Search found 32 matches

by Eshwar Venkat 1F
Sat Mar 16, 2019 10:49 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: 15.67
Replies: 1
Views: 115

Re: 15.67

Make sure you're using the right equation, k = A*e^(-Ea/RT). Also remember that you need to convert your activation energy from kJ to J so that you can properly divide by the rate constant (R). I was able to get the right answer.
by Eshwar Venkat 1F
Sat Mar 16, 2019 4:38 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: dS = nC ln(T2/T1) [ENDORSED]
Replies: 1
Views: 94

Re: dS = nC ln(T2/T1) [ENDORSED]

There is no difference between the two formulas. For some reason, the textbook writes the formula like that, and in the subsequent examples, makes the substitution Cv = n*Cv,m. Just remember that entropy is an extensive property, and is thus dependent on the number of moles.
by Eshwar Venkat 1F
Sat Mar 16, 2019 4:32 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Standard Free Energy of Formation Units
Replies: 1
Views: 117

Re: Standard Free Energy of Formation Units

I can't offer any insight as to whether it is kJ/mol or just kJ, but for what it's worth, the molecules in the problem we did in class yesterday all had stoichiometric coefficients of 1, so the moles could cancel out to leave us with units of kJ, and the answer would be the same.
by Eshwar Venkat 1F
Fri Mar 15, 2019 9:21 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Order of States of Matter
Replies: 1
Views: 31

Re: Order of States of Matter

You always need to have solids on the outside of the diagram, but aside from that, I don't believe there is any order in which you need to list liquids, aqueous molecules, and gases. You would just need to follow the other rules for writing a cell diagram, and use commas when dealing with molecules ...
by Eshwar Venkat 1F
Mon Mar 11, 2019 3:35 pm
Forum: Second Order Reactions
Topic: Question 15.13 Part B (Sixth Edition)
Replies: 2
Views: 51

Re: Question 15.13 Part B (Sixth Edition)

Since each reactant has a first-order process, rate is directly proportional to the concentration. Thus, if you double the concentration of H2, the rate will double.
by Eshwar Venkat 1F
Tue Mar 05, 2019 10:21 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: 6th edition 15.19
Replies: 2
Views: 52

6th edition 15.19

For parts c and d, the answers listed in the book are k=2.85*10^12 L^4 mmol^-4 s^1, and Rate = 1.13*10^-2 mmol L^-1 s^1. When I did my calculations, the units and values did not match up, as the only way to get these values is if you convert all the initial concentrations and reactions rates to unit...
by Eshwar Venkat 1F
Mon Mar 04, 2019 8:13 pm
Forum: General Rate Laws
Topic: Order of Reaction
Replies: 6
Views: 75

Re: Order of Reaction

Remember that each reactant has an order with respect to itself as well, which we determine using reaction rate values. Thus, the order of a reaction is the sum of all the exponents (orders) of the reactants.
by Eshwar Venkat 1F
Wed Feb 27, 2019 8:47 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6th edition 14.13 d [ENDORSED]
Replies: 5
Views: 124

Re: 6th edition 14.13 d [ENDORSED]

I understand that Au (s) is serving as a conducting electrode in this case, but isn't the oxidation reaction that's occurring in this problem
Au+(aq) --> Au3+ (aq) + 2e- ? Is it not necessary to include Au+ when writing out the cell diagram for the anode side?
by Eshwar Venkat 1F
Tue Feb 26, 2019 8:11 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6th edition 14.13 d [ENDORSED]
Replies: 5
Views: 124

6th edition 14.13 d [ENDORSED]

For 14.13 d), I was confused as to why the cell diagram was written as is. I thought that the Au+ ion was being reduced to Au (s) and oxidized to Au3+. So why isn't the cell diagram Au+(aq) | Au3+(aq) || Au+(aq) | Au(s). Does it have something to do with the fact that Au (s) is also a pure conductor?
by Eshwar Venkat 1F
Sun Feb 24, 2019 7:24 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Standard Reduction Potential - Intensive
Replies: 1
Views: 26

Re: Standard Reduction Potential - Intensive

It is significant because when you are balancing redox reactions, which usually involves multiplying the entire oxidation or reduction reaction by a constant, you do not multiply the standard reduction potentials by that constant.
by Eshwar Venkat 1F
Sun Feb 24, 2019 7:21 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Endothermic Reactions Spontaneous
Replies: 2
Views: 47

