lnk(k'/k)=E/R((1/T)-(1/T'))

ln(k'/k)=(38kJ/mol)/(.08314kJ/K*mol)((1/298K)-(1/310K))-.59

k=1.5x10^10 L/mol*s

Solve for k'

k'=2.7 x 10^10 L/mol*s

## Search found 32 matches

- Mon Mar 12, 2018 3:17 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.63
- Replies:
**4** - Views:
**281**

- Mon Mar 12, 2018 2:31 am
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: types of systems
- Replies:
**3** - Views:
**431**

### Re: types of systems

Open:living plant, gas in a car engine

Closed: reusable ice pack, mercury in a thermometer, coolant in a refrigerator coil

Isolated:bomb calorimeter, high quality thermos

Closed: reusable ice pack, mercury in a thermometer, coolant in a refrigerator coil

Isolated:bomb calorimeter, high quality thermos

- Tue Mar 06, 2018 10:02 am
- Forum: Second Order Reactions
- Topic: 15.19
- Replies:
**2** - Views:
**198**

### Re: 15.19

Compare rate 1&3 to find the order of concentration B because A conc and C conc remain constant between these two experiments.

Rate 3/rate 1=50.8/8.7=5.85

B conc3/B conc1=3.02/1.25=2.416

2.416^x=5.85

X=2

Therefore, the concentration B is a second order reaction.

Rate 3/rate 1=50.8/8.7=5.85

B conc3/B conc1=3.02/1.25=2.416

2.416^x=5.85

X=2

Therefore, the concentration B is a second order reaction.

- Tue Mar 06, 2018 9:50 am
- Forum: First Order Reactions
- Topic: Book Problem 15.21
- Replies:
**5** - Views:
**547**

### Re: Book Problem 15.21

First start with the equation: ln(A/A(initial)=-kt

Then plug in k(7.6x10^-3/min) and t (5 hours but use 300 min)

Second you will now have: ln(A/A(initial)=-2.28

Cancel out on by using e: (A/A(initial)=e^-2.28

A(initial)= 20mg

Solve for A which is 2.0 mg

Then plug in k(7.6x10^-3/min) and t (5 hours but use 300 min)

Second you will now have: ln(A/A(initial)=-2.28

Cancel out on by using e: (A/A(initial)=e^-2.28

A(initial)= 20mg

Solve for A which is 2.0 mg

- Tue Mar 06, 2018 9:43 am
- Forum: First Order Reactions
- Topic: 15.21 SOS!!!!!
- Replies:
**2** - Views:
**197**

### Re: 15.21 SOS!!!!!

It is a property of natural logs. When you subtract two natural logs, you can use division.

For example, lnA-lnB= ln(A/B)

So we are given lnA=-kt+lnA(initial)

Then you can change it depending on what you are solving for:

So lnA-lnA(initial)=-kt which is converted to ln(A/A(initial))=-kt

For example, lnA-lnB= ln(A/B)

So we are given lnA=-kt+lnA(initial)

Then you can change it depending on what you are solving for:

So lnA-lnA(initial)=-kt which is converted to ln(A/A(initial))=-kt

- Wed Feb 28, 2018 8:21 pm
- Forum: Second Order Reactions
- Topic: 15.39
- Replies:
**2** - Views:
**156**

### Re: 15.39

t=((1/[A])-(1/[A]"sub0"))/k=((1L/.080mol)-(1L/0.10mol))/0.015L/mol min=1.7 x 10^2 min

[A]=0.15 mol A/L- [(0.19 mol/L)-(1 mol A/ 2 mol B)] = 0.055 (mol A)/L=0.38[A]"sub0"

[A]=0.15 mol A/L- [(0.19 mol/L)-(1 mol A/ 2 mol B)] = 0.055 (mol A)/L=0.38[A]"sub0"

- Wed Feb 28, 2018 8:12 pm
- Forum: First Order Reactions
- Topic: 15.37
- Replies:
**2** - Views:
**160**

### Re: 15.37

(a)t1/2=.693/k=.693/2.81 x 10^-3 min^-1=247 min (b) t=(ln([so2cl2]"sub0"/[So2Cl2]"subt")/k=ln10/(2.81 x 10^-3 min^-3)=819 min (c)[A]"subt"=[A]"sub0"e^-kt // masses and concentrations are proportional (b/c the vessel is sealed), therefore (mass left)"subt&...

- Wed Feb 28, 2018 1:29 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Test 3
- Replies:
**2** - Views:
**164**

### Re: Test 3

15.1-15.6, assuming it's the same for whether you have discussion early in the week or later.

- Thu Feb 22, 2018 3:08 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Galvanic vs. Electrolytic Cell
- Replies:
**3** - Views:
**191**

### Re: Galvanic vs. Electrolytic Cell

We mainly learned about Galvanic cells: they are cells in which chemical energy released by a spontaneous redox reaction are converted to electrical energy. For the overall cell reaction, deltaG should be less than 0 to indicate that the reaction is spontaneous. In an electrolytic cell, electrical e...

