## Search found 31 matches

Sun Mar 17, 2019 6:06 pm
Forum: Calculating Work of Expansion
Topic: Conversion Units for Work Expansion
Replies: 1
Views: 272

### Conversion Units for Work Expansion

When converting units a problem relating to work of expansion given L•atm, are there certain conditions that would require us to use (101.325 J/ L•atm) and (8.314 J•K•mol^-1/ 0.08206 L•atm•K^-1*mol^-1)? Or can we just use any one of the conversion units?
Sun Mar 17, 2019 5:57 pm
Forum: Balancing Redox Reactions
Topic: redox in basic solutions
Replies: 1
Views: 273

### Re: redox in basic solutions

After getting the skeletal equations for the oxidation and reduction half-reactions, you balance O atoms by adding H2O molecules. To balance the H atoms on each side add, for each H atom needed, one H2O molecule to the side of each equation that needs hydrogen and one OH- ion to the opposite side. F...
Sun Mar 17, 2019 5:54 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Le Chatlier Principle
Replies: 10
Views: 1301

### Re: Le Chatlier Principle

Even though a system is at equilibrium, the reaction does not stop. Instead, it continues just as fast in the direction from reactants to products (the forward direction), as it does from products to reactants (the reverse direction). The addition of a catalyst speeds up the forward and back reactio...
Sun Mar 17, 2019 5:48 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: What is pKa and Ka exactly?
Replies: 11
Views: 9273

### Re: What is pKa and Ka exactly?

Yes, it would. The weaker the base, the smaller the value of Kb and the greater the value of pKb. Essentially, a large Kb value indicates the high level of dissociation of a strong base. A lower pKb value indicates a stronger base.
Tue Mar 12, 2019 2:37 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Order of reaction
Replies: 3
Views: 392

### Order of reaction

The text states that the rate law for a reaction is determined experimentally and cannot be inferred in general from the chemical equation of the reaction. So how can you determine the order of a reaction?
Tue Mar 12, 2019 5:06 am
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Problem 7A17
Replies: 2
Views: 284

### Re: Problem 7A17

The number value 0.457 that you're referring to from the solutions is probably the initial rate for experiment 4, except the data gives the initial rate as 457 mol •L^-1•s^-1. In order to find the rate constant of the reaction, you have to take the initial rate and divide it by the product of the re...
Tue Mar 12, 2019 4:56 am
Forum: Balancing Redox Reactions
Topic: Balancing redox rxns in basic solution
Replies: 2
Views: 135

### Re: Balancing redox rxns in basic solution

After getting the skeletal equations for the oxidation and reduction half-reactions, you balance O atoms by adding H2O molecules. To balance the H atoms on each side add, for each H atom needed, one H2O molecule to the side of each equation that needs hydrogen and one OH- ion to the opposite side. F...
Tue Mar 12, 2019 4:52 am
Forum: Balancing Redox Reactions
Topic: Basic solution reaction
Replies: 2
Views: 120

### Re: Basic solution reaction

To get the reduction half-reaction for NO2-(aq) + Al(s) ----> NH3(g) + AlO2-(aq), first you get the skeletal equation (NO2-(aq) ----> NH3(g)). Then you balance O atoms by adding 2 H2O molecules. To balance the H atoms on each side add, for each H atom needed, one H2O molecule to the side of each equ...
Tue Mar 12, 2019 4:37 am
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: System and surroundings
Replies: 4
Views: 233

### Re: System and surroundings

The change in the system= the negative change of the surrounding in regards to reversible reactions is conceptually based on the fact that the amount lost and gained in both systems must be equal to 0. Essentially, if a system loses 10 units, the surrounding must gain 10 units so that the amount of ...
Tue Mar 12, 2019 4:31 am
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: entropy and enthalpy
Replies: 5
Views: 223

### Re: entropy and enthalpy

There is no direct relationship between entropy and enthalpy. Enthalpy (H) is the amount of energy released or absorbed during a chemical reaction while entropy (S), defines the degree of randomness or disorder in a system. However, the free energy (G) regroups both terms as G= H-TS, where at a cons...
Tue Mar 12, 2019 4:25 am
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Residual Entropy
Replies: 2
Views: 143

