Search found 32 matches

by Aditya Pimplaskar 1J
Thu Mar 15, 2018 5:26 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Activation Energy
Replies: 1
Views: 106

Re: Activation Energy

In an endothermic process, the potential energy of the products is higher than that of the reactants. Its reverse process will be exothermic. In the endothermic process, the reaction must build up activation energy that is even greater than the difference in the two potentials. On the other hand, th...
by Aditya Pimplaskar 1J
Thu Mar 15, 2018 4:44 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Slow/Fast Step
Replies: 2
Views: 98

Re: Slow/Fast Step

The elementary slow step will be the one that reflects the rate law. Sometimes this is given directly in the slow step in terms of concentrations of reactants. In other cases you may have to use the pre-equilibrium method to substitute out a concentration of an intermediate.
by Aditya Pimplaskar 1J
Thu Mar 15, 2018 4:36 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Max work and delta G
Replies: 1
Views: 86

Re: Max work and delta G

The work that we refer to in these cases is non-expansionary (no volume change). There is a mathematical derivation on this Chemistry Community post!
viewtopic.php?t=5141
by Aditya Pimplaskar 1J
Mon Mar 05, 2018 11:03 pm
Forum: General Rate Laws
Topic: 15.3
Replies: 3
Views: 146

Re: 15.3

Unique rate of reaction takes the form:

For the reaction aA -> bB + cC (where a, b, and c are stoichiometric coefficients),
Rate =

Hope this helps!
by Aditya Pimplaskar 1J
Mon Mar 05, 2018 10:58 pm
Forum: Balancing Redox Reactions
Topic: Reducing Power
Replies: 2
Views: 113

Re: Reducing Power

Oftentimes, this is most easily done by looking at standard reduction potentials. The more negative the standard reduction potential, the less likely the element is to be reduced; this makes it more likely to be a reducing agent. Therefore, the more negative the standard reduction potential, the hig...
by Aditya Pimplaskar 1J
Mon Mar 05, 2018 10:30 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Example 15.2
Replies: 1
Views: 82

Re: Example 15.2

Yes, you can plug in the data from any experiment to the general rate law to figure out k (once you have found the orders of each reactant).
by Aditya Pimplaskar 1J
Wed Feb 28, 2018 11:17 pm
Forum: Zero Order Reactions
Topic: Real Life Example
Replies: 2
Views: 116

Re: Real Life Example

Zero order reactions are ones where reactant concentration does not affect reaction rate. As Dr. Lavelle said, this can happen when a catalyst or enzyme is needed for a reaction. When the catalyst/enzyme is saturated, an increase in concentration of reactant/substrate (reactants for enzymes) won't c...
by Aditya Pimplaskar 1J
Wed Feb 28, 2018 11:01 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: When do we calculate for k
Replies: 3
Views: 123

Re: When do we calculate for k

You can calculate k using integrated rate laws or half life formulas if you are given information about the other variables that are involved (including final and initial concentrations and time/half life). Algebraically isolate k in the formula appropriate for the reactant order and solve.
by Aditya Pimplaskar 1J
Wed Feb 28, 2018 10:57 pm
Forum: Second Order Reactions
Topic: 15.39
Replies: 2
Views: 114

Re: 15.39

The 0.19 moles/L of substance B is equated to 0.095 moles/L of substance A by stoichiometric ratios (1 mole A: 2 moles B). However, the 0.095 moles/L of substance A is the decrease in substance A, leaving 0.055 moles/L of substance A as the final concentration. The 0.37 is the ratio of final concent...
by Aditya Pimplaskar 1J
Wed Feb 28, 2018 10:53 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: 15.17 a [ENDORSED]
Replies: 3
Views: 136

Re: 15.17 a [ENDORSED]

To find the order of each reactant, find trials where the concentration of only one of the reactants changes. Compare the change in reactant concentration with the change in rate. If the rate changes by the same factor as the concentration, it is order 1. If it changes by a square of the change in c...
by Aditya Pimplaskar 1J
Tue Feb 20, 2018 9:38 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: 14.51
Replies: 1
Views: 85

Re: 14.51

E 0 must be zero in this case as the electrodes used on both sides are the same (Ag). The only thing that allows for the flow of electrons in the case of the same electrode would be a difference in concentration. Yet, in the standard case E is calculated when both sides are in standard 1M solution. ...
by Aditya Pimplaskar 1J
Tue Feb 20, 2018 9:32 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 4.13 part b
Replies: 1
Views: 104

Re: 4.13 part b

Though I2 is given as a solid in the reaction, it isn't used as the electrode as it is limited in its ability to conduct electron flow. We use platinum, which is more conductive, for our electrodes in this case.
by Aditya Pimplaskar 1J
Tue Feb 20, 2018 9:26 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Standard Hydrogen Electrode (SHE)?
Replies: 1
Views: 49

Re: Standard Hydrogen Electrode (SHE)?

