Search found 33 matches
- Thu Mar 15, 2018 1:05 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Coefficients when writing cell diagrams
- Replies: 3
- Views: 1386
Re: Coefficients when writing cell diagrams
We only have to write the species that would take place in the redox reaction - its simply a summary of the cell structure. There is no need to write the coefficients in the cell diagram. However, if you were given a cell diagram and told to write the half reactions, you must balance the equations a...
- Tue Mar 13, 2018 2:02 pm
- Forum: Balancing Redox Reactions
- Topic: Test 2 #1
- Replies: 6
- Views: 866
Test 2 #1
Given the reaction:
C (s) + O2 (g) --> CO2 (g)
Why is the oxidation state of C in CO2 4+ and why is O2 the species that is reduced, when their oxidation states as reactants and products are essentially the same?
C (s) + O2 (g) --> CO2 (g)
Why is the oxidation state of C in CO2 4+ and why is O2 the species that is reduced, when their oxidation states as reactants and products are essentially the same?
- Tue Mar 13, 2018 1:58 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Test 2 #6b
- Replies: 2
- Views: 369
Test 2 #6b
The following redox couple forms a galvanic cell which generates a current under standard conditions.
Pt2+/Pt and AgF/Ag,F-
b) Identify the reducing agent.
For this problem I put down AgF as the reducing agent, but the correct answer was Ag. Why is Ag the reducing agent and AgF not?
Pt2+/Pt and AgF/Ag,F-
b) Identify the reducing agent.
For this problem I put down AgF as the reducing agent, but the correct answer was Ag. Why is Ag the reducing agent and AgF not?
- Sat Mar 10, 2018 2:41 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Molecularity and Coefficients
- Replies: 3
- Views: 1008
Molecularity and Coefficients
Does the stoichiometric coefficient in a reaction affect the molecularity? For example, if there was a reaction such as 2 A --> B + C, would the reaction be unimolecular or bimolecular?
- Thu Mar 08, 2018 10:19 am
- Forum: Second Order Reactions
- Topic: Half Life for First Order vs Second Order
- Replies: 5
- Views: 1761
Re: Half Life for First Order vs Second Order
I believe this difference arises because for first order reactions, concentrations and rates are directly proportional. On the other hand, for second order reactions, concentrations and rates have such a relationship that if you double the concentration, you would be raising the rate by a factor of ...
- Thu Mar 08, 2018 10:15 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: To determine k
- Replies: 2
- Views: 572
Re: To determine k
Based on the chemical equation, A is a reactant and B is a product. If product B is being formed, then it must mean that reaction A is being consumed. We want to find the equivalent of reactant A being consumed in relation to the product B. By finding the molar ratio between A and B, we can see that...
- Tue Mar 06, 2018 2:13 pm
- Forum: First Order Reactions
- Topic: 15.25
- Replies: 4
- Views: 633
Re: 15.25
Adding on to this, we want to use the exponential (non-linear form) when we are discussing a reactant that is decomposing.
- Thu Mar 01, 2018 10:26 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.9
- Replies: 3
- Views: 549
Re: 15.9
The units of k depend on the overall order of reaction. Page 620 of the textbook shows which units of k correspond to first, second, and third order reactions respectively. Generally, units of k will match up with the units concentration so that the units of rate will be concentration / time (mol / ...
- Thu Mar 01, 2018 10:22 am
- Forum: Second Order Reactions
- Topic: Example?
- Replies: 4
- Views: 599
Re: Example?
Another example is the reaction of nitrogen dioxide with fluorine gas.
NO2 (g) + F2 (g) → 2NO2F(g)
The respective rate would be rate = k[NO2][F2]. Since both NO2 and F2 are of the first order, the overall order is 1 + 1 = 2.
NO2 (g) + F2 (g) → 2NO2F(g)
The respective rate would be rate = k[NO2][F2]. Since both NO2 and F2 are of the first order, the overall order is 1 + 1 = 2.