### Re: Residual Entropy

At 0 K, NO has a greater residual entropy than BF3 because NO can be oriented as both NO and ON; BF3, no matter how you rotate the molecule, will look the same, because the F's are symmetric around the B. Therefore, NO has higher residual entropy, because it takes upon more orientations.
Tue Mar 12, 2019 4:23 am
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: Entropy and Reduction in moles of Gas
Replies: 2
Views: 288

### Re: Entropy and Reduction in moles of Gas

When the number of moles of gas is reduced, the standard residual entropy value is smaller than the standard molar entropy because residual entropy is the measure of how many different arrangements a molecule can have whereas, molar entropy is a measure of a molecule's chaos. With the moles of gas r...
Tue Mar 12, 2019 4:14 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Initial Concentration and Equilibrium Concentration
Replies: 3
Views: 190

### Re: Initial Concentration and Equilibrium Concentration

Although adding a reactant increases the concentration of products at equilibrium, the increase in product concentration won't necessarily mean that the equilibrium product concentration becomes higher than that for the reactants. If K is bigger than one, the equilibrium concentration of the product...
Tue Mar 12, 2019 4:09 am
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: pKa and pKb
Replies: 3
Views: 463

### Re: pKa and pKb

The pKa gives you the acidity value of a substance and the Ka value indicates that the amount of acid that actually dissociates. While the pkb gives you the basicity value of a substance and the Kb value determines the level of dissociation of a base. For example, if given the pKa, you can use 10^-p...
Tue Mar 12, 2019 4:03 am
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Salt Solution
Replies: 5
Views: 379

### Re: Salt Solution

Spectator ions are ions that are present in a solution but don't take part in a solution's chemical reaction. When reactants dissociate into ions, some of the ions may combine to form a new compound. Therefore, because they don't take on the role of an electron donator or acceptor, they will not aff...
Tue Mar 12, 2019 4:01 am
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Acid Base
Replies: 3
Views: 377

### Re: Acid Base

If given the pKa, you can use 10^-pKA to get the Ka. Essentially to solve for the [H+] or x value, you set the Ka equal to [products]/[reactants]. To find the pH, you then take the -log of [H+]. The pKa gives you the acidity value of a substance and the Ka value indicates that the amount of acid tha...
Tue Mar 12, 2019 3:54 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Equilibrium Constants and Solubility
Replies: 2
Views: 131

### Re: Equilibrium Constants and Solubility

If given the equilibrium constant, the higher the value, the higher a substance's solubility is. The equilibrium constant refers to the product of the concentration of the ions that are present in a saturated solution of an ionic compound, it is given the name solubility product constant. Therefore ...
Tue Mar 12, 2019 3:48 am
Forum: Calculating Work of Expansion
Topic: Reversible vs irreversible
Replies: 6
Views: 441

### Re: Reversible vs irreversible

When there is constant pressure and the reaction is irreversible, the equation w=-Pex∆V is derived from P=F/A, in which to find the force of the equation you can use the relationship and get F=PexA. The work needed to drive the piston out through a distance d is, therefore, work= PexA •d (work=force...
Tue Mar 12, 2019 3:35 am
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: reversible and irreversible processes
Replies: 9
Views: 486

### Re: reversible and irreversible processes

A reversible process is one that can be reversed by an infinitely small change in a variable. Whereas an irreversible process is an expansion against an external pressure that differs by a finite amount from the pressure of a system. To get a better understanding of a reversible change, you can cons...
Tue Mar 12, 2019 3:25 am
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Closed vs Isolated
Replies: 5
Views: 433

### Re: Closed vs Isolated

An open system can exchange both matter and energy with the surroundings whereas a closed system has a fixed amount of matter but it can exchange energy with the surroundings. Examples of open systems can be automobiles or the human body. For the human body, we actively interact with our environment...
Sat Feb 23, 2019 5:25 pm
Forum: Balancing Redox Reactions
Topic: Balancing / Skeletal Equations
Replies: 4
Views: 211

