## Search found 44 matches

- Mon Mar 19, 2018 12:59 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Delta U
- Replies:
**3** - Views:
**403**

### Re: Delta U

delta U is the change between the initial and final values of U, meaning that the pathway it took doesn't matter, as long as any path begins and ends on the same values.

- Sat Mar 17, 2018 12:02 pm
- Forum: General Rate Laws
- Topic: Test #3 Question 3
- Replies:
**1** - Views:
**334**

### Re: Test #3 Question 3

We first balance the equation into 2AB + C2 ---> 2ABC

Because C2 is consumed at a rate of 0.250 M/s, we can use the ratios to see that ABC is formaed at 0.500 M/s

We can then use

24.00 M-0.35 M = 23.65 M increase, and

23.65/(0.500 M/s)=47.3 seconds

Because C2 is consumed at a rate of 0.250 M/s, we can use the ratios to see that ABC is formaed at 0.500 M/s

We can then use

24.00 M-0.35 M = 23.65 M increase, and

23.65/(0.500 M/s)=47.3 seconds

- Thu Mar 15, 2018 1:02 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.13 d
- Replies:
**2** - Views:
**181**

### Re: 14.13 d

We split the reaction into the anode and cathode reactions: Anode: Au(s)--->Au3+ (aq) + 3e- Cathode: Au+(aq)+e- ----> Au(s) We see that both the anode and cathode have solid Au, so there shouldn't be Pt in the cell diagram. Thus, from our anode and cathode, our cell diagram should be Au(s)| Au3+(aq)...

- Wed Mar 14, 2018 12:16 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.101 [ENDORSED]
- Replies:
**2** - Views:
**149**

### Re: 8.101 [ENDORSED]

Because the ratio of SO2 to O2 is 2 to 1, when we use all 0.030 moles of SO2 for the reaction, we also use 0.015 moles of O2 as well. Thus,

there is still 0.015 moles of O2 left over, meaning SO2 is the limiting reactant.

there is still 0.015 moles of O2 left over, meaning SO2 is the limiting reactant.

- Mon Mar 12, 2018 12:50 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.67b
- Replies:
**1** - Views:
**115**

### Re: 15.67b

For part b, we aren't actually comparing the temperatures 298 K and 350K. We are actually simply doing part a again with T=350 K. Therefore,

we should have T=350 for both k1 and k2, and have

k1/k2 = Ae^(-75kJ/mol/RT)/Ae^(-125kJ/mol/RT)

Substituting R=8.314 and T=350 K, we get

e^17.1824 = 3x10^7

we should have T=350 for both k1 and k2, and have

k1/k2 = Ae^(-75kJ/mol/RT)/Ae^(-125kJ/mol/RT)

Substituting R=8.314 and T=350 K, we get

e^17.1824 = 3x10^7

- Thu Mar 08, 2018 7:58 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: to find k
- Replies:
**4** - Views:
**317**

### Re: to find k

I think that normally, the question will have the data be in such a way that k will generally be the same throughout several experiments with different concentrations. I'm guessing that if there is a noticeable difference in k values from each experiment, we might average them out.

- Wed Mar 07, 2018 9:21 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.23
- Replies:
**1** - Views:
**119**

### Re: 15.23

Since this is a first order reaction, we can use the first order half life equation

t_(1/2)=ln 2/k

We know that the half-life is 1000s, so what we need to do is calculate for k

k=ln 2 /t_(1/2)=0.693/1000 s = 6.93x10^-4 s^-1

t_(1/2)=ln 2/k

We know that the half-life is 1000s, so what we need to do is calculate for k

k=ln 2 /t_(1/2)=0.693/1000 s = 6.93x10^-4 s^-1

- Tue Mar 06, 2018 10:27 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Homework 15.23
- Replies:
**3** - Views:
**204**

### Re: Homework 15.23

We also know that because the reaction is a first order reaction, we can utilize the equation:

ln [A] = ln [A_0] - kt, and plug in the equations for A_0 = .153 mol/L and A=.153-2*.034=.085 mol/L, and t=115 s

ln [A] = ln [A_0] - kt, and plug in the equations for A_0 = .153 mol/L and A=.153-2*.034=.085 mol/L, and t=115 s

- Mon Mar 05, 2018 3:35 pm
- Forum: Second Order Reactions
- Topic: 15.35
- Replies:
**2** - Views:
**182**

### Re: 15.35

The reason is because the reaction is a second-order reaction, meaning we must use the equation 1/[A]=kt+1/[A_0]. We notice that that for 16*[A]=[A_0], we have 16/[A_0]=kt+1/[A_0] kt=15/[A_0]. Compared to the half -life 2[A]=[A_0] 2/[A_0]=kt+1/[A_0] kt=1/[A_0]. Thus, we see that for second-order rea...