- Thu Feb 22, 2018 2:55 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Finding Q
- Replies:
**2** - Views:
**287**

### Re: Finding Q

Using the Nernst equation with Ln and log should be interchangeable, but I had this same issue with this problem. I also got 3.27x10^6 rather than just 10^6.

There is an added constant for log. For ln: E = E-RT/nF(lnQ) but for log: E = E-2.303RT

There is an added constant for log. For ln: E = E-RT/nF(lnQ) but for log: E = E-2.303RT

- Thu Feb 22, 2018 2:51 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: n
- Replies:
**3** - Views:
**215**

### Re: n

N is the number of moles of electrons being transferred in a chemical reaction. Make sure to balance the equation for the chemical reaction before seeing how many moles of electrons are exchanged.

- Fri Feb 16, 2018 11:59 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Question 14.19
- Replies:
**4** - Views:
**273**

### Re: Question 14.19

Looking at the cell:

Cu/Cu2+ electrode is the anode (oxidation is occurring)

M2+/M electrode is the cathode

Then use the formula: E=E(cathode)-E (anode)

-.689V=E(cathode)-(+.34V)

E(cathode)=-0.349V

Cu/Cu2+ electrode is the anode (oxidation is occurring)

M2+/M electrode is the cathode

Then use the formula: E=E(cathode)-E (anode)

-.689V=E(cathode)-(+.34V)

E(cathode)=-0.349V

- Fri Feb 16, 2018 11:38 pm
- Forum: Balancing Redox Reactions
- Topic: Oxidizing vs. Reducing Agent
- Replies:
**7** - Views:
**358**

### Re: Oxidizing vs. Reducing Agent

Oxidizing Agent: its the species that is being reduced as it removes the electron from a substance

Some examples include: O2,O3,Fe3+

Reducing Agent: the species that is being oxidized as it supplies electrons to a substance

Examples include: H2, H2S

Some examples include: O2,O3,Fe3+

Reducing Agent: the species that is being oxidized as it supplies electrons to a substance

Examples include: H2, H2S

- Fri Feb 16, 2018 6:51 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Galvanic/Voltaic
- Replies:
**2** - Views:
**114**

### Re: Galvanic/Voltaic

They are both an electrochemical cell in which a spontaneous chemical reaction is used to generate an electric current.

Galvanic and Voltaic are two different names for the same electrochemical cell, both converting chemical energy into electrical energy.

Galvanic and Voltaic are two different names for the same electrochemical cell, both converting chemical energy into electrical energy.

- Sun Feb 11, 2018 10:20 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.15
- Replies:
**12** - Views:
**955**

### Re: 11.15

Nevermind, I reread the question

- Sun Feb 11, 2018 10:09 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.15
- Replies:
**12** - Views:
**955**

### Re: 11.15

If deltaG is positive, how is the reaction spontaneous?

- Wed Feb 07, 2018 9:48 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Calculating Degeneracy
- Replies:
**3** - Views:
**252**

### Re: Calculating Degeneracy

W=# of possible orientations^# of molecules.

The problem with either give you a specific amount of molecules or referring to the example 9.8 where it says one mole, you can use Avogadro's number to account for the amount of molecules in the one mole.

The problem with either give you a specific amount of molecules or referring to the example 9.8 where it says one mole, you can use Avogadro's number to account for the amount of molecules in the one mole.

- Wed Feb 07, 2018 9:34 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.21
- Replies:
**3** - Views:
**187**

### Re: 9.21

a) Entropy of a solid made of 64 molecules in which they are all aligned in the same direction. First you need to solve for W (orientation^#molecules). therefore, W=1^64=1. Then use the Boltzmann formula: S=klnW. (k=1.381x10^-23J/K #this is a constant that is given). Snce we know ln1=0, S=0. b) Entr...

- Wed Feb 07, 2018 9:20 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.41
- Replies:
**1** - Views:
**127**

### Re: 9.41

In order to get entropy (-enthalpy(kJ/mol)/Temperature(K)) you need to change the grams to mol in order to cancel the moles in the enthalpy. Then your resulting entropy will be J/K.

- Wed Jan 31, 2018 3:45 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies:
**7951** - Views:
**1214616**

### Re: Post All Chemistry Jokes Here

The optimist sees the glass half full.

The pessimist sees the glass half empty.

The chemist see the glass completely full, half in the liquid state and half in the vapor state.

The pessimist sees the glass half empty.

The chemist see the glass completely full, half in the liquid state and half in the vapor state.

- Wed Jan 31, 2018 3:41 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Thermodynamics 1st 2nd 3rd Law
- Replies:
**2** - Views:
**252**

### Re: Thermodynamics 1st 2nd 3rd Law

1ST: Law of Conservation of Energy-energy can't be created nor destroyed; it can only be transferred from one form to another. Equation: deltaU=q+w 2ND: Entropy of any isolated system always increases, going to maximum entropy. 3RD: Entropy of a system approaches a constant value as temperature appr...