### Re: Balancing / Skeletal Equations

Sometimes when balancing the skeletal equation, one species may have oxygen on one side of the reaction and not the other. Essentially, you would need to add H2O to balance the O atoms and balance the H atoms by adding H+ where necessary.
Sat Feb 23, 2019 5:20 pm
Forum: Balancing Redox Reactions
Topic: HW help
Replies: 1
Views: 76

### Re: HW help

To balance out a redox equation, first, start by identifying the species that is being reduced and oxidized from the changes in oxidation numbers. Then write the skeletal equations for the half-reactions and balance all elements except for H+ and O. You may notice that in the skeletal equation one s...
Tue Jan 29, 2019 9:19 pm
Forum: Phase Changes & Related Calculations
Topic: 4B.13a
Replies: 1
Views: 97

### Re: 4B.13a

With the equation w=-Pex(∆V), the resulting unit is L•atm. 1 L•atm= 101.325 Joules. Essentially, you have to multiply your answer by 101.325 Joules in order to have the answer written in SI units.
Tue Jan 29, 2019 9:13 pm
Forum: Calculating Work of Expansion
Topic: Irreversible and reversible expansion, 6th edition 8.11
Replies: 2
Views: 116

### Re: Irreversible and reversible expansion, 6th edition 8.11

An irreversible process is one by which the expansion against an external pressure differs by a finite (measurable) amount from the pressure of the system. Whereas, a reversible process is one that can be reversed by infinitesimal contributions within a variable. Even though the final volumes are th...
Tue Jan 29, 2019 8:55 pm
Forum: Phase Changes & Related Calculations
Topic: 4B.13b
Replies: 1
Views: 103

### Re: 4B.13b

The equation w=-nRT(ln(V2/V1)), is the standard integral of the original equation, dw=-(nRT/V)dV. A reversible process is one that can be reversed by an infinitely small change (so small that the change is not measurable) in a variable. Therefore, the total work done within a process is the sum (int...
Tue Jan 29, 2019 8:42 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Help on 8.37a.
Replies: 1
Views: 99

### Re: Help on 8.37a.

The standard way of reporting the values for enthalpy of vaporization is always reported as per one mole. Essentially you divide the heat by the number of moles, to get the enthalpy per mole.
Tue Jan 29, 2019 4:03 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Work
Replies: 4
Views: 264

### Work

In comparing the work of when a gas expands versus the work of isothermal expansion, I noticed a change carried out reversibly always does more work than a change carried out irreversibly but the values for the reversible process are technically lower. Why is it that the more negative the value, the...
Wed Jan 09, 2019 9:05 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5G.9 Moles and Partial Pressure in Equilibrium
Replies: 1
Views: 39

### 5G.9 Moles and Partial Pressure in Equilibrium

In the problem 5G.9: "A sample of ozone, O3, amounting to 0.10 mol, is placed in a sealed container of volume 1.0 L and the reaction 2 O3(g) S 3 O2(g) is allowed to reach equilibrium. Then 0.50 mol O3 is placed in a second container of volume 1.0 L at the same temperature and allowed to reach e...
Wed Jan 09, 2019 8:58 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: ICE Table
Replies: 4
Views: 78

### Re: ICE Table

Normally, when the concentration of the products isn't given for equilibrium problems and only reactants are presented initially, it is assumed products have very low concentrations that would make solving the problems to be more complex. Essentially, to make calculations easier and less complicated...
Wed Jan 09, 2019 8:46 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5H.1 part B
Replies: 2
Views: 64

### Re: 5H.1 part B

In reaction presented, N2(g) + 3H2(g) -> 2 NH3(g) at 400, K = 41. The reaction of part B, 1/2 N2(g) + 3/2 H2(g) ->NH3(g), changed in the amount of moles. However, the ratio of the moles of the equation of part B is proportional to the ratio of the moles of the equation of part A by one-half. We assu...
Wed Jan 09, 2019 8:27 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Number 5G.9 Question
Replies: 1
Views: 71

### Re: Number 5G.9 Question

I think that you have to start by looking at the equation for the equilibrium constant, K, of the problem presented. Notice that if you were to write out the expression for the reaction, you would get K=[O2]^3/[O3]^2. Essentially, you would also get the ratio: [O2]^3/[O3]^2. Now we know that the equ...