Search found 48 matches
- Thu Mar 14, 2019 11:54 pm
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
- Replies: 179
- Views: 22929
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
Why is it that in #3 in the reaction mechanisms worksheet we're able to use the pre-equilibrium assumption to substitute and replace [C]? Wouldn't the approximation fail since the first and second steps are both fast, so there's no bottleneck to stop [C] form being consumed instead of forming A and ...
- Wed Mar 13, 2019 6:57 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Frequency Factor, A
- Replies: 1
- Views: 229
Re: Frequency Factor, A
We were told that the value of A would be given, and if not, to assume that it equals 1.
- Mon Mar 11, 2019 9:40 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Pre-Equilibrium Approach
- Replies: 1
- Views: 243
Re: Pre-Equilibrium Approach
Professor Lavelle told us that we would not have to use the direct computation method in the class, but the slides did mention that assuming you perform the steady-state approximation correctly, the final result should be equivalent to the result you get by doing the pre-equilibrium method. Your cho...
- Mon Mar 11, 2019 9:37 pm
- Forum: General Rate Laws
- Topic: Decreasing Instantaneous Rate
- Replies: 2
- Views: 258
Re: Decreasing Instantaneous Rate
Because the rate law expression is Rate = k[A] (as a first order example), rate is proportional to the concentration of A (or the reactant). As the reaction proceeds, the concentration of A decreases, and as a result, the rate also decreases because there are less reactants available for reaction. H...
- Mon Mar 11, 2019 9:35 pm
- Forum: First Order Reactions
- Topic: Integrated Rate Laws (7b.1/ 15.21)
- Replies: 3
- Views: 334
Re: Integrated Rate Laws (7b.1/ 15.21)
Using the balanced mock chemical reaction, you can use stoichiometry to convert from [B] to [A]. Then, using the concentration of A that you solve in this manner, you can use the integrated rate law of the first order. Hope this helps.
- Mon Mar 11, 2019 9:34 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: 15.89
- Replies: 1
- Views: 222
Re: 15.89
A reaction profile is just a diagram (2D plot) of energy as a function of time/progression of the reaction. I would say that it's a good idea to know how to draw one (Professor Lavelle drew a few in class today) as it lets you visualize the values of the activation energy, whether or not the reactio...
- Tue Mar 05, 2019 8:18 pm
- Forum: Zero Order Reactions
- Topic: Order Reactions
- Replies: 4
- Views: 429
Re: Order Reactions
Yes, since reactions of higher orders are more uncommon. The textbook and our constant/formula sheet seems to only cover reactions of order 0, 1, and 2, only.
- Tue Mar 05, 2019 8:15 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Activation Energy
- Replies: 2
- Views: 331
Re: Activation Energy
The dependence of k on temperature and activation energy can be seen in the Arrhenius equation, given as k = Ae^{-E_{a}/RT} . The constant A, is a parameter found through experimentation, similar to the activation energy Ea. As you can see, as the activation energy increases, k approaches 0. As k de...
- Tue Mar 05, 2019 8:04 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Rate Constant
- Replies: 3
- Views: 370
Re: Rate Constant
The concentrations and orders of the reactants represent the reaction mechanism (how many molecules must collide for the reaction to proceed, which is more clear when we go into elementary reactions), whereas the rate constant k depends on the temperature and activation energy. The higher the value ...
- Tue Mar 05, 2019 8:02 pm
- Forum: First Order Reactions
- Topic: First Order Reaction
- Replies: 3
- Views: 318
Re: First Order Reaction
[A] is assumed to represent the concentration of A at some time t, or the time that elapsed since the reaction began.
- Mon Feb 25, 2019 10:12 pm
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
- Replies: 179
- Views: 22929
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
Can someone please explain #4 on the electrochem worksheet? I don't understand why the answer is Ag+ not Fe2+ Since you want copper to spontaneously reduce, then you need to check the oxidation values of elements/options given. In order to have a spontaneous reaction with copper reducing, the cell ...
- Mon Feb 25, 2019 8:15 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Gibbs Units
- Replies: 2
- Views: 405
Re: Gibbs Units
Since Gibbs free energy is in terms of joules, it would be more accurate to change the units from V to J/C. You always want to make sure you cancel your units, not just out of necessity, but also because it's a very convenient way to check your work and make sure you're using the correct formula and...
- Mon Feb 25, 2019 8:12 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 5G.15 (7th Ed.)/ 11.17 (6th ed.)
