Search found 33 matches
- Mon Mar 12, 2018 10:50 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Intermediates in Rate law
- Replies: 3
- Views: 4109
Re: Intermediates in Rate law
For the overall rate law you should not include intermediates (because they are produced and then used up by the reaction). The rate law should only include reactants. Example: Step 1: A + B → C Step 2: C → D Overall: A + B → D (C is the intermediate) rate law = k[A][B] (do not include [C] in the ra...
- Mon Mar 12, 2018 10:46 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Rate-determining slowest step
- Replies: 5
- Views: 1181
Re: Rate-determining slowest step
The rate determining step is the slowest step in the reaction mechanism. You can determine which of the elementary steps is the slow step by comparing their activation energies. The step with the largest activation energy is the rate determining/slow step because it requires the largest input of ene...
- Mon Mar 12, 2018 10:41 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: What is A?
- Replies: 4
- Views: 533
Re: What is A?
Frequency factor (A) represents the frequency of collisions between reactant molecules with the correct orientation. Therefore, increasing A increases k (the rate constant), which increases the rate of the reaction.
- Mon Mar 05, 2018 4:25 pm
- Forum: Second Order Reactions
- Topic: Limiting Step
- Replies: 7
- Views: 1093
Re: Limiting Step
The reaction rate is always dependent on the slow step, which is the limiting step. The slow step has the highest activation energy out of all the steps, and therefore has the slowest reaction rate because more energy is needed to overcome the activation energy barrier.
- Mon Mar 05, 2018 4:19 pm
- Forum: Zero Order Reactions
- Topic: Half life and rate order
- Replies: 4
- Views: 603
Re: Half life and rate order
Each rate order has different corresponding half-life equation.
Zero order reaction: half-life= [A]o/2k
*1st order reaction: half-life=ln2/k
2nd order reaction: half-life=1/k[A]o
*Note that the 1st order half-life does not depend on the initial concentration of the reactant.
Zero order reaction: half-life= [A]o/2k
*1st order reaction: half-life=ln2/k
2nd order reaction: half-life=1/k[A]o
*Note that the 1st order half-life does not depend on the initial concentration of the reactant.
- Mon Mar 05, 2018 4:17 pm
- Forum: Zero Order Reactions
- Topic: slope
- Replies: 9
- Views: 4761
Re: slope
Using A → B Zero order reaction: graph [A] vs. time, this plot will produce a negative slope. Therefore, slope=-k because the rate constant k must be a positive value. 1st order reaction: graph ln[A] vs. time, this plot will produce a negative slope. Therefore, slope=-k because the rate constant k m...
- Mon Feb 26, 2018 2:45 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: units of the rate of reaction
- Replies: 7
- Views: 1017
Re: units of the rate of reaction
The units of the rate of the reaction (k) depends on the order of the reaction and the corresponding rate law.
0 order: rate=k, units (1/s)
1st order: rate=k[R], units (L/mol x s)
2nd order: rate=k[R]^2, units (L^2/mol^2 x s)
0 order: rate=k, units (1/s)
1st order: rate=k[R], units (L/mol x s)
2nd order: rate=k[R]^2, units (L^2/mol^2 x s)
- Mon Feb 26, 2018 2:41 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Diamond and graphite
- Replies: 8
- Views: 3523
Re: Diamond and graphite
The reaction of diamond --> graphite is spontaneous. Spontaneous does not imply that the reaction is fast, it means the reaction will eventually occur (because energy of the reactants > energy of the products). Therefore, a diamond is kinetically stable with respect to graphite because there is a la...
- Mon Feb 26, 2018 2:33 pm
- Forum: First Order Reactions
- Topic: Calculating slope
- Replies: 4
- Views: 669
Calculating slope
For a first order reaction, why does the slope = -k? Why do we need a negative sign in front of k?
- Mon Feb 19, 2018 2:51 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Concentration cell vs. Galvanic and Electrolytic cells [ENDORSED]
- Replies: 1
- Views: 2159
Concentration cell vs. Galvanic and Electrolytic cells [ENDORSED]
What is the difference between a concentration cell, galvanic cell, and electrolytic cell? Would a concentration cell be considered a galvanic or electrolytic cell?
