## Search found 14 matches

- Sat Mar 17, 2018 3:26 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: A?
- Replies:
**6** - Views:
**416**

### Re: A?

Yep, I think the fact that it is constant (in general) is the reason it can be subtracted out of the equation when combining two Arrhenius equations to predict k1 using k2, T2, and T1.

- Sat Mar 17, 2018 3:19 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Midterm Number 5
- Replies:
**2** - Views:
**235**

### Re: Midterm Number 5

I think it's because you want the overall change in U to be 0.

From Step 1, we know that deltaU = w = -158 J.

So now we need the deltaU from Step 2 to offset that change; basically, deltaU1 + deltaU2 = 0. This means that deltaU2 = -deltaU1, and so for the second step, deltaU = -(-158J) = +158J.

From Step 1, we know that deltaU = w = -158 J.

So now we need the deltaU from Step 2 to offset that change; basically, deltaU1 + deltaU2 = 0. This means that deltaU2 = -deltaU1, and so for the second step, deltaU = -(-158J) = +158J.

- Sat Mar 17, 2018 3:16 pm
- Forum: *Enzyme Kinetics
- Topic: Catalysts in a reaction
- Replies:
**7** - Views:
**467**

### Re: Catalysts in a reaction

Catalysts go from reactant to product, while intermediates go from product to reactant.

- Wed Feb 14, 2018 11:21 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Problem 11.111
- Replies:
**2** - Views:
**141**

### Re: Problem 11.111

Got it, thanks for the explanation! I didn't know to use 298K for the temperature when doing the homework problem. Plugging that in for T in the expression I got leads to the same answer of -194 kJ/mol.

- Wed Feb 14, 2018 2:40 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Problem 11.111
- Replies:
**2** - Views:
**141**

### Problem 11.111

Problem 11.111 is as follows: A certain enzyme-catalyzed reaction in a biochemical cycle has an equilibrium constant that is 10 times the equilibrium constant of the next step in the cycle. If the standard Gibbs free energy of the first reaction is 200. kJ/mol, what is the standard Gibbs free energy...

- Thu Feb 08, 2018 3:12 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: 9.76
- Replies:
**2** - Views:
**134**

### Re: 9.76

I was wondering if we need to know this as well!

- Thu Feb 08, 2018 3:03 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Delta G positive
- Replies:
**2** - Views:
**104**

### Re: Delta G positive

I think in terms of spontaneity, a reaction with a positive Delta G is non-spontaneous in the forward direction (which means the reverse reaction is spontaneous) and a reaction with a negative Delta G is spontaneous in the forward direction (reverse reaction is non-spontaneous).

- Thu Feb 08, 2018 2:49 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: DG = -nFE
- Replies:
**3** - Views:
**197**

### Re: DG = -nFE

I believe n refers to electrons transferred. Basically, you find the number of electrons in each balanced half-reaction and if they match, that is n. If they don't, you use the lowest common multiple as n.

- Thu Feb 01, 2018 8:02 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: sign of q
- Replies:
**7** - Views:
**209**

### Re: sign of q

It might help to think of it using the familiar concept q(system)=-q(surroundings). A system's surroundings gain the heat that a system loses. Because a gain in heat is an increase or positive value, the system must have the opposite sign when it loses heat. So if heat is leaving a system, q will be...

- Thu Feb 01, 2018 7:57 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: homework 9.19
- Replies:
**4** - Views:
**139**

### Re: homework 9.19

From a previous CC post: "The water is heated up to 100 C because that is the boiling point of water where water turns from liquid to gas. There is a change because the question asks for the standard enthalpy of vaporization of water at 85 C, which implies that it was a liquid that was heated u...

- Thu Feb 01, 2018 7:53 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Homework 9.5 [ENDORSED]
- Replies:
**11** - Views:
**410**

### Re: Homework 9.5 [ENDORSED]

I think Sfinal - Sinitial works in general because that is the definition of delta S. In this case, I think it can be explained by the concept above; because the 800. K reservoir is losing heat, so its entropy when you calculate overall entropy change should be negative.

- Sat Jan 27, 2018 1:19 pm
- Forum: Calculating Work of Expansion
- Topic: Reversible vs Irreversible
- Replies:
**2** - Views:
**123**

### Re: Reversible vs Irreversible

I don't think it's possible to assume either one of those; reversible reactions are not necessarily isothermal, and irreversible reactions are definitely not necessarily isobaric.

- Thu Jan 25, 2018 9:52 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Gibbs Free Energy and Maximum Non-Expansion Work
- Replies:
**1** - Views:
**519**

### Re: Gibbs Free Energy and Maximum Non-Expansion Work

I found this from a previous CC post: G is actually defined as the maximum non-expansion work under constant T and P. Non-expansion means not related to a volume change, for example, electrical work, which is also what we call "useful work". This can be proven mathematically using differen...

- Thu Jan 25, 2018 9:32 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.49
- Replies:
**5** - Views:
**187**

### Re: 8.49

I think we should use Kelvin in problems where it makes a difference because that is the SI unit for temperature.

For problems where the equation just asks for a change in temperature, using Kelvin or Celsius will yield the same result.

For problems where the equation just asks for a change in temperature, using Kelvin or Celsius will yield the same result.