## Search found 31 matches

Sat Mar 16, 2019 4:34 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Midterm Question 6
Replies: 6
Views: 527

### Re: Midterm Question 6

For Question 7, I wrote out an equation and balanced it out. I got CH4 + 2O2 -> 2H2O + CO2. I then drew out diagrams for each molecule and calculated the delta H for each. CH4 was 1648kJ, O2 was 992kJ, H2O was 1852 kJ, and CO2 was 1486kJ. I finally did (delta H of CH4 + O2) - (delta H of H2O + CO2) ...
Sat Mar 16, 2019 4:27 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Midterm Question 6
Replies: 6
Views: 527

### Re: Midterm Question 6

For Question 6, I first found delta V using (delta)nRT/P. Using delta V, I plugged the values given into delta U = delta H -PdeltaV. delta H = 4 * -2878 kJ/mol and PdeltaV = (-1atm)(282L)(8.314 J/Kmol )(1/8.206*10^-2Latm/Kmol)(1kJ/1000J) = 28.6kJ. Using those values I got -11541 kJ.
Sat Mar 16, 2019 4:17 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Test 1 problem 5
Replies: 6
Views: 439

### Re: Test 1 problem 5

I used the pKa to find pKb.
I used pKb to find Kb.
I then made an ice table and used that to find the concentration of OH-.
I used the concentration of OH- to find the pOH.
Finally, I used pOH to convert to pH.
Wed Mar 13, 2019 5:03 pm
Forum: Second Order Reactions
Topic: 6th edition 15.39
Replies: 2
Views: 327

### 6th edition 15.39

For part A of question 15.39 from the 6th edition, how does the coefficient in front of the reactant affect the equation? If it was just A -> B + C would it still be the same answer?
Mon Mar 11, 2019 1:24 pm
Forum: General Rate Laws
Topic: 6th edition 15.17
Replies: 1
Views: 107

### 6th edition 15.17

For 15.17 from the 6th edition, why did we use the 2nd and 4th experiment to find the order of A? Why couldn't we use the 1st and 2nd or 3rd and 4th? And how do we know C is independent from the rate.
Mon Mar 11, 2019 1:21 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: 1/2 dNO2/dt
Replies: 1
Views: 76

### 1/2 dNO2/dt

During Lavelle's lecture today, why was there a 1/2 in front of dNO2/dt in step 2?
Mon Mar 11, 2019 1:20 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: k in Lavelle's Lecture
Replies: 1
Views: 104

### k in Lavelle's Lecture

During the lecture today, why did k = 2k2K = 2k1k2/k'1? I don't get how he derived this formula.
Fri Mar 01, 2019 1:14 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: delta G = 0 ??
Replies: 5
Views: 762

### Re: delta G = 0 ??

delta G equals to 0 when the equation is at equilibrium. At equilibrium, the amount of energy released by reactants is used by the products or vice versa. As a result, there is no change in free energy because all the energy released is absorbed.
Fri Mar 01, 2019 1:08 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Significant Figures
Replies: 2
Views: 118

### Re: Significant Figures

Based on Lavelle's sig fig sheet, integers and exact numbers do not affect sig figs.
Tue Feb 26, 2019 9:43 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Pt
Replies: 2
Views: 93

### Pt

Under what circumstance would you include Pt in the cell diagram and when do you not include it?
Tue Feb 26, 2019 7:44 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 9.65 6th edition writing chemical equations
Replies: 1
Views: 91

### 9.65 6th edition writing chemical equations

For question 9.65 from the 6th edition, how do you know if the compound is decomposing for forming. For example, how do you know its P + 5/2Cl2 -> PCL5 instead of PCl5 -> P + 5/2Cl2?
Tue Feb 26, 2019 7:40 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 9.65 6th edition
Replies: 1
Views: 85

### 9.65 6th edition

When you balance the equations for 9.65 in the 6th edition, why are the balanced coefficients in fractions?
Tue Feb 26, 2019 7:37 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.15 a 6th edition
Replies: 1
Views: 101

### 14.15 a 6th edition

For question 14.15a, how do you which equation is the cathode or anode?
Fri Feb 15, 2019 1:17 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Liquid and moles
Replies: 8
Views: 534

### Re: Liquid and moles

The equilibrium constant only applies for gases therefore solids and liquids are disregarded.
Fri Feb 15, 2019 1:05 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Ideal gases
Replies: 4
Views: 179

### Re: Ideal gases

If you were given the specific heat capacity I would use the given value but it should equal 5/2 R if pressure is constant and 3/2R if volume is constant.
Fri Feb 15, 2019 12:58 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: 6th edition 8.25
Replies: 4
Views: 2805