- Wed Jan 31, 2018 3:33 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 8.41
- Replies:
**3** - Views:
**198**

### Re: 8.41

The explanations above are great. Here is the math to maybe make it more clear. heat (ice cube)=(50g/18.02 gmol^-1)(6.01 x 10^3 J/mol) +(50g)(4.184JC^-1g^-1)(Tfinal-0)=1.67 x 10^4J +(209J/C)(Tfinal-0) heat (water)=(400g)(4.184J/Cg)(Tfinal-45)=1.67 x 10^3 J/C (Tfinal-45) then you set them equal to ea...

- Wed Jan 24, 2018 12:57 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Problem 8.51
- Replies:
**2** - Views:
**296**

### Re: Problem 8.51

When using the enthalpy of formation equation, you should get -13168 kJ/mol. This is the amount of energy released per mole of reaction. In order to get the amount of energy released per mole of TNT consumed, you divide the total amount of energy released (-13158kJ/mol) by 4 (from looking at the bal...

- Wed Jan 24, 2018 9:54 am
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Closed vs. Isolated
- Replies:
**8** - Views:
**506**

### Re: Closed vs. Isolated

Examples of these are: Closed: an reusable ice pack-->matter cannot be exchanged with the surroundings because it is sealed, but energy can when it is refrozen or melted. Isolated: insulated thermos-->matter cannot be exchanged with the surroundings because it is sealed. energy can also not be excha...

- Wed Jan 24, 2018 9:48 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.113
- Replies:
**3** - Views:
**229**

### Re: 8.113

the state gr means graphite. It is the most stable form of carbon under standard conditions.

- Thu Jan 18, 2018 1:59 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Help understanding what is being asked in 8.19
- Replies:
**2** - Views:
**195**

### Re: Help understanding what is being asked in 8.19

for part b, where they ask "what percentage of the heat is used to raise the temperature of the water?" you take the amount of heat used to raise the temperature of the water and divide it by the total heat. Then multiply it be 100 to get the percent. HEAT: q=mCdeltaT ((1.30 X 10^5J)/(1.45...

- Thu Jan 18, 2018 1:41 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Difference between heat(q), internal energy(U), and enthalpy(H)
- Replies:
**4** - Views:
**2137**

### Re: Difference between heat(q), internal energy(U), and enthalpy(H)

Heat(Q): the transfer of energy between a system temp. and its surrounding temperature. It can go in either direction, but is transferred from high temperature to low temperature. An object, itself, does not possess heat. Internal energy (U): the total amount of energy stored in a system. For a clos...

- Thu Jan 18, 2018 1:23 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Q. 8.39
- Replies:
**2** - Views:
**169**

### Re: Q. 8.39

Step 1: (melting the ice at 0 degrees C) deltaH=amount of moles of water x the enthalpy of fusion at standard state deltaH=(80g/18.02g/mol)x(6.01kJ/mol)=26.7 kJ Step 2: (raising the temperature of the liquid water from 0 degrees C to 25 degrees C) deltaH=m(specific heat capacity)(deltaT) deltaH=(80g...

- Sat Jan 13, 2018 11:14 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.51
- Replies:
**1** - Views:
**136**

### Re: 8.51

Yes, you use enthalpies of formation to find the enthalpy of reaction for the reaction. 28(-393.51kJ/mol)+10(-241.82kJ/mol)-4(067kJ/mol)=-13168kJ/mol (the energy released per mole of reaction) Since it is asking "per mole," 1/4 of the energy (3292kJ/mol) will be released per mole of TNT co...

- Sat Jan 13, 2018 10:55 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Problem 8.31
- Replies:
**3** - Views:
**173**

### Re: Problem 8.31

Gas has two values for molar heat capacity because one accounts for the constant volume process (Cv) and the other is the constant pressure process (Cp). Cv is the change in internal energy (U) with respect to to change in temperature at a fixed volume whereas Cp is the change in enthalpy (H) with r...

- Sat Jan 13, 2018 10:48 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.23
- Replies:
**5** - Views:
**293**

### Re: 8.23

Yes, using kJ/C would be ideal because it is the same units that are given in the problem. To solve this problem, you simply calculate the heat capacity of the calorimeter by dividing the amount of energy supplied (22.5kJ) by the change in temperature (23.97C-22.45C). Unless it says to calculate the...

- Sat Jan 13, 2018 10:38 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Molar heat capacity of gas
- Replies:
**3** - Views:
**138**

### Re: Molar heat capacity of gas

Gas has two values for molar heat capacity because one accounts for the constant volume process (Cv) and the other is the constant pressure process (Cp). Cv is the change in internal energy (U) with respect to to change in temperature at a fixed volume whereas Cp is the change in enthalpy (H) with r...