- Thu Mar 01, 2018 10:14 am
- Forum: General Rate Laws
- Topic: Differential vs integrated
- Replies: 2
- Views: 392
Re: Differential vs integrated
Differential rate law is a function of change in concentration whereas integrated rate law is a function of time. One way to see the two rate laws is that differential rate law describes the mechanism of a reaction whereas integrated rate law relates the initial concentration to a measured concentra...
- Sun Feb 25, 2018 1:46 pm
- Forum: General Rate Laws
- Topic: initial rate law
- Replies: 5
- Views: 614
Re: initial rate law
For ease of analysis, we want to study specifically the initial concentrations of reactants, because as the reaction progresses, the presence of products could affect the interpretation of the reaction rate.
- Thu Feb 22, 2018 10:11 pm
- Forum: General Rate Laws
- Topic: Doubling Concentration in Second-Order Reaction
- Replies: 2
- Views: 2057
Doubling Concentration in Second-Order Reaction
Hi, can someone explain what the book means when it states "Doubling the concentration of the reactant in any second-order reaction increases
the reaction rate by a factor of 22 = 4" on page 619 of the textbook?
the reaction rate by a factor of 22 = 4" on page 619 of the textbook?
- Thu Feb 22, 2018 4:25 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Units of Reaction Rate
- Replies: 7
- Views: 928
Units of Reaction Rate
Are the units of concentration of reactant or product (i.e. mol / L) included when finding the reaction rate?
- Thu Feb 22, 2018 10:38 am
- Forum: Balancing Redox Reactions
- Topic: Gibbs free energy of half reactions
- Replies: 4
- Views: 737
Re: Gibbs free energy of half reactions
Yes. I believe this is required in homework problem 14.33).
- Sat Feb 17, 2018 8:38 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Delta G=-nFE
- Replies: 5
- Views: 2686
Re: Delta G=-nFE
The n in the equation is the moles of electrons transferred. To find the electrons transferred, you'll have to write the two half-reactions and balance them. A good example of this is homework problem 14.9.a): Given the overall equation 2 Ce 4+ (aq) + 3 I - (aq) --> 2 Ce 3+ (aq) + I 3 - (aq), the h...
- Sat Feb 17, 2018 2:09 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Electrolytes
- Replies: 1
- Views: 242
Re: Electrolytes
I believe the electrolyte is both the solutions with which the electrodes are in, as well as the solution that makes up the salt bridge in a galvanic cell. By definition, an electrolyte is something that can behave as an ionically conducting medium.
- Thu Feb 15, 2018 9:55 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Diagrams
- Replies: 3
- Views: 482
Cell Diagrams
For cell diagrams, how do you know when the anode should be on the left side and cathode on the right side versus cathode on the left side and anode on the right side?
- Thu Feb 08, 2018 9:27 pm
- Forum: Van't Hoff Equation
- Topic: Exothermic and decrease in tempterature
- Replies: 4
- Views: 702
Re: Exothermic and decrease in tempterature
Adding on to the above, an endothermic reaction would indicate heat as a reactant and an exothermic reaction would indicate heat as a product. With that said, you could relate this general relationship to Le Chatelier's principle. Say you had a reaction A + B --> C + Heat, then increasing the temper...
- Thu Feb 08, 2018 9:22 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: internal energy for isothermal expansion
- Replies: 2
- Views: 422
Re: internal energy for isothermal expansion
I think this is because internal energy is a state function dependent on temperature, and since the temperature stays the same, there should be no change in internal energy. We can also relate this relationship to q = -w. If the change internal energy is 0, then 0 = q + w and thus, q = -w.
- Thu Feb 08, 2018 9:18 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Spontaneity
- Replies: 9
- Views: 1294
Re: Spontaneity
When delta G is negative, the process is spontaneous. When delta G is positive, the forward process is not spontaneous, but the reverse process is spontaneous.
- Sat Feb 03, 2018 6:55 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: formula for standard entropy
- Replies: 5
- Views: 791
Re: formula for standard entropy
Adding on to this, an easy way to relate these equations is to remember the simple form of (Final - Initial). These equations are usable because state functions can be added or subtracted.