- Thu Mar 01, 2018 11:35 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Units for the Different Order Reactions [ENDORSED]
- Replies:
**2** - Views:
**207**

### Re: Units for the Different Order Reactions [ENDORSED]

Depending on the order of the reaction, we see that the units for k will be M/s divided by M^n, where n is the sum of the orders of the compounds the the reaction.

Take for example Rate=k*[A]^2 * [B]. Combined, the total order is 2+1=3, so we have the unit of k be M/s * 1/M^3, or 1/(M^2*s)

Take for example Rate=k*[A]^2 * [B]. Combined, the total order is 2+1=3, so we have the unit of k be M/s * 1/M^3, or 1/(M^2*s)

- Tue Feb 27, 2018 9:15 pm
- Forum: General Rate Laws
- Topic: Elementary Reactions
- Replies:
**4** - Views:
**232**

### Re: Elementary Reactions

Elementary reactions are essentially reactions where the reactants react to form the products directly, rather than forming any intermediate products.

- Tue Feb 27, 2018 9:09 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell potential
- Replies:
**5** - Views:
**263**

### Re: Cell potential

I think that, if it's not specified, we would orient the reaction so that the potential is positive.

- Mon Feb 26, 2018 6:00 pm
- Forum: First Order Reactions
- Topic: Equation in the Book vs. in Lecture
- Replies:
**2** - Views:
**168**

### Re: Equation in the Book vs. in Lecture

Since we can derive one equation from the other, there doesn't seem to be any impact if we use one or the other. It just depends on which one you feel more comfortable using.

- Sun Feb 25, 2018 12:01 am
- Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
- Topic: 14.55
- Replies:
**3** - Views:
**397**

### Re: 14.55

This problem is actually electrolysis, which is different from a Galvanic cell. In electrolysis at a cathode, the more positive potential is preferred, which is why Ni is chosen istead of So4

- Wed Feb 21, 2018 12:22 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: pg. 578 example
- Replies:
**1** - Views:
**114**

### Re: pg. 578 example

i think that in the case of Hydrogen, we alter the other electrode half-reaction i such a way that its Standard electric potential becomes positive. In this case, the original cell has Zinc as the cathode, but because it has a negative E value -0.76, we reverse the reaction to have a positive E valu...

- Tue Feb 20, 2018 2:30 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Surface Area of Electrodes
- Replies:
**1** - Views:
**97**

### Re: Surface Area of Electrodes

Increasing surface area generally won't increase E and E°, since they depend on the substances making up the electrodes, not how much there is.

- Mon Feb 19, 2018 8:06 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Example 14.4
- Replies:
**1** - Views:
**74**

### Re: Example 14.4

Pt means platinum, and we include it in cell diagrams when there aren't any solid metals on either side. It's also not included in the reaction equation because it is inert.

- Fri Feb 16, 2018 3:38 pm
- Forum: Balancing Redox Reactions
- Topic: Basic vs Acidic
- Replies:
**2** - Views:
**174**

### Re: Basic vs Acidic

When we're balancing the solution for acidic, we use H+ and H2O, whereas in a basic solution, we balance the sides with OH0 and H2O.

- Wed Feb 14, 2018 12:01 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: #4 on practice midterm
- Replies:
**3** - Views:
**229**

### Re: #4 on practice midterm

Because delta_U=0 due to the initial and final states being the same, we have w=-q, so essentially, we just need to find the energy of the work, which we can then use to find heat. We can do this by splitting the problem into 3 parts: 1. Calculate work done compressing 50.0 L to 20.0 L 2. Calculate ...

- Tue Feb 13, 2018 11:58 am
- Forum: Calculating Work of Expansion
- Topic: Work
- Replies:
**1** - Views:
**148**

### Re: Work

Yes. The formula for work is W=-P*Delta_V, so if V decreases (compresses), Work is done on the system, so Work is positive. Likewise, if V increases, Work is negative.

