Search found 51 matches
- Thu Mar 12, 2020 5:50 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Enaught in Concentration Cells
- Replies: 4
- Views: 462
Re: Enaught in Concentration Cells
No, remember that to find the change in gibbs free energy, you need the cell potential. You calculate the cell potential through the equation E = E˚ - (RT/nF) lnQ. This equation incorporates your standard cell potential value but it is not the only value that determines the cell potential. Then you...
- Thu Mar 12, 2020 5:39 pm
- Forum: General Rate Laws
- Topic: Fast and Slow Step Reactions
- Replies: 5
- Views: 448
Re: Fast and Slow Step Reactions
The slow step can be thought of as a funnel. Water can only flow through as fast as the neck of the funnel allows. Similarly, a reaction can only occur as quickly as the slow step proceeds. Therefore, it can also be thought of as the rate determining step. The fast steps are much faster than the slo...
- Thu Mar 12, 2020 5:36 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Slow step of reaction?
- Replies: 5
- Views: 432
Re: Slow step of reaction?
The slow step of a reaction can also be determined experimentally as the textbook does mention that some techniques such as the stopped-flow technique or spectrometry can be used to measure the rate of fast reactions.
- Thu Mar 12, 2020 5:29 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: q and delta H
- Replies: 4
- Views: 447
Re: q and delta H
q = delta H when pressure is kept constant.
- Thu Mar 12, 2020 5:28 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: activation energy units
- Replies: 3
- Views: 528
Re: activation energy units
To add on, the units of activation energy are joules per mole since this cancels out with the constant R.
- Fri Mar 06, 2020 10:40 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 6N.5B
- Replies: 1
- Views: 233
Re: 6N.5B
The Q value should be [Mn2+]2[Cl2]5/[H+]16[MnO4-]2[Cl-]10. When simplified, you should get 1*1066/[Cl-]10. So you should divide 1*1066 by 1*1076.
- Fri Mar 06, 2020 10:29 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 6.73
- Replies: 2
- Views: 295
Re: 6.73
If you look at the standard cell potentials for the half reactions, you'll see that Al3+ + 3e- --> Al has a Eo of -1.66 V and O2 + 2 H2O + 4 e- --> 4 OH- has a Eo of 0.40 V. So you know that Al has to be the anode and O2 has to be the cathode.
- Fri Mar 06, 2020 10:10 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Reducing Mass of Electrode
- Replies: 6
- Views: 613
Re: Reducing Mass of Electrode
Halving the mass of an electrode will not affect the cell potential because cell potential is an intensive property, meaning it is not affected by the quantity of a species present.
- Fri Mar 06, 2020 10:08 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Purpose of Electrode
- Replies: 9
- Views: 723
Re: Purpose of Electrode
The electrode is a metallic conductor that is in contact with an electrolyte solution. This allows electrons to flow between the cathode and the anode so that the reaction can proceed. Platinum is sometimes used as an electrode because in reactions that only occur with aqueous species, there is no s...
- Fri Mar 06, 2020 10:04 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Signs of Enaught
- Replies: 7
- Views: 609
Re: Signs of Enaught
A negative standard cell potential means that the reduction of that species is not spontaneous. This means that it is spontaneous when oxidized. Cell potential of a galvanic cell should be positive because this means that the reaction is spontaneous and does not need any external sources for the rea...
- Fri Feb 28, 2020 10:40 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 6M.1
- Replies: 4
- Views: 360
Re: 6M.1
As mentioned before, Ecell=Ecathode-Eanode.
But since the voltage is negative, we know that copper is actually supposed to be oxidized; it was placed on the wrong side. Since E standard for copper is +0.34 V and E cell is -0.689 V, -0.689 V= Ecathode - 0.34 V. We solve and get -0.349 V.
But since the voltage is negative, we know that copper is actually supposed to be oxidized; it was placed on the wrong side. Since E standard for copper is +0.34 V and E cell is -0.689 V, -0.689 V= Ecathode - 0.34 V. We solve and get -0.349 V.
- Fri Feb 28, 2020 10:33 am
- Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
- Topic: Preventing Corrosion
- Replies: 2
- Views: 222
Re: Preventing Corrosion
When you galvanize a metal, you coat it in zinc because zinc is a stronger reducing agent than other metals such as iron. As a result, when there is a scratch and the metal under the zinc is exposed, the zinc oxidizes instead of the metal underneath, protecting the metal underneath from corrosion.