- Fri Feb 02, 2018 12:33 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.7
- Replies: 7
- Views: 737
Re: 9.7
I don't think we have to necessarily know the details of how we get the fractions, but pages 277 and 281 in the textbook explain what those values are and how they come to be.
- Fri Feb 02, 2018 12:10 pm
- Forum: Calculating Work of Expansion
- Topic: Calculating Work of Expansion
- Replies: 4
- Views: 681
Re: Calculating Work of Expansion
On exams we probably just need to use the equation. I think Dr. Lavelle shows the integral during lecture just so we can see how the equation comes to be.
- Sat Jan 27, 2018 9:17 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Ideal Behavior
- Replies: 3
- Views: 411
Ideal Behavior
When a problem states assume ideal behavior (such as in 9.11), what exactly are we assuming?
- Thu Jan 25, 2018 9:19 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Question 9.7
- Replies: 2
- Views: 1800
Re: Question 9.7
The question wants you to calculate ∆S, or the entropy change, when temperature increases at a) constant pressure and b) constant volume. We can find ∆S using the formula ∆S = n*C*ln(T 2 /T 1 ), where n is moles and C is heat capacity. Because we are using the heat capacity of ideal gas, we will ref...
- Wed Jan 24, 2018 10:00 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bond enthalpies vs Standard Enthalpies of Formation
- Replies: 2
- Views: 266
Re: Bond enthalpies vs Standard Enthalpies of Formation
When calculating bond enthalpies, reactants appear positive and products appear negative because by definition, breaking bonds requires energy whereas forming bonds release energy. Standard enthalpies of formation are experimentally found. By nature of the equation H° = ∑nH f °(products) - ∑nH f °(r...
- Fri Jan 19, 2018 5:58 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Clarification About DeltaU=q+w
- Replies: 1
- Views: 223
Re: Clarification About DeltaU=q+w
I believe so. If a system is being cooled, then heat would be leaving the system (-q) and if the system is expanding, then work is being done by the system (-w).
- Mon Jan 15, 2018 8:53 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 3 Methods
- Replies: 4
- Views: 475
Re: 3 Methods
In terms of signs for bond enthalpies, breaking bonds always require energy (endothermic, and thus a positive value) and forming bonds always releases energy (exothermic, and thus a negative value). Therefore, it does appear that we are subtracting the enthalpies of the products from the enthalpies ...
- Mon Jan 15, 2018 7:45 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: U = q + w assumptions
- Replies: 5
- Views: 732
Re: U = q + w assumptions
And if external pressure is constant, you would be able to substitute q with ∆H in the equation and w with P*∆V!
- Sat Jan 13, 2018 6:48 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Intensive vs Extensive [ENDORSED]
- Replies: 4
- Views: 411
Re: Intensive vs Extensive [ENDORSED]
Adding on to the above, because intensive properties are independent of sample size, they do not change when the size of the sample is changed, and therefore, are more useful. Also take note that some intensive properties are the ratio of two extensive properties, such as specific heat capacity, whi...
- Fri Jan 12, 2018 7:42 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat capacity
- Replies: 4
- Views: 361
Re: Heat capacity
I believe heat capacity is an extensive property where as specific heat capacity and molar heat capacity are intensive properties. And to elaborate, I think molar heat capacity at a constant pressure is considered a state function.
- Fri Jan 12, 2018 7:14 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.73.b
- Replies: 3
- Views: 384
8.73.b
For homework question 8.73b we are given the chemical equation CH 4 (g) + 4 Cl 2 (g) --> CCl 4 (g) + 4 HCl (g). In the solutions manual, it shows that we are breaking 4 mols of a C-H bond and we are forming 4 mols of a H-Cl bond. Can someone explain to me why there isn't only one mol being broken an...
- Tue Jan 09, 2018 4:23 pm
- Forum: Phase Changes & Related Calculations
- Topic: Expansion Work
- Replies: 3
- Views: 639
Re: Expansion Work
From what I understand, the main difference is that expansion work involves a change in volume whereas non-expansion work does not involve a change in volume (page 262 also defines both terms as such). On the same page, an example given for expansion work is an expanding gas inside a cylinder that p...