- Mon Feb 12, 2018 5:02 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 2nd and 3rd law of Thermodynamics
- Replies:
**2** - Views:
**164**

### Re: 2nd and 3rd law of Thermodynamics

The main difference between the two laws is what they have to say about entropy. The second law states how a system favors going from lower to higher entropy. The third law states how temperature affects entropy by having entropy approach 0 as temperature approaches 0 (for perfect crystals)

- Fri Feb 09, 2018 11:30 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy and Equilibrium
- Replies:
**1** - Views:
**113**

### Re: Gibbs Free Energy and Equilibrium

Yes, ∆G=-RTln(k) is the standard Gibbs free energy of reaction, whereas the ∆G=0 as equilibrium is the Gibbs free energy of reaction. The difference is that Standard free energy is the calculated by finding the difference of products and reactants in their standard states, whereas the ∆G=0 equilibri...

- Thu Feb 08, 2018 10:10 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.23
- Replies:
**1** - Views:
**104**

### Re: 9.23

The double bond for COF2 actually restricts the configuration for the molecule. However, COF2 still ultimately has more configurations than BF3 due to the presence of oxygen and 2 fluorine molecules around the central carbon atom, meaning COF2 has higher molar entropy.

- Tue Feb 06, 2018 8:26 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: How to approach 9.19
- Replies:
**2** - Views:
**179**

### Re: How to approach 9.19

The calculate entropy, we need to calculate the entropy of raising liquid water from 85 C to 100 C, turning into vapor, and then cooling the vapor back to 85 C. The molar heat capacities will help in calculating entropy in temperature change, and you add in the entropy of vaporization to get the tot...

- Tue Feb 06, 2018 8:22 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.19
- Replies:
**3** - Views:
**223**

### Re: 8.19

0.38 J/g C is the specific heat of copper, meaning the total heat supplied includes heating both the water and the copper kettle.

- Tue Feb 06, 2018 12:34 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Boltzmann Equation
- Replies:
**2** - Views:
**96**

### Re: Boltzmann Equation

Kb is a constant, at 1.381 x 10^-23 J/K. When calculating W for 1.00 mole where there are n states for each molecule, you raise n to the power of 6.02x10^23 (Avagadro's Number), so for 1.00 mole, you can

have Entropy S= (6.02x10^23) x Kb ln n, where there are n states for the molecule.

have Entropy S= (6.02x10^23) x Kb ln n, where there are n states for the molecule.

- Fri Feb 02, 2018 10:02 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Entropy change at higher temperature
- Replies:
**2** - Views:
**137**

### Re: Entropy change at higher temperature

The equation delta_S=delta_q/T means there is a change in entropy due to a change in heat, rather than the overall entropy. Therefore, what this equation really means is that, at higher temperatures, a change in heat correlates to less of a change in entropy. Adding more heat does means more entropy...

- Thu Feb 01, 2018 9:17 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.7
- Replies:
**3** - Views:
**156**

### Re: 9.7

Cp means the heat capacity of a gas at constant pressure. We use this equation as part of the formula for calculating entropy in the problem, along with C_v, which is heat capacity for constant volume.

- Wed Jan 31, 2018 6:44 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.13
- Replies:
**2** - Views:
**134**

### Re: 9.13

I think the ideal conditions part means that the nitrogen gas is behaving as an ideal gas.

I don't know about the 1.00 moles of gas, though, but I don't think it has to do with the ideal conditions part.

I don't know about the 1.00 moles of gas, though, but I don't think it has to do with the ideal conditions part.

- Tue Jan 30, 2018 2:58 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.27 D
- Replies:
**3** - Views:
**160**

### Re: 9.27 D

I think we can use the equation PV=nRT. Given a constant mole of Argon and Temperature, we see that Pressure and volume are inversely proportional, so because 2.00 atm is higher than 1.00 atm, the 2.00 atm situation will have a smaller volume.

- Fri Jan 26, 2018 12:00 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.41
- Replies:
**6** - Views:
**330**

### Re: 8.41

It is because the ice melts at 0.0 C in the problem, and becomes water. We therefore use the specific heat of water rather than ice.

- Wed Jan 24, 2018 7:36 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Relating Degeneracy and Volume
- Replies:
**2** - Views:
**114**

### Re: Relating Degeneracy and Volume

I think it may be because the volume has doubled, and in V1, the positions available consist of the entire volume of the container. V2 is double the volume of V1, meaning there is twice the volume available for gas molecules. I also think we assume that the gas is an ideal gas, so the volume occupie...

- Tue Jan 23, 2018 5:33 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Monatomic Gas
- Replies:
**5** - Views:
**304**

### Re: Monatomic Gas

R is the gas constant, which is 8.314 J/(mol*K)

- Mon Jan 22, 2018 6:26 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Recognizing standard states
- Replies:
**2** - Views:
**174**

### Re: Recognizing standard states

I would also like to know as well, though right now, I think we may be given a table on the information if a problem requires us to use this knowledge.