Re: Endothermic Reactions Spontaneous

As we know, reactions where deltaG (deltaH - T*deltaS) < 0 are spontaneous. Endothermic reactions can be spontaneous if the entropy is negative, or, if the entropy is also positive, the temperature is high enough such that T*deltaS > deltaH.
by Eshwar Venkat 1F
Sun Feb 24, 2019 7:18 pm
Forum: Balancing Redox Reactions
Topic: 1/2 Rxns: Elements vs. Compounds
Replies: 3
Views: 44

Re: 1/2 Rxns: Elements vs. Compounds

Usually, it is good practice to be writing down the entire compound to keep track of the elements and any possible electron transfers, because there can be more advanced cases where there are multiple oxidations/reductions within a single reaction.
by Eshwar Venkat 1F
Sun Feb 24, 2019 7:16 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Temperature's Effect on Gibbs Free Energy
Replies: 2
Views: 35

Re: Temperature's Effect on Gibbs Free Energy

If a reaction is endothermic (deltaH > 0) and has a positive entropy, it can only be spontaneous if the temperature is sufficiently large such that T*deltaS > deltaH. Also, if a reaction is exothermic (deltaH < 0) and has a negative entropy, it can only be spontaneous if the temperature is sufficien...
by Eshwar Venkat 1F
Sun Feb 24, 2019 6:51 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Cell diagrams
Replies: 1
Views: 32

Re: Cell diagrams

You are always supposed to write a cell diagram in the order of reactants to products. When dealing with metals, the anode is where the oxidation of a solid metal to its ions takes place, and the cathode is where the reduction of another metal's ions into that metal's solid state takes place. In the...
by Eshwar Venkat 1F
Sun Feb 24, 2019 6:46 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 14.13 6th Edition
Replies: 1
Views: 51

Re: 14.13 6th Edition

Your solid inert conductor, which is Pt (s) in both of these cases, is what is supposed to be written on the outside. You always have to write in the order of reactants|products. In the case of the first example, the I-(aq) is being oxidized to form I2 (s), which is why the reaction is written that ...
by Eshwar Venkat 1F
Sun Feb 24, 2019 6:31 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Positional disorder vs. thermal disorder
Replies: 2
Views: 53

Re: Positional disorder vs. thermal disorder

Positional disorder refers to the change in entropy from the rearranging of molecules, while thermal disorder refers to the thermal motion of molecules, which increases with temperature. In this case, positional disorder is dominant because KNO3, when dissolved in water, breaks apart into K+ and NO3...
by Eshwar Venkat 1F
Wed Feb 13, 2019 3:15 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: 6th edition 8.93
Replies: 1
Views: 52

Re: 6th edition 8.93

First, you have to write out the balanced equation for the combustion reaction, in which C6H6 reacts with O2 to form CO2 and H2O. Since this is an isothermal reaction, you can use the formula w (work) = P*deltaV. Using the ideal gas law, you can also figure out that P*deltaV = delta n * R*T, and sub...
by Eshwar Venkat 1F
Wed Feb 13, 2019 3:00 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 4.11
Replies: 1
Views: 49

Re: 4.11

You are actually supposed to use both equations, the reason being that both irreversible expansion and heating of the system occur. The change in entropy of the gas is the sum of the answers you get from plugging into both equations.
by Eshwar Venkat 1F
Wed Feb 13, 2019 1:54 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Extensive vs Intensive Properties
Replies: 2
Views: 80

Re: Extensive vs Intensive Properties

Heat and work aren't properties because they are not state functions. However, internal energy (deltaU) and enthalpy (deltaH, which is often confused for heat) are state functions and extensive properties.
by Eshwar Venkat 1F
Wed Feb 06, 2019 10:50 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 6th edition 6.41
Replies: 1
Views: 37

Re: 6th edition 6.41

Assuming no heat is lost, heat from the water will be transferred to the ice cube. Remember that it is an ice cube (solid), at 0 degrees Celsius. The ice will melt and then increase in temperature, which is why you need to add the heat of fusion to the equation for heat.
by Eshwar Venkat 1F
Tue Feb 05, 2019 3:52 pm
Forum: Calculating Work of Expansion
Topic: Work on Phases of Matter
Replies: 2
Views: 68

