## Search found 30 matches

- Sat Mar 17, 2018 4:27 pm
- Forum: *Enzyme Kinetics
- Topic: Enzyme
- Replies:
**5** - Views:
**1497**

### Re: Enzyme

A homogeneous catalyst is a catalyst in the same phase as the reactant. A heterogeneous catalyst is a catalyst that is in a different phase than the reactant.

- Sat Mar 17, 2018 1:51 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Rule when writing cell diagrams?
- Replies:
**5** - Views:
**338**

### Re: Rule when writing cell diagrams?

In a cell diagram, the left side should be the anode, and the right side should be the cathode.

- Sat Mar 17, 2018 1:49 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Determining Slow Step
- Replies:
**2** - Views:
**2177**

### Re: Determining Slow Step

The two important components to look for when determining reaction mechanisms are 1) that the sum of elementary steps = overall reaction and 2) that the mechanisms must agree with the experimentally determined rate law. The elementary step that has a rate law that matches the overall reaction's rate...

- Fri Mar 09, 2018 10:29 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Negative Order
- Replies:
**7** - Views:
**4469**

### Re: Negative Order

A negative order means the concentration appears in the denominator of the rate law. Increasing the concentration of the species (usually a product) slows down the reaction.

- Fri Mar 09, 2018 10:27 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: unit of concentration [ENDORSED]
- Replies:
**4** - Views:
**392**

### Re: unit of concentration [ENDORSED]

I think the important thing to remember is to make sure that all your units match.

- Fri Mar 09, 2018 10:22 pm
- Forum: General Rate Laws
- Topic: 15.23 b
- Replies:
**3** - Views:
**246**

### Re: 15.23 b

If you make both sides of the equation ln([A]t/[A]0) = -kt negative, you will have: ln([A]0/[A]t) = kt. Making the natural log negative flips the content because:

ln([A]t/[A]0) = ln[A]t - ln[A]0

ln([A]t/[A]0) = ln[A]t - ln[A]0

- Sat Mar 03, 2018 9:55 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.15 [ENDORSED]
- Replies:
**5** - Views:
**349**

### Re: 15.15 [ENDORSED]

In general, rate = k * [A]^a * [B]^b ...

Thus, overall order = a + b + ...

Thus, overall order = a + b + ...

- Sat Mar 03, 2018 9:49 pm
- Forum: General Rate Laws
- Topic: half-life
- Replies:
**8** - Views:
**412**

### Re: half-life

The greater the rate constant, the shorter is the half life. You can use half life to calculate the amount of reactant remaining in a reaction.

At any stage of a reaction, the concentration of A remaining after n half-lives is: (1/2)^n * [A]

At any stage of a reaction, the concentration of A remaining after n half-lives is: (1/2)^n * [A]

- Sat Mar 03, 2018 9:43 pm
- Forum: General Rate Laws
- Topic: 15.3 partb ?
- Replies:
**3** - Views:
**186**

### Re: 15.3 partb ?

When you are not looking at the unique rate of the reaction, you must specify the species to which the rate refers. The species consumed or produced have rates that are related to the stoichiometry of the reaction.

- Fri Feb 23, 2018 2:35 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: 9.5
- Replies:
**4** - Views:
**352**

### Re: 9.5

ΔS(total) = -q/T(1) + q/T(2)

The first reservoir loses heat, so the q should be negative. The second reservoir gains the heat that the first reservoir loses, so the q should be positive.

The first reservoir loses heat, so the q should be negative. The second reservoir gains the heat that the first reservoir loses, so the q should be positive.

- Fri Feb 23, 2018 2:32 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.111
- Replies:
**3** - Views:
**242**

### Re: 11.111

Since ΔG(1) is given, you can solve for k(1). The relationship between k(1) and k(2) is given, with k(2)=k(1)/10. You can solve for ΔG(2) using k(2). For this problem, you should use ΔG=-RT ln(k).

- Fri Feb 23, 2018 2:20 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 14.27
- Replies:
**6** - Views:
**385**

### Re: 14.27

Gibbs Free Energy is a state function, so you can add intermediate components to reach your desired final value. Cell potential is NOT a state function, so you can't use the same addition method. However, you can relate Gibbs Free Energy and cell potential through ΔG=-nFE.

- Thu Feb 15, 2018 11:40 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.15 Question
- Replies:
**2** - Views:
**207**

### Re: 9.15 Question

ΔH(reverse process)=-ΔH(forward process), so

ΔH(freezing)=-ΔH(fusion)

ΔH(freezing)=-ΔH(fusion)

- Thu Feb 15, 2018 11:29 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess's Law
- Replies:
**5** - Views:
**359**

### Re: Hess's Law

Enthalpy is a state function, meaning you can calculate the desired value from the final and initial conditions, and you can also add components to reach the desired value. Hess's Law uses the fact that enthalpy is a state function to calculate ΔH.

- Thu Feb 15, 2018 11:27 am
- Forum: Phase Changes & Related Calculations
- Topic: Specific heat of water or ice?
- Replies:
**6** - Views:
**462**

### Re: Specific heat of water or ice?

To determine which specific heat capacity to use, you need to look at the phase of the substance. More specifically, if you are working on a problem in which you are calculating the heat required to change ice at -10 degrees C to liquid water at 0 degrees C, you would first calculate the heat associ...

- Fri Feb 09, 2018 2:20 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: 9.37
- Replies:
**3** - Views:
**252**

### Re: 9.37

For this problem, you are finding ΔS(rxn) by using ΔS(m), the standard molar entropy (which is the entropy of 1 mol of substance). Use ΔS(rxn)=ΣnΔS(m)(products)-ΣnΔS(m)(reactants) I think ΔS(rxn) refers to the entropy of the entire reaction, and ΔS(f) refers to the entropy of formation of a substance.

