## Search found 30 matches

Wed Mar 14, 2018 8:48 pm
Forum: Balancing Redox Reactions
Topic: 14.15 a?
Replies: 2
Views: 189

### Re: 14.15 a?

They have a charge of +1 and -1 respectively, so that the overall charge will be 0.
Wed Mar 14, 2018 8:08 pm
Forum: Balancing Redox Reactions
Topic: 14. 85
Replies: 1
Views: 135

### Re: 14. 85

Look at their cell potentials. If the cell potential is bigger it will be a stronger oxidizing agent, while if it is lower it will be a stronger reducing agent.
Tue Mar 13, 2018 10:23 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: 15. 89
Replies: 1
Views: 173

### Re: 15. 89

We know Step 2 is the slow step instead of Step 3 because N2O2 is an intermediate, which is why we would have to use the pre-equilibrium method and substitute in [NO]^2 for [N2O2]. If step 3 were the slow state N2O would have to be replaced by H2 and N2O2, which does not match with the rate law given.
Sun Mar 11, 2018 5:23 pm
Forum: Balancing Redox Reactions
Topic: Ranking elements
Replies: 8
Views: 581

### Re: Ranking elements

You would look at the standard potential of the reaction. The higher standard potential value, the stronger it is as an oxidizing agent. While, the smaller (or more negative) the standard potential is, the stronger it is as a reducing agent.
Fri Mar 09, 2018 10:22 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: 15.63
Replies: 3
Views: 178

### Re: 15.63

For the final answer I got 2.7x10^10. I think there is a typo in the solutions manual in which it uses the value of R as 0.08314 kJ*K-1*mol-1, when it should be 0.008314 kJ*K-1*mol-1. In order to solve this problem I used the equation: ln(k2/k1) = Ea/R x (1/t1 -1/t2) and solved for k2.
Fri Mar 09, 2018 10:05 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Intermediate and rate law [ENDORSED]
Replies: 3
Views: 245

### Re: Intermediate and rate law[ENDORSED]

Intermediates aren't included in the rate law because within the whole mechanism they are produced and used up, so there is no net effect on the rate from the intermediate, so there is no reason to include something that doesn't affect the rate in the rate law.
Fri Mar 09, 2018 10:04 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Kinetic information [ENDORSED]
Replies: 3
Views: 201

### Re: Kinetic information[ENDORSED]

It can't prove the mechanism because there is no way to tell what else is happening. The rate law is taken only from the slow steps, and doesn't account for what is happening in the faster steps.
Sun Mar 04, 2018 3:33 pm
Forum: First Order Reactions
Topic: Integrated Rate Law
Replies: 1
Views: 152

### Re: Integrated Rate Law

I think the only difference would be changing the sign. Since in these reactions the concentration of reactants is decreasing while the concentration of products is increasing.
Sat Mar 03, 2018 2:02 pm
Forum: General Rate Laws
Topic: Integrated Rate Law Units
Replies: 4
Views: 320

### Integrated Rate Law Units

When using the integrated rate law equations do we have to use the reactant in terms of concentration, or can we use any units as long as they match? In problem 15.23 we use mg instead of mol/L.
Sun Feb 25, 2018 6:11 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Positive Slope???
Replies: 4
Views: 321

### Re: Positive Slope???

I believe that this is only true if it is a forward reaction going from reactants to products or a products favoring reaction, since in such cases products are produced, so the amount of products should be increasing leading to a positive slope of the tangent.
Sun Feb 25, 2018 5:27 pm
Forum: *Free Energy of Activation vs Activation Energy
Topic: What is Free Energy of Activation
Replies: 5
Views: 1450

### Re: What is Free Energy of Activation

The free energy of activation refers to the deltaG(Gibbs Free Energy) of the reaction.
Mon Feb 19, 2018 6:46 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: 14.33 (b)
Replies: 4
Views: 452

### Re: 14.33 (b)

You flip the sign of deltaG because the deltaG given is for the formation of Tl3+. The question asks for the standard potential of the Tl3+ /Tl couple, which when written like that indicates that Tl3+ is the reactant and Tl is the product, making it the flipped version of the formation of Tl3+. That...
Sun Feb 18, 2018 2:49 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.21 cell notation
Replies: 1
Views: 103

### Re: 14.21 cell notation

I think when it comes to looking at cell notation you have to think more about if an oxidation or reduction is taking place. Since, the left side of the notation refers to the anode, the reaction would have to be an oxidation. Knowing this we would form the reaction accordingly: Ag(s) +I-(aq) --> Ag...
Sun Feb 18, 2018 2:32 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Standard Cell Potentials
Replies: 5
Views: 228