- Mon Feb 19, 2018 2:48 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Concentration Cells
- Replies: 5
- Views: 725
Re: Concentration Cells
A concentration cell consists of two half-cells with the same electrodes but with different concentrations. This difference in ion concentration drives the flow of electrons from the more concentrated solution to the lower concentrated solution, which creates a voltage as the cell reaches equilibriu...
- Mon Feb 19, 2018 2:43 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 14.51
- Replies: 1
- Views: 281
14.51
Question 14.51 asks us to consider a cell Ag(s) / Ag+(aq, 5.0mmolxL-1)// Ag+(aq, 0.15molxL-1) / Ag(s) and asks if this cell can do work. Why can we assume that the cell can do work because [Ag+] anode < [Ag+] cathode and the ratio is less than 1 and E>0 (as stated in the solutions manual)?
- Fri Feb 16, 2018 4:46 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Anode/ Cathode
- Replies: 6
- Views: 800
Re: Anode/ Cathode
Generally, the anode is on the left and the cathode is on the right. If this is not the case, however, you can identify which electrode is the anode and which is the cathode by looking at the half reactions. The oxidation half reaction takes place at the anode (electrons are lost), the reduction hal...
- Fri Feb 16, 2018 4:44 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Standard cell reduction potential
- Replies: 1
- Views: 282
Standard cell reduction potential
Why don't we flip the sign of the anode or multiply the standard cell reduction potential by any coefficients (ex. in a case where we multiply the entire half oxidation reaction by 3 to cancel out the 3 electrons gained the he half reduction reaction)? Does this have to do with cell potential being ...
- Fri Feb 16, 2018 4:37 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.5 part a
- Replies: 1
- Views: 273
14.5 part a
We are asked to write the cell diagram for the reaction: AgBr(s) <--> Ag+(aq) + Br-(aq). I am confused about the order in which the ions and molecules are written for the anode portion of the cell diagram: Ag(s) | AgBr(s) | Br-(aq) || Ag+(aq) | Ag(s) The half-reaction for the anode is Ag+(aq) + Br-(...
- Fri Feb 09, 2018 9:46 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Open, closed, or isolated test 1 question
- Replies: 10
- Views: 1597
Re: Open, closed, or isolated test 1 question
A 20mL vial with an organic solvent is an open system. An open system can exchange both energy and matter with the surroundings. Solvent (matter) can be taken out of the vial, the key word is that the vial is "open." The solvent can be heated up or cooled down, in which energy is exchange ...
- Fri Feb 09, 2018 9:43 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: How to calculate W [ENDORSED]
- Replies: 6
- Views: 782
How to calculate W [ENDORSED]
How do you calculate W in the equation S = kBlnW? What does W represent?
- Fri Feb 09, 2018 9:38 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: ΔG=0
- Replies: 2
- Views: 448
ΔG=0
What does it mean when ΔG=0? Why does ΔG=0 at the boiling point of a substance?
- Sat Feb 03, 2018 3:44 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Spontaneous
- Replies: 14
- Views: 1960
Re: Spontaneous
When △G is negative, the reaction is spontaneous. A spontaneous reaction is one occurs on its own without being driven by an outside force. The products are favored over the formation of the reactants in a spontaneous reaction. Using the equation △G=△H-T△S: Positive △S and negative △H = spontaneous ...
- Sat Feb 03, 2018 3:37 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: When to use Q versus K [ENDORSED]
- Replies: 6
- Views: 2604
Re: When to use Q versus K [ENDORSED]
Q is the reaction quotient and expresses the relative ratio of products to reactants at a given instant. Whereas, K is the equilibrium constant and expresses the ratio of products to reactants at equilibrium (when delta G=0). Use ΔG= ∆G°+ RTlnQ when the system is not at equilibrium. Use ΔG= ∆G°+ RTl...
- Sat Feb 03, 2018 3:32 pm
- Forum: Van't Hoff Equation
- Topic: How to calculate K [ENDORSED]
- Replies: 3
- Views: 867
How to calculate K [ENDORSED]
How would you calculate K if you are given ΔG°? I used the equation ΔG°=-RTlnK and isolated lnK so that lnK=ΔG°/-RT. How do you isolate the K term from lnK?
- Sat Jan 27, 2018 4:54 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bondy Enthalpy Accuracy
- Replies: 3
- Views: 635
Bondy Enthalpy Accuracy
When calculating standard reaction enthalpies, why is using bond enthalpies the least accurate method?
