0 mL of 0.175 M KOH(aq) af...

Kim Tran 1J

Le Chatelier’s principle states that "when a stress is applied to a system in dynamic equilibrium, the equilibrium tends to adjust to minimize the effect of the stress" (textbook). So if any changes were to be made to a reaction like increasing or decreasing the concentration, pressure, et...

Kim Tran 1J

Based on the plot you know the initial kPa and the kPa at equilibrium. So you can draw out an ICE box. A B C I 28 0 0 C (-10) (+5) (+10) E 18 5 10 Hence knowing the initial and equilibrium you can determine the change, which would be 10, 5, and 10. You can simplify this ratio to 2:1:2, thus you get ...

Kim Tran 1J

Temperature permanently affects the rate constant of the forward and reverse reactions. Increasing/decreasing the temperature would add/remove heat changing the rate of collision of molecules of the whole reaction. But for changes in pressure, heat, or concentration, the system eventually will retur...

Kim Tran 1J

Yes, you would use the QC equation. Since the problem only gives you moles and the equilibrium constant equation uses concentration (i.e. molarity which is mol/L), you would need to divide the given moles by 3.00 L in order to be able to calculate the QC. And you would not include the solid as a par...

Kim Tran 1J

You can use the I.C.E. box to help visualize/solve this. Initial molar concentration (I) Change in molar concentration (C) Equilibrium molar concentration (E) N2 H2 NH3 I 0.250 0.500 0 C -0.075 -0.225 +0.15 E 0.175 0.275 0.15 Given the change in the concentration of NH3, you can find the change in N...

Kim Tran 1J

When K is larger than 1, there are more products at equilibrium, so equilibrium favors products ("equilibrium sits to the right"). When K is smaller than 1, there are more reactants at equilibrium, so equilibrium favors reactants ("equilibrium sits to the left"). This is because ...

Kim Tran 1J

One of the sig fig rules is your answer should have no more significant figures than the least accurately known number. So, 0.031 which has 2 sig figs is the "least accurately known number" and is the number of sig figs you should base your answer on.

Kim Tran 1J

Yes, you are correct.

acid: C6H5NH3+ because it donates a H+ to H2O
base: H2O because it accepts a H+ from C6H5NH3+
conjugate acid: H3O+ because it gains H+
conjugate base: C6H5NH2 because it loses H+

Kim Tran 1J

Can sulfur form more than an octet? I have seen SO2 drawn with two double bonds, but the textbook draws sulfur with a single bond and a double bond. Would the SO2 structure with two double bonds between O and S be more stable? Which would be the correct lewis structure to draw for SO2 to be the most...

Kim Tran 1J

tetrakis(trimethylphosphine)cobalt(III) sulfate

If the ligand is polydentate (able to attach at more than one binding site), then for 4 ligands you would say tetrakis- instead of tetra-
you can figure out the charge for cobalt by solving for the oxidation state.

Kim Tran 1J

The central atom is the atom with the lower ionization energy. Ionization energy increases across a period, so N would have a lower ionization energy then O. Also if you were to calculate the formal charges for the structure with N as the central atom and with O as the central atom, the one with N a...

Kim Tran 1J

In instances where you have structures with VSEPR formulas of AX2E3 or AX2E4, the molecular shape will be linear.; there are 2 bonded atoms and lone pairs present. However, the lone pair electrons are arranged in such a way that their repulsions cancel out to make the shape linear.

Kim Tran 1J

The question asks to predict the shapes and estimate the bond angles. For BH2- the e- arrangement is tetrahedral and the molecular shape is bent/angular (AX2E2) so I thought the bond angle for H-B-H would be less than 109.5 degrees. But the solution manual says the H-B-H angle is slightly less than ...

Kim Tran 1J

Electron arrangement is the arrangement of the electron groups, so it includes the lone pairs as part of the structure.
Molecular shape is the actual structure of the molecule, so it refers to the arrangement of atoms, not including lone pairs.

Kim Tran 1J

You count the number of bonded pairs and lone pairs around the central atom to find the total regions of electron density. Double and triple bonds count as one region of electron density.

Kim Tran 1J

S character refers to the percentage of s in a hybridization. So for example, sp3 = 25% s-character and 75% p-character because s makes up 1/4 of the hybridization. The s character will decrease as the number of orbitals increase. If there is more s character, the bond will be stronger.

Kim Tran 1J

The Lewis structure will have Cl in the middle (with 4 lone e-) with one O single bonded (with 6 lone e- around it) to Cl and the other O double bonded (with 4 lone e-) to Cl. Then add brackets around the Lewis structure and put the negative sign. And yes, chlorine can have an expanded octet so it c...

Kim Tran 1J

Bent and angular are synonyms and are used interchangeably to refer to the same types of structure.

Kim Tran 1J

Linear structures will always be 180 degrees. But trigonal planar can be 120 degrees or if it is trigonal planar bent then it will be less than 120 degrees. Tetrahedral is 109.5 degrees but if bent or trigonal pyramidal then it will be less than 109.5 degrees. So if the molecular geometry is not lin...