A standard hydrogen electrode is one that is used as a reference for other electric potentials. By setting this as the baseline, we can find other standard potentials by checking the electron flow across a cell with hydrogen as one of the components and the substance in question as the other.
by Aditya Pimplaskar 1J
Tue Feb 13, 2018 9:32 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Bonds formed - negative?
Replies: 3
Views: 143

Re: Bonds formed - negative?

Breaking bonds requires energy, so for broken bonds we use a positive bond enthalpy. On the other hand, forming bonds releases energy as it is a lower energy state; so we use negative bond enthalpies.
by Aditya Pimplaskar 1J
Tue Feb 13, 2018 8:42 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Spontaniety
Replies: 2
Views: 160

Re: Spontaniety

No. Spontaneity does not say anything about speed. It is just about the Gibb's free energy. Speed has more to do with the kinetics of the reaction.
by Aditya Pimplaskar 1J
Tue Feb 13, 2018 8:40 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Heat
Replies: 1
Views: 88

Re: Heat

Heat is transferred from hotter bodies to colder bodies and depends on the other thermodynamic phenomena that are occurring. In turn, it matters the path that the heat follows. Heat is not something that a body inherently has; it's defined as transferred energy, which depends on math.
by Aditya Pimplaskar 1J
Fri Feb 09, 2018 3:52 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Delta G= Wmax
Replies: 8
Views: 409

Re: Delta G= Wmax

W max refers to the maximum amount of work a system can do. Delta G is the maximum work that a system can do at a given temperature and pressure. So we can equate the two. I also found this previous post on Chemistry Community from Dr. Lavelle: https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=5141
by Aditya Pimplaskar 1J
Fri Feb 09, 2018 3:43 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Delta G Standard
Replies: 2
Views: 251

Re: Delta G Standard

To address the second part of your question, standard ΔG refers to ΔG at 25 degrees C and 1 atm with 1M reactants and products. We know that standard ΔG= -RTln(K). Standard ΔG will provide information about the relative concentrations of products and reactants at equilibrium. It will only be equal t...
by Aditya Pimplaskar 1J
Fri Feb 09, 2018 3:27 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: What is a Faraday? [ENDORSED]
Replies: 3
Views: 139

Re: What is a Faraday? [ENDORSED]

To add on to the previous reply, Faraday's constant is derived by dividing Avogadro's number (number of electrons in 1 mole) by the number of electrons that make up one Coulomb. This yields the 96485 C/mol from the previous response and also provides an explanation for its units.
by Aditya Pimplaskar 1J
Thu Feb 01, 2018 9:22 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 9.7
Replies: 3
Views: 118

Re: 9.7

This refers to page 281 in the textbook. The Cp=5/2 R refers to molar heat capacity at a constant pressure for atoms. This comes from Cp = Cv + R, as Cv for atoms is 3/2R. We can use the molar heat capacity for atoms of ideal gas in this problem as it plugs into ΔS = n*Cp*ln(T2/T1). This formula der...
by Aditya Pimplaskar 1J
Thu Feb 01, 2018 8:48 pm
Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
Topic: Homework 9.21
Replies: 3
Views: 131

Re: Homework 9.21

Since the molecules are all aligned in the same direction, the number of possible states is 1. In turn we use the formula (# of possible states)^# of particles. This yields 1^64 as there are 64 molecules.
by Aditya Pimplaskar 1J
Thu Feb 01, 2018 6:30 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Negative delta s value
Replies: 4
Views: 582

Re: Negative delta s value

A negative ΔS value will correspond to a spontaneous process when temperature is low and ΔH is negative. This is because of the relationship ΔG = ΔH - TΔS. We know that ΔG needs to be negative in order for the reaction to be spontaneous as that means that the entropy in the system is increasing (By ...
by Aditya Pimplaskar 1J
Fri Jan 26, 2018 12:37 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: State properties of heat and enthalpy
Replies: 2
Views: 137