- Sat Jan 27, 2018 4:50 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Adiabatic system
- Replies: 6
- Views: 754
Re: Adiabatic system
Adiabatic systems are isolated systems, in which both energy/heat and matter are not transferred to the surroundings. Therefore q=0 and delta U=w (using the equation delta U = q + w) adiabatic process is one that occurs without transfer of heat or matter between a thermodynamic system and its surrou...
- Sat Jan 27, 2018 4:45 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Formation of Snow in Clouds [ENDORSED]
- Replies: 5
- Views: 7303
Re: Formation of Snow in Clouds [ENDORSED]
Formation of snow in clouds is an exothermic process because heat is being transferred from the system (liquid water in the sky) to the surroundings, and the surroundings therefore heat up. In order for liquid water to be condensed into a solid form as snow, heat must be removed from the liquid wate...
- Mon Jan 22, 2018 3:58 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Heat versus work
- Replies: 6
- Views: 808
Re: Heat versus work
Heat and work are both forms of energy. Heat is the energy associated with the random motion of particles, while work is the energy of ordered motion in a specific direction. First Law: Heat and work contribute to the total energy of the system. ΔU = q + w Second Law: Work can be transformed into he...
- Mon Jan 22, 2018 3:51 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: the sign of q
- Replies: 5
- Views: 461
Re: the sign of q
If the system loses/releases heat and warms the surroundings then q is negative (exothermic), if the system gains/absorbs heat and cools the surroundings then q is positive (endothermic). Therefore, in your example if an reaction " required 5.5 kJ of heat. How would we know if that value was po...
- Mon Jan 22, 2018 3:47 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: ΔU = ΔH for biological reactions
- Replies: 2
- Views: 322
ΔU = ΔH for biological reactions
Why does ΔU = ΔH in most biological reactions if ΔU = q + w? Why don't we account for work when calculating the change internal energy (ΔU).
- Mon Jan 15, 2018 9:56 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess's Law
- Replies: 3
- Views: 336
Re: Hess's Law
Hess's law states that the overall reaction enthalpy change is the sum of all the reaction enthalpies, regardless of the multiple steps into which the reaction can be divided. We are able to add the reaction enthalpies because enthalpy is a state function, in which the overall reaction enthalpy chan...
- Mon Jan 15, 2018 9:48 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Expansion Work vs. Nonexpansion Work
- Replies: 2
- Views: 1543
Expansion Work vs. Nonexpansion Work
What is the difference between expansion work and nonexpansion work? Why does the equation for expansion work have a negative sign in front of the external pressure? What is the equation for nonexpansion work?
Equation for expansion work: w = -Pex ΔV
Equation for expansion work: w = -Pex ΔV
- Mon Jan 15, 2018 9:37 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Type of Systems
- Replies: 2
- Views: 294
Re: Type of Systems
There are three types of systems (the region in which we are interested): open, closed, or isolated. Open system : can exchange matter and energy with the surroundings (everything else besides the system). An example of an open system is boiling soup in an open saucepan. The saucepan is an open syst...
- Fri Jan 12, 2018 5:43 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.61
- Replies: 4
- Views: 321
Re: 8.61
Reverse the first equation and multiply it by 2 to get HBr on the product side with a coefficient of 2. Reverse the second equation to get the 2 NH3's to cancel out. Leave the last equation as is. You must also reverse the sign and multiply the first equations ΔH° by 2 and reverse the sign of the se...
- Fri Jan 12, 2018 5:33 pm
- Forum: Phase Changes & Related Calculations
- Topic: Heat Capacity vs. Specific Heat Capacity
- Replies: 3
- Views: 4458
Heat Capacity vs. Specific Heat Capacity
Why is heat capacity considered an extensive property and specific heat capacity considered an intensive property? Why is specific heat capacity more useful in calculations than heat capacity, does this have to do with the fact that one is intensive and the other is extensive, respectfully?
- Fri Jan 12, 2018 5:26 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Standard Enthalpies of Formation of Diatomic Molecules
- Replies: 6
- Views: 4842
Re: Standard Enthalpies of Formation of Diatomic Molecules
The standard enthalpy of formation for diatomic molecules such as O2 (g) equals 0 because no change occurs in the reaction O2 (g) --> O2 (g). The definition of the standard enthalpy of formation is that the standard enthalpy of formation of an element in its most stable form equals 0. O2 (g) is the ...