- Sat Jan 20, 2018 6:05 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: 8.5
- Replies:
**3** - Views:
**227**

### Re: 8.5

We are measuring the internal energy of the system. Compressing the gas with 340 kJ of work. Remember that the Work done in a a system is w= - P*delta_V. The sign is negative, and because the system is compressed, delta_V is negative, meaning work done is positive, since compressing the system does ...

- Fri Jan 19, 2018 12:40 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Homework 87
- Replies:
**2** - Views:
**105**

### Re: Homework 87

I think that phase changes are only positive if the process is endothermic (melting and vaporization). The reason they are all positive is that the problem asks to convert ice into gas, which requires that the ice absorb heat, making the enthalpy positive.

- Thu Jan 18, 2018 12:49 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Homework Problem 8.41
- Replies:
**1** - Views:
**107**

### Re: Homework Problem 8.41

The heat needed to melt the ice cube and raise the melted cube's temperature comes from the surrounding water. Therefore, in actuality, we should calculate the heat needed to melt the ice cube and m*c*deltaT for ice, and equate it with the heat lost by the water in order to melt the ice cube. The eq...

- Mon Jan 15, 2018 10:23 pm
- Topic: 8.45
- Replies:
**2** - Views:
**151**

### Re: 8.45

It's because we are calculating in terms of S8 moles rather than Carbon (We're calculating using the number of S8 moles rather than C). That means for 1 mole of S8 reacting in the equation, enthalpy change is 358.8 kJ. Our units is thus +358.8 kJ/1 mole S8, which we use to calculate the answer, rath...

- Mon Jan 15, 2018 1:02 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Example 8.3
- Replies:
**2** - Views:
**139**

### Re: Example 8.3

It's because both Celsius and Kelvin have the same "scale" (An increase of 1 Celsius is equal to 1 Kelvin).By finding the difference between the two temperatures, the conversion cancels out. If we convert 20 C and 100 C to Kelvin, we get 293.15 K and 373.15 K respectively, and finding the ...

- Sat Jan 13, 2018 5:12 pm
- Topic: Sig Figs for Question 8.45 (a)
- Replies:
**5** - Views:
**253**

### Re: Sig Figs for Question 8.45 (a)

I've also gotten the answer of 448.5 kJ, rounded to 449 in my calculations. It doesn't seem to be due to be a rounding error, as you get the answer 448.5 kJ from only one equation. In addition, this situation also appears to occur in other answers I've gotten here and there. How accurate do we need ...

- Wed Jan 10, 2018 5:55 pm
- Forum: Phase Changes & Related Calculations
- Topic: Heat Capacity vs Specific Heat Capacity
- Replies:
**3** - Views:
**209**

### Re: Heat Capacity vs Specific Heat Capacity

Heat capacity and specific heat capacity differ in the fact that heat capacity only accounts for temperature and energy, whereas specific heat capacity also takes into account mass. As such, the units for heat capacity is Joules/Kelvin (or Celsius), whereas the units for specific heat capacity is of...

- Wed Jan 10, 2018 5:52 pm
- Forum: Phase Changes & Related Calculations
- Topic: Reaction in a Flask
- Replies:
**5** - Views:
**455**

### Re: Reaction in a Flask

Endothermic reactions absorb heat, and in most cases, this heat is taken from the surroundings. In this case, the flask is considered the surroundings of the reaction (that's why the exothermic reaction, which releases heat, warms the flask/surroundings). The endothermic reaction absorbs the heat fr...

- Tue Jan 09, 2018 7:17 pm
- Forum: Phase Changes & Related Calculations
- Topic: Phase Changes [ENDORSED]
- Replies:
**3** - Views:
**204**

### Re: Phase Changes [ENDORSED]

The most important phase changes would probably be between solids and liquids (melting and freezing) along with phase changes between liquids and gases (evaporation and condensation). I think sublimation (solid to gas) and deposition (gas to solid) i might also be talked about in class, though I'm n...

- Mon Jan 08, 2018 11:38 pm
- Forum: Phase Changes & Related Calculations
- Topic: Supercooling and Superheating
- Replies:
**2** - Views:
**209**

### Re: Supercooling and Superheating

For superheating, boiling typically occurs when the vapor pressure of a liquid equals the pressure of its surroundings. Some liquids may require additional pressure to overcome surface tension. Superheating basically occurs because of the surface tension and surrounding pressure both opposing the va...