Kim Tran 1J

Angstrom is not in SI units but it can be converted to SI units. 1 Angstrom = 1*10^-10m

Kim Tran 1J

Why is the electronegativity of fluorine higher than hydrogen? but hydrogen has a high electronegativity right?

Kim Tran 1J

The CN bond in H3CNH2 is longer. If you were to draw out the Lewis structures, CN in HCN forms a triple bond, while CN in H3CNH2 forms a single bond. As the number of bonds increases between atoms, the length of the bond becomes shorter. So a single bond would be longer than a triple bond. Bond leng...

Kim Tran 1J

I think it has to do with when filling electrons, the orbitals with the lowest energy/principal quantum number would fill up first so 6s would fill up first, then, 4f, then 5d would fill up last. [Xe]6s^2 4f^14 5d^4 But then you would have to rewrite the electron configuration in order of energy, wh...

Kim Tran 1J

Orbitals are filled in the lower energy orbitals first, then fill higher energy orbitals. The 3d orbitals have a slightly higher energy than the 4s orbitals. So because the 4s orbitals has the lower energy, it gets filled first. When 3d orbitals are filled, 4s is no longer lower in energy. Hence ele...

Kim Tran 1J

True: Electrons in an s-orbital are more effective than those in other orbitals at shielding other electrons from the nuclear charge because an electron in an s-orbital can penetrate to the nucleus of the atom. This is because the s orbitals are located closer to the nucleus of an atom, so the attra...

Kim Tran 1J

You would need to know that the s-orbital only has 1 orbital. Each orbital can only hold 2 electrons.
So if there is only 1 orbital, it can only hold a maximum of 2 electrons.

Kim Tran 1J

To find how many electrons in an atom based on its quantum numbers, you need to determine how orbitals there are, meaning you have to first find l. You can find l given n. l = 0,1,2,n-1 n=2 so l=0,1 If l=0, then ml=0, so there is 1 orbital. If l=1, then ml= -1,0,+1, so there are 3 orbitals. Each orb...

Kim Tran 1J

You will use n (principal quantum number) to find l. l will be a range of values starting from 0 up to n-1. If n=3, l=0,1,2 (because from n-1, you know the last value l can be is 2) If n=2, l=0,1 Once you know l, you will use l to find ml. ml= -l ,..., +l. If l=0, ml=0 if l=1, ml= -1, 0, +1 if l=2, ...

Kim Tran 1J

Helium has the electron configuration 1s2 so it should belong to the s-block, but helium is placed with the p-block because helium is a noble gas (it has a full electron shell) so it is placed with the other noble gases in group 18. The periodic table is also sorted based on chemical properties, hen...

Kim Tran 1J

Yes. Electrons with lower values of l are better at shielding. The lower the number of l, the closer it is to the nucleus. The electrons closer to the nucleus block (or shield) the outer electrons from getting close to the nucleus. The electron density of s- or p- orbital is much higher than the f- ...

Kim Tran 1J

The cartesian axes are the x-, y-, and z- axes to represent three dimensions. This helps to describe the orientation of orbitals. So for what we would use cartesian axes for in this class, then the answer to your question would yes. It would be used to refer to the axes of the 3D orbitals.

Kim Tran 1J

You would use the Rydberg Equation when calculating the wavelength of light emitted when an electron moves between energy levels. In terms of what we have learned so far in class, this equation can be applied to find the wavelength of the lines in the hydrogen emission spectrum (refer to the Balmer ...

Kim Tran 1J

In part (a) of this question, we determined that BrF3 is the excess reactant and ClO2 is the limiting reactant. So to find the moles of product we use ClO2 (the limiting reactant) and mole ratios (refer to the given chemical equation) to calculate the amount of product that can form: 12 mol ClO2 (6 ...

Kim Tran 1J

You have to solve for the limiting reactant using mole ratios: 12 mol ClO2 (6 mol ClO2F/6 moleClO2) = 12 mol ClO2F 5 mol BrF3 (6 mol ClO2F/2 mol BrF3) = 15 mol ClO2F ClO2 is the limiting reactant because it produced the lesser amount of product (only 12 mol ClO2F). Hence, BrF3 is the excess reactant...

Kim Tran 1J

The question provides you with the molarity and moles of solute, and asks you to find the volume. So you will be using the formula for molarity to solve for the volume: rearrange M=n/V to V=n/M to solve for the volume. M= 0.278 M (or 0.278 mol/L) n= 0.00450 mol (the question gives you 4.50 mmol but ...

Kim Tran 1J

A would be the limiting reactant because in the given equation there is a ratio of 2 mol A : 1 mol B.
But since there is only 1 mol A, then you would only need 0.5 mol B.
So all 1 mol of A is consumed, however there is still 0.5 mol B excess since not all of B was needed to react with A.

Kim Tran 1J

"Formula units" are another way to say "atoms" or "molecules."
So 6.022 x 10^23 atoms = 6.022 x 10^23 formula units

Kim Tran 1J

Answer the following 4 questions for the balanced equation: 2A + 1B ---> 3C Same equation. 1 mol B reacts with 2 mol A, but there is only 1 mol A present, so it is the limiting reagent. 33. What is the maximum amount of product that can be produced? So I understood the 3 previous questions to this p...

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