Re: Work on Phases of Matter

The energy of a system can change by doing work on liquids and solids as well. Of course, this is assuming that the external pressure remains constant.
by Eshwar Venkat 1F
Tue Jan 29, 2019 9:42 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: HW Q
Replies: 2
Views: 53

Re: HW Q

With heat capacity problems, we're looking for the change in temperature. This means that as long as both temperature values are given in Celsius, there is no need to convert between Celsius and Kelvin and the value of the constant R can be plugged in as is.
by Eshwar Venkat 1F
Tue Jan 29, 2019 9:37 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Applied Exercises 8.111
Replies: 1
Views: 40

Re: Applied Exercises 8.111

You would multiply all of the mole ratios of the second reaction by 2, and then add both reactions, which would cancel out the SO2 term.
by Eshwar Venkat 1F
Thu Jan 24, 2019 12:02 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Ka and Kb values
Replies: 1
Views: 48

Re: Ka and Kb values

Since you are given an initial concentration value of NaC2H3O2, which dissociates into Na+ and C2H3O2-, you need to write out a reaction where the acetate ion reacts with water to form acetic acid and the hydroxide ion. This means that you need to use Kb to find the pOH and then convert to the pH.
by Eshwar Venkat 1F
Wed Jan 23, 2019 11:45 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Problem 12.45 (6th edition)
Replies: 2
Views: 115

Re: Problem 12.45 (6th edition)

The trend is that the higher the pKa value, the weaker the acid, and the stronger the base, the weaker its conjugate acid will be. As a result, the base strengths for this problem correspond to the increasing pKa values for the conjugate acids.
by Eshwar Venkat 1F
Wed Jan 23, 2019 11:39 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: K for gaseous equations
Replies: 2
Views: 49

Re: K for gaseous equations

It depends on whether the equilibrium values are partial pressure ones or concentration ones, which would determine if the constant is Kp or Kc.
by Eshwar Venkat 1F
Wed Jan 16, 2019 11:21 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Le Chatelier's Principle
Replies: 11
Views: 222

Re: Le Chatelier's Principle

If you increase the partial pressure of a reactant, the reaction shifts towards the products. If you decrease the partial pressure of a reactant, the reaction shifts towards the reactants.
by Eshwar Venkat 1F
Wed Jan 16, 2019 11:18 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Effect of Pressure [ENDORSED]
Replies: 2
Views: 52

Re: Effect of Pressure [ENDORSED]

When the number of moles on both sides is equal, increasing pressure will have no overall effect on the equilibrium.
by Eshwar Venkat 1F
Tue Jan 15, 2019 10:02 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 11.59
Replies: 3
Views: 58

Re: 11.59

To check, you can plug the two x values back into the ICE table. In cases like this, usually when you plug in the larger x value, it results in the equilibrium concentrations of both reactants being less than 0, which cannot happen.
by Eshwar Venkat 1F
Thu Jan 10, 2019 2:50 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Calculating Equilibrium Constant from Percent of Reactant
Replies: 2
Views: 36

Re: Calculating Equilibrium Constant from Percent of Reactant

You are supposed to use an ICE table for this problem. It says only 18.3% of BrCl gas remains, which means that 81.7% of the initial concentration will be your change "x" in the table, and 18.3% will be in the E row for BrCl gas.
by Eshwar Venkat 1F
Thu Jan 10, 2019 2:42 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Difference between C and P?
Replies: 4
Views: 57

Re: Difference between C and P?

If they just give the formula, I think it's safe to write the formula using partial pressure when gases are present, and use concentration brackets when aqueous compounds are present. However, if they give concentration numbers for gases, you can calculate K using those values.
by Eshwar Venkat 1F
Wed Jan 09, 2019 5:20 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: 11.41 in 6th edition
Replies: 1
Views: 38

Re: 11.41 in 6th edition

Yes, the reaction starts with no NH3 or CO2. For the equilibrium values, you need to convert the 17.4 mg of CO2 to moles and divide by the volume of the container to get molarity, and multiply that by 2 to get the molarity of NH3.

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