Re: State properties of heat and enthalpy

Heat transfer depends on the path in which the heat travels. Since we are talking about heat as a flow of energy, the path that this flow takes matters. This makes heat a path function. However, enthalpy, which is a change in a quantity of heat (absorbed/released), we ultimately only have to conside...
by Aditya Pimplaskar 1J
Fri Jan 26, 2018 12:35 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Formula confusion [ENDORSED]
Replies: 2
Views: 67

Re: Formula confusion [ENDORSED]

S = Kb ln2 is just the formula for when the number of positions available doubles (W2 = 2*W1). We see this when volume doubles.
S = Kb lnW is just the general equation linking the degeneracy (number of accessible states) with entropy.
by Aditya Pimplaskar 1J
Thu Jan 25, 2018 9:50 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Entropy of an Ideal Gas
Replies: 2
Views: 119

Re: Entropy of an Ideal Gas

Note that the formula in the previous reply (ΔS = n*R*ln(V2/V1)) applies when you plug in moles of gas (n). In a more general sense, ΔS = Kb*ln(W2/W1). When you multiply Kb by Avogadro's number, you end up with R. ΔS ends up q/T, where at a constant temperature, change in entropy is linked to a chan...
by Aditya Pimplaskar 1J
Fri Jan 19, 2018 10:41 am
Forum: Phase Changes & Related Calculations
Topic: q water=-q ice equation
Replies: 1
Views: 76

Re: q water=-q ice equation

This is because the heat absorbed by the ice cube, which is colder than the water (heat is the transfer of energy due to a temperature difference: higher temperature to lower temperature), is exchanged from the water. The q value for the ice is positive because it absorbs heat, while the q value for...
by Aditya Pimplaskar 1J
Fri Jan 19, 2018 10:23 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 8.29
Replies: 4
Views: 127

Re: 8.29

I agree with the previous replies: more bonds (greater molecular complexity) contributes to a higher heat capacity. To add to previous responses, this property stems from the fact larger, more complex molecules can move (vibrate, rotate, etc.) in more ways. More options for motion leads to more ener...
by Aditya Pimplaskar 1J
Fri Jan 19, 2018 12:16 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Delta U?
Replies: 7
Views: 324

Re: Delta U?

Just like the previous replies state, ΔU is the change in internal energy of the system. To add on to what the previous replies said, you can think of U as the kinetic energy + potential energy of the particles that your sample is comprised of. Internal energy is a state function, like enthalpy. It ...
by Aditya Pimplaskar 1J
Fri Jan 19, 2018 12:10 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Enthalpy [ENDORSED]
Replies: 3
Views: 97

Re: Enthalpy [ENDORSED]

Additionally, it can be helpful to know that change in enthalpy (ΔH) is equal to q p or the heat given off/absorbed in a chemical reaction at a constant pressure. There is a derivation for this that comes from the equation: ΔH = ΔU + PΔV. ΔH = q p becomes useful in a lot of the exercises from the te...
by Aditya Pimplaskar 1J
Fri Jan 12, 2018 1:19 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Catalysts and Enthalpies
Replies: 2
Views: 119

Re: Catalysts and Enthalpies

I agree! Catalysts would decrease the activation energy of a particular reaction, which could represent a different pathway in getting from your initial point to your final point. However, given enthalpy is a state function, altering the path does not alter the value of enthalpy, so enthalpy remains...
by Aditya Pimplaskar 1J
Fri Jan 12, 2018 12:24 pm
Forum: Phase Changes & Related Calculations
Topic: ∆Hsub= ∆Hfus+ ∆Hvap
Replies: 4
Views: 353

Re: ∆Hsub= ∆Hfus+ ∆Hvap

To add on to the previous replies, we know that state functions do not consider the path taken to get from point A to point B. In this case, even though the path for sublimation is different from the path where fusion is followed by vaporization, since the initial (solid) and final (vapor) states ar...
by Aditya Pimplaskar 1J
Fri Jan 12, 2018 12:08 pm
Forum: Phase Changes & Related Calculations
Topic: Constant Temperature
Replies: 4
Views: 191

Re: Constant Temperature

One way to look at it is that the heat isn't contributing to the kinetic energy of the molecules, as the heat is going towards breaking intermolecular forces. Since average kinetic energy of the water molecules is unchanged when the phase change is happening, temperature stays constant. On the other...

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