### Standard Cell Potentials

If a half-reaction is multiplied by a value is the value of the Standard Cell Potential of the reaction affected?
Sun Feb 18, 2018 2:15 pm
Forum: Balancing Redox Reactions
Topic: 14.3 part d
Replies: 1
Views: 161

### Re: 14.3 part d

For the balanced reduction half reaction I got: Cl2 +2e- --> 2Cl- I got this by simply balancing the number of Chlorines and accounting for gaining 2 electrons since there are 2 Cl-. For the balanced oxidation half reaction I got: 2H2O + Cl2 --> 2HClO + 2H+ 2e- I got this by first balancing the numb...
Sun Feb 11, 2018 10:57 pm
Forum: Calculating Work of Expansion
Topic: different ways to calculate w
Replies: 4
Views: 283

### Re: different ways to calculate w

Generally you would use w=-nrtln(v2/v1) when it is an isothermal, reversible reaction.
Sat Feb 10, 2018 4:43 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: O2 Microstates
Replies: 2
Views: 346

### Re: O2 Microstates

The O2 molecules are identical, so that's why it would be only 1 microstate instead of 2
Sat Feb 10, 2018 12:41 am
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Delta G for diatomic molecules [ENDORSED]
Replies: 2
Views: 243

### Re: Delta G for diatomic molecules[ENDORSED]

It is only 0 if you are looking at the Gibbs Free Energy of formation due to the standard states of Br, Cl, N, O, and F being diatomic. It is the same for standard enthalpy of formation.
Sun Feb 04, 2018 8:07 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: combustion
Replies: 4
Views: 295

### Re: combustion

Yes, because enthalpy is a state function.
Sun Feb 04, 2018 8:01 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: 9.35
Replies: 2
Views: 167

### Re: 9.35

I believe it may be due to Container A having a gas, while Container C is just vibrating. Gases are able to move around more, so their entropy will be higher.
Sat Feb 03, 2018 5:32 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Value of G
Replies: 4
Views: 221

### Re: Value of G

I don't think there is a lowest value that G can be. And in many instances G is negative. If you look at the table of Gibbs Free Energy of formation in the back of the textbook, many of the values are below 0.
Sat Jan 27, 2018 10:49 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Constant Volume and Constant Pressure
Replies: 3
Views: 485

### Re: Constant Volume and Constant Pressure

If the volume is changing you would use: n*R*ln(V2 / V1)

If the pressure is changing you would use: n*R*ln(P1/P2)

where n = moles, V = volume, and P= pressure.
Sat Jan 27, 2018 10:45 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Ch. 9 Problems, 5
Replies: 3
Views: 174

### Re: Ch. 9 Problems, 5

I think as long as you are consistent within the problem it shouldn't matter whether you put your answer in kJ/K or J/K.
Sat Jan 27, 2018 10:31 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 9.7
Replies: 1
Views: 139

### 9.7

How do we know which values for heat capacity we are supposed to use for this question?
Sun Jan 21, 2018 1:54 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.65, where did that equation come from?
Replies: 3
Views: 250

### Re: 8.65, where did that equation come from?

In the textbook standard enthalpy of formation is defined as: "The standard enthalpy of formation of a substance is the standard reaction enthalpy per mole of formula units for the formation of a substance from its elements in their most stable form" (294). The most important sentence from...
Fri Jan 19, 2018 4:06 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Done By The System vs. Done On the System
Replies: 3
Views: 159

### Re: Done By The System vs. Done On the System

If work of the system is positive, that means that work is being done on the system, while if work of the system is negative that means the system is doing the work.
Fri Jan 19, 2018 11:36 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Calorimeters
Replies: 1
Views: 126

### Calorimeters

How do calorimeters work?
Sun Jan 14, 2018 7:13 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Which q is negative?
Replies: 2
Views: 145

### Re: Which q is negative?

I'm pretty sure if water is absorbing heat its q should be positive. In the textbook it states, "If energy enters a system as heat, the internal energy of the system increases and q is positive; if energy leaves the system as heat, the internal energy of the system decreases and q is negative&q...
Thu Jan 11, 2018 1:53 pm
Forum: Phase Changes & Related Calculations
Topic: Hess Law [ENDORSED]
Replies: 5
Views: 328

### Re: Hess Law[ENDORSED]

You would just reverse the sign.
Wed Jan 10, 2018 3:22 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Exothermic reaction in bond enthalpy example in lecture
Replies: 5
Views: 244

### Re: Exothermic reaction in bond enthalpy example in lecture

It is exothermic because new bonds are being formed, so energy is being released.