Search found 31 matches

Fri Mar 15, 2019 7:20 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6th Edition 11.7
Replies: 1
Views: 53

6th Edition 11.7

How can you tell, by looking at the different flasks, which one is at equilibrium? Is it when the number of moles of diatomic molecules is half that of it dissociated particles?
Fri Mar 15, 2019 7:05 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: q cal versus q reaction
Replies: 1
Views: 173

Re: q cal versus q reaction

When a reaction takes place within a calorimeter, a device in which heat transfers is monitored by recording the change in temperature produced by a process taking place within it, any heat lost from the reaction is absorbed by the calorimeter. Thus -q = q(cal).

(I hope this helps).
Fri Mar 15, 2019 10:13 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: free expansion
Replies: 2
Views: 226

Re: free expansion

Work happens when a system is pushing against an opposing force. Free expansion occurs in a vacuum, where the opposing external pressure is zero. Because there is no opposing force, no work is done during free expansion.
Fri Mar 15, 2019 10:06 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Heat Capacity at Constant Volume and Pressure
Replies: 2
Views: 203

Re: Heat Capacity at Constant Volume and Pressure

Also, an additional tip to help you better understand which relation to use is knowing the relationship between Cp,m and Cv,m: Cp,m= Cv,m + R. Keeping in mind that molar heat capacity increases with molecular complexity, and knowing that you simply have to add R to the molar head capacity at constan...
Fri Mar 15, 2019 10:03 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: ∆H and q [ENDORSED]
Replies: 6
Views: 249

Re: ∆H and q[ENDORSED]

At constant pressure with no nonexpansion work, delta H is equal to q.
Fri Mar 15, 2019 9:56 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Replies: 4
Views: 262

The reacting gases (the gases that are actually part of the equilibrium constant) still occupy the same individual molar concentration/partial pressure within the reaction vessel so there is no change to the equilibrium constant.
Fri Mar 15, 2019 9:54 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Inert Gas
Replies: 2
Views: 259

Re: Inert Gas

When adding an inert gas into the reaction vessel, the reacting gases continue to occupy the same volume, so their individual molar concentrations/partial pressures remain unchanged despite this addition of inert gas. Thus, the equilibrium constant with the addition of an inert gas remains unchanged.
Fri Mar 15, 2019 9:45 am
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Arrhenius Equation
Replies: 2
Views: 75

Re: Arrhenius Equation

Yes, adding to the response above, as long as the k value corresponds to the correct temperature value (k1 with T1, and so on), it should be fine.
Fri Mar 15, 2019 9:40 am
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: catalyst vs intermediate
Replies: 2
Views: 63

Re: catalyst vs intermediate

When looking at the steps in a reaction mechanism, you can also distinguish the two by looking at where they end up getting placed. Intermediates aren't written in the final chemical equation because they get used up in the reaction- they become products in one step, and a reactant in the follow ste...
Wed Mar 13, 2019 12:14 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: collision theory
Replies: 2
Views: 49

Re: collision theory

The textbook gives a really good analogy to collision theory: Imagine playing pool (with billiard balls). Say you hit the ball with a low speed. All that happens is that the balls collide and bounce apart from each other. However, if you hit the ball with a high enough speed, the balls colliding sma...
Tue Mar 12, 2019 5:10 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Slow or Fast Step
Replies: 3
Views: 94

Slow or Fast Step

How can you tell whether or not a step in a reaction mechanism is slow or fast? Or is this told ahead of time?
Tue Mar 12, 2019 5:02 pm
Forum: Reaction Mechanisms, Reaction Profiles
Replies: 2
Views: 86

According to the textbook, the steady-state approximation assumes that any intermediate formed during a reaction remains at a constant, low concentration. The pre-equilibrium approach assumes that the rate of consumption of the intermediate in the slow step is insignificant relative to its rates of ...
Fri Mar 08, 2019 12:50 pm
Forum: General Rate Laws
Topic: Mechanism
Replies: 2
Views: 72

Re: Mechanism

Reaction mechanisms are basically the sequence of steps that a reaction must undergo on the molecular level to go from reactants to products. It has to show the order and rate in which bonds form/break at each step and all steps must add up to give the overall balanced chemical equation for the reac...
Fri Mar 08, 2019 12:29 pm
Forum: First Order Reactions
Topic: Units
Replies: 2
Views: 66

Re: Units

The different units for k:

For 0th order: molA/L*s
For 1st order: 1/s
For 2nd order: L/molA*s
Sun Feb 17, 2019 10:39 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Calculating Standard Entropy of Fusion
Replies: 2
Views: 72

Calculating Standard Entropy of Fusion

Why does one use freezing temperature to calculate the entropy of fusion at a substance's melting point? (Reference: 6th ed-Self Test 9.8B)
Sun Feb 17, 2019 10:04 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Standard Entropy of Fusion at Freezing Point (Self-Test 9.8A- 6th Ed.)
Replies: 1
Views: 71

Standard Entropy of Fusion at Freezing Point (Self-Test 9.8A- 6th Ed.)