- Fri Feb 28, 2020 10:28 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: EMF
- Replies: 1
- Views: 184
Re: EMF
In the textbook, it mentions that the cell potential is still widely called the electromotive force, but since it isn't actually a force, the term is becoming obsolete.
- Fri Feb 28, 2020 10:13 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Galvanic Cells
- Replies: 10
- Views: 599
Re: Galvanic Cells
When Ecell equals zero the reaction has reached equilibrium. This means that E(cathode) is equal to E(anode) and there is no electron flow, meaning the battery has died out.
- Fri Feb 28, 2020 10:10 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Salt Bridge
- Replies: 4
- Views: 919
Re: Salt Bridge
Salt bridges do not provide electrons, they maintain charge balance by transferring ions so that the electrical circuit is completed and the reaction will not stop due to charge buildup. The ions are typically chosen so that they do not affect the reaction. Usually, the cation goes toward the cathod...
- Fri Feb 21, 2020 11:55 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: relevance of salt bridge
- Replies: 9
- Views: 624
Re: relevance of salt bridge
A salt bridge will allow ion transfer without charge buildup (which stops electron transfer from one reaction beaker to another) because it allows the two solutions to stay neutral.
- Fri Feb 21, 2020 11:52 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: measuring electron transfer
- Replies: 2
- Views: 190
Re: measuring electron transfer
We cannot measure the electron transfer of one cell. Therefore, we compare the electron transfer of all cells (half reactions) with respect to the standard hydrogen electrode.
- Fri Feb 21, 2020 11:49 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: property of E
- Replies: 6
- Views: 430
Re: property of E
E is an intensive property because the voltage difference is the same and does not depend on how many times the reaction occurs. As a result, E will not change even when you multiply the reaction by a coefficient.
- Fri Feb 21, 2020 11:44 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: How to write Cell Diagrams
- Replies: 5
- Views: 456
Re: How to write Cell Diagrams
The basic structure of a cell diagram is (left) anode || (right) cathode, where || represents a salt bridge. You want to put the substance being oxidized on the left and and the substance being reduced on the right so that the voltage is positive. Within the left or right side, you want to write out...
- Fri Feb 21, 2020 1:57 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 5J.15
- Replies: 3
- Views: 294
Re: 5J.15
At different temperatures, delta G changes, which means that you cannot use the standard Gibbs free energy values at 100 degrees C, as those are only the values at 25 degrees C. This means in order to find the delta G at a different temperature, you would need to use the equation that relates delta ...
- Fri Feb 21, 2020 1:51 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Standards
- Replies: 2
- Views: 214
Re: Standards
A standard quality is like a reference point. Essentially, it is a measurement under certain conditions, like how standard enthalpy is measured at 25 degrees C and 1 atm pressure. You use standard values when an equation specifies it.
- Fri Feb 14, 2020 1:44 pm
- Forum: Ideal Gases
- Topic: ideal gas
- Replies: 5
- Views: 369
Re: ideal gas
An ideal gas is an approximation of gases that helps to model how they behave. By definition, ideal gas molecules do not attract or repel each other and the molecules themselves take up zero volume.
- Fri Feb 14, 2020 1:42 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: K
- Replies: 12
- Views: 736
Re: K
You generally include gases and aqueous solutions and exclude solids and liquids in the calculation of K. This is because solids and liquids are usually present in excess so a change to its concentration results in a negligible change.
- Fri Feb 14, 2020 1:41 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Standard Enthalpies of Formation
- Replies: 10
- Views: 587
Re: Standard Enthalpies of Formation
The standard enthalpy of formation of elements in their standard state (such as O2 (g) or C (s)) is zero.
- Thu Feb 13, 2020 7:53 pm
- Forum: Calculating Work of Expansion
- Topic: Work
- Replies: 5
- Views: 317
Re: Work
Leslie Almaraz 4G wrote:Is irreversible always sudden?
Generally, irreversible expansions are quicker than reversible ones because they do not have be changed by an infinitesimally small amount at a time. However, this doesn't always mean that they are very sudden.
- Thu Feb 13, 2020 7:50 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Textbook question 4J.17
- Replies: 2
- Views: 191
Re: Textbook question 4J.17
You know that reaction B is spontaneous everywhere because Δ H is calculated to be negative and Δ S is calculated to be positive. By using the Gibbs free energy formula, Δ G = Δ H − T Δ S, we will get a negative number minus a positive number for any temperature, which will only result in negative a...