### Re: 6th edition 8.25

In a calorimeter, no heat is lost to the surroundings so the heat released from the reaction is absorbed by the calorimeter. q reaction is negative because heat is lost and q calorimeter is positive because heat is absorbed.
Fri Feb 15, 2019 12:52 pm
Forum: Biological Examples (*DNA Structural Transitions, etc.)
Topic: Structures of the Human Body and Systems
Replies: 5
Views: 502

### Re: Structures of the Human Body and Systems

The human body as a whole would be considered an open system.
Fri Feb 15, 2019 12:45 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: enthalpy, entropy, Gibbs Free Energy signs
Replies: 3
Views: 135

### Re: enthalpy, entropy, Gibbs Free Energy signs

In a spontaneous reaction, enthalpy is negative, entropy is positive and Gibbs Free Energy is negative. In a nonspontaneous reaction, enthalpy is positive, entropy is negative and Gibbs Free Energy is positive.
Thu Feb 07, 2019 10:28 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Change in entropy and its relation with temperature
Replies: 2
Views: 126

### Change in entropy and its relation with temperature

For problem 4F.3 from the 7th edition, why is there a greater change of entropy at a lower temperature?
Sun Feb 03, 2019 9:15 pm
Forum: Calculating Work of Expansion
Topic: Bond Formation
Replies: 3
Views: 140

### Re: Bond Formation

Bond formation is exothermic. Energy is released when bonds are formed. When a new bond is formed, there is a net lowering of energy and heat is released into the environment. It costs energy to break a bond so bond breaking is endothermic.
Sun Feb 03, 2019 9:09 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Units
Replies: 7
Views: 264

### Re: Units

It is J/C*mol because C-1 = 1/C and mol-1 = 1/mol.
Sun Feb 03, 2019 9:06 pm
Forum: Calculating Work of Expansion
Topic: Heat Capacity vs. Specific Heat Capacity
Replies: 3
Views: 154

### Re: Heat Capacity vs. Specific Heat Capacity

Specific heat capacity is the amount of heat needed to raise the temperature of an object per unit mass of an object. In other words, specific heat capacity is the heat capacity per unit mass of a material.
Sat Jan 26, 2019 7:26 pm
Forum: Phase Changes & Related Calculations
Topic: Phase change enthalpy signs
Replies: 2
Views: 149

### Re: Phase change enthalpy signs

When you freeze something, the heat from the object is absorbed into the surroundings and as a result, the total heat in the system is lost. When heat is lost, enthalpy becomes more negative. When you melt something, heat is added into the object and the total heat of the system is increased. Enthal...
Sat Jan 26, 2019 7:21 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Signs in bond enthalpies
Replies: 4
Views: 147

### Re: Signs in bond enthalpies

Enthalpy is the total energy in the system. The reactants require energy to break the bonds so the sign is positive when energy goes into the system and because energy is released when bonds form in the products, energy goes out and the sign is negative.
Sat Jan 26, 2019 7:13 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Enthalpy Units
Replies: 7
Views: 303

### Re: Enthalpy Units

Enthalpy is the amount of heat released or absorbed at constant pressure. Whether or not it's kJ or kJ * mol^-1, it doesn't really matter. Just remember to use the units that best fits the question asked.
Wed Jan 16, 2019 2:05 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6th Edition 12.59
Replies: 2
Views: 224

### Re: 6th Edition 12.59

They're all bases because the question says "given that the pka of its conjugate acid". "its" refers to the molecules listed in each subsection (a, b, c, d) and because the question says conjugate acid, then "its" are bases.
Wed Jan 16, 2019 1:50 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Number 115 Chapter 11 HW
Replies: 2
Views: 112

### Re: Number 115 Chapter 11 HW

But because there is an equal number moles of gases on each side of the reaction, compressing the system should have little to no effect.
Wed Jan 16, 2019 1:46 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Pka and Acidty
Replies: 2
Views: 323

### Re: Pka and Acidty

The larger the pKa the weaker the acid.
Wed Jan 09, 2019 9:39 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Question 5G.1 part d
Replies: 1
Views: 55

### Re: Question 5G.1 part d

This statement is true because the concentration of the reactants over the products at equilibrium should equal the equilibrium constant. The constant for the reaction at equilibrium will not change, therefore, as you increase the concentration of the reactants, the concentration of the products sho...
Wed Jan 09, 2019 9:34 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Equilibrium constant
Replies: 1
Views: 49

### Re: Equilibrium constant

To find k you use the concentrations of the products and the reactants. The reason there isn't any units is because concentrations are an approximation of chemical activity and chemical activity does not have any units. We use concentration instead of chemical activity because chemical activity is t...
Mon Jan 07, 2019 2:52 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Homework Problem 11.3 for 6th Edition
Replies: 3
Views: 139

### Homework Problem 11.3 for 6th Edition

For Question 11.3 from the 6th edition textbook, the question asks to find the equilibrium constant of the following equations. According to the solution manual, the equilibrium constant involves the partial pressure of the molecules. Is it partial pressure because we do not know the molarity of the...