- Fri Feb 09, 2018 2:10 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Finding entropy of vaporization from temperatures NOT = to boiling point [ENDORSED]
- Replies:
**3** - Views:
**287**

### Re: Finding entropy of vaporization from temperatures NOT = to boiling point [ENDORSED]

To find entropy of vaporization at temperatures NOT = to boiling point: example of finding entropy of vaporization of water at 85 degrees C (9.19) 1) increase T of water (from 85 degrees C to 100 degrees C) 2) find entropy of vaporization at boiling point (100 degrees C for water) 3) cool/decrease T...

- Mon Feb 05, 2018 12:43 am
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: determining number of microstates
- Replies:
**3** - Views:
**218**

### determining number of microstates

In general, how would you calculate the number of microstates of a more complicated molecule? Does anyone have an example?

- Fri Feb 02, 2018 8:58 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Problem 9.19 about Standard Entropy
- Replies:
**6** - Views:
**321**

### Re: Problem 9.19 about Standard Entropy

You should break this problem into 4 steps: 1) heat liquid water (ΔS=C * ln(T2/T1)) 2) vaporize water at boiling point (ΔS(vap) is given in the problem) 3) cool water vapor back to 85 degrees C (ΔS=C * ln(T2/T1)) 4) sum all the entropies from the above steps to find ΔS(total) You need to calculate e...

- Fri Feb 02, 2018 8:50 pm
- Forum: Calculating Work of Expansion
- Topic: Free expansion and irreversible expansion
- Replies:
**2** - Views:
**180**

### Re: Free expansion and irreversible expansion

No work is done in free expansion; an example of free expansion is a gas pushing against a piston and into a vacuum. Free expansion is an irreversible process, but it is not the same as irreversible expansion. For an irreversible expansion, P(external) does not equal P(internal). Reversible expansio...

- Fri Feb 02, 2018 8:44 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Free expansion
- Replies:
**2** - Views:
**200**

### Re: Free expansion

No work is done in free expansion because there is no external pressure; for example, free expansion occurs when a gas pushes against a piston and expands into a vacuum (in which no work is done). For a gas that is expanding by pushing against a piston and an external pressure (no free expansion), t...

- Sat Jan 27, 2018 7:34 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.40
- Replies:
**4** - Views:
**252**

### Re: 8.40

I also got about the same answer (using E(total) = E(raising temperature) + E(phase change)).

- Sat Jan 27, 2018 7:25 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: 8.65
- Replies:
**3** - Views:
**234**

### Re: 8.65

For this problem, first use Hess's Law: multiply the first thermochemical equation given by a factor of 2: 4NO(g) + 2O2(g) --> 4NO2(g) ΔH=(-114.1 kJ) * 2 Combine the two thermochemical equations given, and you'll see that the 4NO2 will cancel out and will not be in the final thermochemical equation:...

- Sat Jan 27, 2018 7:14 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: q isn't a state function
- Replies:
**3** - Views:
**345**

### Re: q isn't a state function

q isn't a state function because it isn't solely dependent on the initial and final states; the value of q depends on the pathway taken to reach the final q. Here is an example of why heat is not a state property: Consider raising the temperature of 50.0g of water from 25.0°C to 50.0°C. One method t...

- Fri Jan 19, 2018 5:04 pm
- Forum: Calculating Work of Expansion
- Topic: Integral from Today's Lecture
- Replies:
**3** - Views:
**204**

### Re: Integral from Today's Lecture

The integral represents the area under the curve, and I think the "infinite number of steps" is referring to infinitesimally small step sizes when calculating the area. This integral shows the relationship of work to pressure and volume through calculus, and I believe for chem 14B right no...

- Fri Jan 19, 2018 4:55 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: P delta V is significant?
- Replies:
**2** - Views:
**351**

### Re: P delta V is significant?

When there's a reaction at constant pressure that involves solids and liquids, a change in volume will be insignificant because V of the reactants will equal (or almost equal) V of the products. ΔV=0, so PΔV=0. Also, there is no expansion work. Thus, with constant pressure, for a reaction that invol...

- Fri Jan 19, 2018 4:45 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.39
- Replies:
**2** - Views:
**151**

### Re: 8.39

For this problem, you need to consider the energy needed for the phase change (melting/fusion) as well as the energy needed for the temperature change. Looking at a heating curve for H2O, solid H2O will melt at 0.0°C. The sum of the energy needed for the phase change and the energy needed for the te...

- Sat Jan 13, 2018 3:55 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat Not a State Property
- Replies:
**6** - Views:
**279**

### Re: Heat Not a State Property

Here is an example of why heat is not a state property: Consider raising the temperature of 50.0g of water from 25.0°C to 50.0°C. One method to raise the temperature is to supply energy as heat through an electric water heater. The heat can be calculated from the specific heat capacity of water: q =...

- Sat Jan 13, 2018 3:39 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: State function vs non state function
- Replies:
**2** - Views:
**2271**

### Re: State function vs non state function

Internal energy is a state function. For example, in the first scenario, you raise the temperature of 50g of water from 25°C to 70°C. In the second scenario, you take the same amount of water at 25°C, raise the temperature to boiling point, vaporize the water, condense it, and eventually cool it to ...

- Sat Jan 13, 2018 2:53 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess' Law
- Replies:
**3** - Views:
**188**

### Re: Hess' Law

Hess's Law states that enthalpy changes are additive; the change of enthalpy at each step of a multi-step reaction can be added to determine the total change in enthalpy. Here is an example for how to use Hess's Law: 1. The first reaction shows one of the reactants from the overall reaction as a rea...