- Fri Feb 07, 2020 10:31 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: using ratios for R
- Replies: 3
- Views: 101
Re: using ratios for R
To add on, heat capacities vary depending on molecular properties. So depending on the type of molecule (i.e. monatomic gas or linear molecule), there will be different properties to account for and thus different ratios to use, as derived through various formulas.
- Fri Feb 07, 2020 10:13 am
- 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: 4G.5
- Replies: 1
- Views: 104
Re: 4G.5
There are 12 orientations because in a cis- isomer, the X's must be adjacent to each other, or 90 degrees from each other in this case. With the octahedral cis-MX2Y4, there are 12 different ways for the two X's to be next to each other.
- Fri Feb 07, 2020 10:02 am
- 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: 3rd Law of Thermodynamics
- Replies: 2
- Views: 114
Re: 3rd Law of Thermodynamics
The third law of thermodynamics states that as a system's temperature approaches absolute zero (0 K), the entropy of the system becomes 0. This is important because it implies that there is an absolute scale for entropy. It also allows us to assume that S=k B ln(w)=0 if a system can be prepared in o...
- Fri Feb 07, 2020 9:56 am
- Forum: Calculating Work of Expansion
- Topic: units
- Replies: 3
- Views: 116
Re: units
1 L*atm is equal to 101.325 J. This conversion is on the constants and equations sheet.
- Fri Feb 07, 2020 9:55 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: negative Delta U
- Replies: 5
- Views: 315
Re: negative Delta U
Internal energy is the total store of energy in a system. When you burn fuel, that energy is released, so the final internal energy of the system will be less than the initial, resulting in a negative delta U.
- Fri Jan 31, 2020 12:47 pm
- Forum: Calculating Work of Expansion
- Topic: Which process does more work?
- Replies: 5
- Views: 1769
Re: Which process does more work?
AKatukota wrote:So for the first one, why does the volume remain 1.20L?
The volume doesn't remain 1.20L, it expands an additional 1.20L.
- Fri Jan 31, 2020 10:35 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess' Law
- Replies: 7
- Views: 263
Re: Hess' Law
Enthalpy is a state property, meaning that it is not dependent on the path taken to obtain that state, and state properties can be added or subtracted.
- Fri Jan 31, 2020 10:31 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: accuracy of bond enthalpies
- Replies: 8
- Views: 409
Re: accuracy of bond enthalpies
Only the bond enthalpies of diatomic molecules are accurate. All other bond enthalpies are calculated from averages of many different molecules, which results in less accurate calculations.
- Fri Jan 31, 2020 10:24 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: H and q
- Replies: 7
- Views: 242
Re: H and q
Delta H refers to the total heat in a system whereas q is the heat that is being transferred. Delta H is equal to q when the pressure is constant, but if the pressure changes, then they are not the same.
- Fri Jan 31, 2020 10:20 am
- Forum: Calculating Work of Expansion
- Topic: Textbook Question 4A.1A
- Replies: 1
- Views: 121
Re: Textbook Question 4A.1A
Using the densities, we can calculate the volume taken up by water. Since there are 100. g of water and the density is 1.00g/cm 3 , we can divide 100. g by 1.00g/cm 3 to find that we have 100. cm 3 water, which is equal to 100. mL. We can do the same for when it is ice. We still have 100. g, but thi...
- Fri Jan 24, 2020 10:18 am
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: pKa vs pH
- Replies: 4
- Views: 239
Re: pKa vs pH
To add on, pH can change depending the concentration of an acid in a reaction, whereas pKa does not change based on concentration because it is the -log of an equilibrium constant.
- Fri Jan 24, 2020 10:13 am
- Forum: Phase Changes & Related Calculations
- Topic: symbol of Heat
- Replies: 3
- Views: 643
Re: symbol of Heat
I believe qp is the heat energy at a constant pressure, and when there is a constant pressure, qp=delta H.
- Fri Jan 24, 2020 10:08 am
- Forum: Calculating Work of Expansion
- Topic: Reversible vs. Irreversible Reactions
- Replies: 2
- Views: 212
Re: Reversible vs. Irreversible Reactions
To add on, often times a reaction is irreversible because the reverse reaction has too high of an activation energy or energy barrier to overcome, so it is not easy for the reaction to proceed in the reverse direction.