Based on Self-Test 9.8A in the 6th ed. textbook, why is it that the standard entropy of fusion for mercury at its freezing point is a positive value? What is happening at the freezing point of mercury that validates finding the standard entropy of fusion (and not of freezing), and thus, using the en...
Fri Feb 15, 2019 4:37 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: (6th edition) Example 9.5: Calculating Change in Entropy for T and V
Replies: 1
Views: 60

(6th edition) Example 9.5: Calculating Change in Entropy for T and V

When you are calculating the change in entropy for a system that was compressed suddenly and irreversibly, followed by change in temperature, you have to calculate both the change in entropy for volume change and the change in entropy for temperature change. For this particular question, why is that...
Thu Feb 14, 2019 8:58 pm
Forum: Calculating Work of Expansion
Topic: Work without change in volume
Replies: 8
Views: 355

Re: Work without change in volume

(If I understand you correctly)

In the case when the system is isothermal (constant temperature) and the system is under constant pressure, work can be defined as the external pressure times the change in volume. If there is no change in volume, no work is done.
Thu Feb 07, 2019 6:15 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Unit for Heat Capacity
Replies: 2
Views: 88

Re: Unit for Heat Capacity

Because each unit of the Kelvin scale (called a Kelvin) is equal to that (a degree) on the Celsius scale, they may have not included in units.
Thu Feb 07, 2019 5:55 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Energy
Replies: 5
Views: 165

Re: Energy

I am not sure by what you mean by the energy of an isolated system being constant at equilibrium specifically, but the definition of an isolated system is a system that cannot exchange either matter or energy with its surroundings. There can be energy transformations within an isolated system, but t...
Fri Feb 01, 2019 12:54 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Isolated System
Replies: 5
Views: 429

Re: Isolated System

An isolated system is defined as a thermodynamic system in which neither energy or matter can be exchanged with the surroundings. Work and heat can be done within the system, but this must be distinguished with whether or not there is a CHANGE in the internal energy within the system. The first law ...
Fri Feb 01, 2019 12:53 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Isolated System
Replies: 5
Views: 429

Re: Isolated System

An isolated system is defined as a thermodynamic system in which neither energy or matter can be exchanged with the surroundings. Work and heat can be done within the system, but this must be distinguished with whether or not there is a CHANGE in the internal energy within the system. The first law ...
Sat Jan 26, 2019 7:05 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: method 1
Replies: 3
Views: 65

Re: method 1

You can cancel them out because they are on opposite sides of the equation! (one is a reactant, the other is product)
Sat Jan 26, 2019 6:43 pm
Forum: Phase Changes & Related Calculations
Topic: Reverse fusion
Replies: 3
Views: 87

Re: Reverse fusion

Fusion happens when solids are heated to such a point that bonds holding molecules together break apart and the solid becomes a liquid. An easy way to put it is to say that fusion means melting. The input of heat makes the change in enthalpy positive, and the reaction endothermic. When you reverse f...
Sat Jan 26, 2019 6:17 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Autoprotolysis
Replies: 4
Views: 281

Re: Autoprotolysis

OH- and H30+'s connection to autoprotolysis is the fact that water is an amphoteric molecule, meaning that it can act as both an acid and a base. This is relevant to autoprotolysis, a process in which a proton is transferred between two identical molecules (in this case water) that can act as both a...
Fri Jan 18, 2019 1:35 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Net Ionic Equations
Replies: 3
Views: 98

Net Ionic Equations

How do you write a net ionic equation so that you can use its results to calculate the equilibrium constant?
Fri Jan 18, 2019 1:02 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Activity
Replies: 2
Views: 70

Re: Activity

In addition to what was said, activities are simple quantities that are introduced to facilitate writing the expression for K. All activities are pure numbers and thus, are unitless. In the simplified version of the equilibrium constant, activity is the numerical value of pressure in bars, or numeri...
Fri Jan 18, 2019 12:53 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Net Ionic Equations for Equilibrium Constants
Replies: 2
Views: 69

Re: Net Ionic Equations for Equilibrium Constants

You should use a net ionic equation when a reaction involves fully dissociated ionic compounds in solution. A net ionic equation automatically gets rid of spectator ions, which are ions that exist in the same form in both reactant and product sides of the chemical reaction (thus, concentrations of s...
Fri Jan 11, 2019 6:33 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Equilibrium constants for ionic compounds
Replies: 1
Views: 27

Equilibrium constants for ionic compounds

For reactions involving fully dissociated ionic compounds in solution, how do you use the activity of each type of ion to write an equilibrium constant for the net ionic equation?
Thu Jan 10, 2019 10:31 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Q vs K
Replies: 5
Views: 111

Re: Q vs K

The same method of calculating K is used to calculate Q. Q, the reactant quotient, is different from K when the reaction has not reached equilibrium, and depending on how much larger of smaller Q is from K, one is able to tell whether the forward or backwards reaction is favored. Like what the perso...
Thu Jan 10, 2019 10:22 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Liquids and Solids in Equilibrium Constants
Replies: 5
Views: 122

Re: Liquids and Solids in Equilibrium Constants

A good example of this can be found in an example given during lecture in which we are required to find Kp for CaC03(s)-->/<-- CaO(s) +CO2(s). The ratio for the equilibrium constant would be Kp= P(partial pressure of)(CO2) because solids or liquids have such an irrelevant change in concentration tha...