- Fri Jan 24, 2020 10:04 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6D.15 b
- Replies: 1
- Views: 115
Re: 6D.15 b
In water, AlCl3 dissociates, and Al3+ acts as an acid and takes the form Al(H2O)63+, as shown in the table 6D.1. In this reaction, Al(H2O)63+ will donate an H+ because it is acting as an acid, which results in H3O+ and Al(H2O)5OH2+.
- Thu Jan 23, 2020 10:03 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Chem. Equilibrium Post Assessment 2 #29
- Replies: 1
- Views: 149
Re: Chem. Equilibrium Post Assessment 2 #29
You could use an ICE table to solve this problem. You know that the equilibrium concentration of BrCl is 3.36 x 10 -5 . You also know that the change from its initial concentration is -2x. We also know that 1.5 x 10 -4 was used up (change in concentration). So 1.5 x 10 -4 = -2x. From this, we can ge...
- Thu Jan 16, 2020 10:15 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Exothermic
- Replies: 5
- Views: 270
Re: Exothermic
Usually when new bonds are formed, it is a lot more stable for the atoms, so the potential energy is lower than it was before. This results in the release of energy.
- Thu Jan 16, 2020 10:10 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 5I.13 c
- Replies: 2
- Views: 160
Re: 5I.13 c
Cl2 is more stable than F2 because it forms less product than F2, as indicated by the fact that its K value is smaller than F2's K value. This means that it is less likely to dissociate, and thus more stable.
- Thu Jan 16, 2020 10:03 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Percentage Ionization
- Replies: 2
- Views: 72
Re: Percentage Ionization
You calculate the percentage of ionization by dividing the concentration of the dissociated species (the conjugate acid or base) by the original concentration of the acid or base, then multiplying by 100%. Conceptually, the percentage of ionization can tell you how much an acid or base has dissociat...
- Thu Jan 16, 2020 9:55 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: acidity and basicity constant
- Replies: 4
- Views: 208
Re: acidity and basicity constant
Yes, the same rules still apply as it is still an equilibrium constant.
- Thu Jan 16, 2020 9:53 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Endothermic vs exothermic
- Replies: 4
- Views: 186
Re: Endothermic vs exothermic
For endothermic reactions, since there is more heat, it will be easier to overcome the activation energy, thus making it easier to form more products. As a result, the products will be favored. For exothermic reactions, the reverse reaction (P-->R) will be endothermic. So using the logic that the re...
- Thu Jan 09, 2020 6:38 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 5I.17
- Replies: 3
- Views: 527
Re: 5I.17
I'm not sure how you got x=0.0114, but if you plug in all the values into the Kc ratio so that it's 1.00*10-5=(2x)2/((0.114-x)(0.114-x)) and solve for x using the quadratic formula, you should get x=1.8*10-4.
- Thu Jan 09, 2020 3:22 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Direction
- Replies: 4
- Views: 151
Re: Direction
If Q<K, then we know that the reaction will favor the forward reaction because there is not enough product for the reaction to be at equilibrium.
If Q>K, then we know the reaction will favor the reverse reaction because there is not enough reactant for the reaction to be at equilibrium.
If Q>K, then we know the reaction will favor the reverse reaction because there is not enough reactant for the reaction to be at equilibrium.
- Thu Jan 09, 2020 3:18 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Q
- Replies: 4
- Views: 164
Re: Q
The reaction quotient is a ratio of the concentration of products to reactants when the system has not yet reached equilibrium. If Q<K, the forward reaction will be favored in order to create more product because there are not enough products in relation to reactants for the system to be at equilibr...
- Thu Jan 09, 2020 2:28 pm
- Forum: Ideal Gases
- Topic: K value
- Replies: 14
- Views: 1175
Re: K value
If the K value is large (K>10 3 ), then the concentration of products is much greater than reactants at equilibrium. This means that the forward reaction is favored and more products are being produced than reactants. If the K value is small (K<10 -3 ), then the concentration of reactants is much gr...
- Thu Jan 09, 2020 1:19 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 5I.33
- Replies: 2
- Views: 153
Re: 5I.33
Since you know the mass of CO2 present at equilibrium, you can find the moles of CO2 at equilibrium by using the molar mass, which will be 0.000395 moles of CO2. Then, you can find the moles of NH3 present at equilibrium by using mole ratios: 0.000395 moles CO2 * (2 moles NH3/1 mole CO2) = 0.00079 m...