- Replies: 1
- Views: 243
Re: 5G.15 (7th Ed.)/ 11.17 (6th ed.)
Since you're given the value of K in this particular problem, rather than look up the Gibbs free energy of formation values, you can just substitute in the -RTlnK for greater convenience in your calculations.
- Mon Feb 25, 2019 8:09 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Value of K in Gibbs free energy equation
- Replies: 3
- Views: 656
Re: Value of K in Gibbs free energy equation
When a chemical reaction is at equilibrium, the Gibbs free energy is, indeed, equal to zero, but that doesn't mean that the Gibbs free energy under standard conditions is (the little circle to the top right is what you're most likely missing). As you can see from the original equation, \Delta G = \D...
- Sun Feb 24, 2019 11:23 pm
- Forum: Balancing Redox Reactions
- Topic: Redox Rxns in Acidic and Basic Solutions
- Replies: 6
- Views: 496
Re: Redox Rxns in Acidic and Basic Solutions
The problem would either have to specify, or there would be an obvious product being formed (H3O+ for acidic, OH- for basic).
- Tue Feb 19, 2019 9:05 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 4I.9
- Replies: 2
- Views: 518
Re: 4I.9
When calculating S for either reaction, you will use the equation delta S = q(rev) / T. Even though the reaction in b. is not reversible, since entropy is a state function, as long as the initial and final states are the same in both reactions, you can just use that same equation. As for the change ...
- Tue Feb 19, 2019 9:02 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: ΔG= ΔH -TΔS
- Replies: 2
- Views: 339
Re: ΔG= ΔH -TΔS
In most cases, if you have to calculate change in entropy and change in enthalpy separately before using the Gibbs Free Equation, entropy tends to be in J/mol whereas enthalpy tends to be in kJ/mol. When calculating each of the constituent thermodynamic values, it doesn't really matter if it's in J/...
- Tue Feb 19, 2019 9:00 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Homework 6th Edition 9.63
- Replies: 3
- Views: 360
Re: Homework 6th Edition 9.63
The idea is that the lower the energy of the substance, the more stable it is as a compound. In cases where the formation is negative from the elements to the substance, this means that energy was lost as the elements reacted to create the compound, or that the compound has lower energy and is there...
- Thu Feb 14, 2019 10:10 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: System and surroundings
- Replies: 4
- Views: 572
Re: System and surroundings
If this is specifically regarding entropy, that is because in reversible reactions, the total change in entropy (in the universe) is equal to zero. As a result, any change in entropy in the surroundings must be equal and opposite to the change in entropy of the system because they add up together to...
- Tue Feb 12, 2019 11:37 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: hw 4f 17
- Replies: 1
- Views: 243
Re: hw 4f 17
In general with state functions, if you're asked to find them at a temperature that's not at it's boiling point, etc., you'll always have three separate steps. The first is to calculate, in this case, the entropy required to reach the "normal" temperature at which a process occurs from the...
- Tue Feb 12, 2019 11:34 pm
- Forum: Calculating Work of Expansion
- Topic: delta U = 3/2nR(delta T)
- Replies: 2
- Views: 2485
Re: delta U = 3/2nR(delta T)
We use this equation to find the change in internal energy when we're given a case that deals with an ideal sample of gases.
- Tue Feb 12, 2019 11:31 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: HW problem 4A. 13
- Replies: 2
- Views: 500
Re: HW problem 4A. 13
The first reaction is intended to give you sufficient information to solve for the calorimeter constant using q = C*deltaT. Then, because the reaction parameters are similar in the subsequent chemical reaction (same amount of solution in the same calorimeter), we can use that same calorimeter consta...
- Fri Feb 08, 2019 9:47 pm
- Forum: Phase Changes & Related Calculations
- Topic: Negative work
- Replies: 11
- Views: 1018
Re: Negative work
If the work is negative, how would the isobaric work equation change? Would the integral now be of a positive quantity instead of a negative one? For an isobaric expansion/compression, the pressure is constant (isobaric: deltaP = 0), so the we would use the equation w = -P*deltaV. If the change in ...
- Thu Feb 07, 2019 10:43 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Internal Energy, Enthalpy, and Heat
- Replies: 2
- Views: 294
Re: Internal Energy, Enthalpy, and Heat
Internal energy is the total store of energy within a system: this would be the sum of the kinetic energy and potential energy of every single particle (electrons, neutrons, protons, etc) within whatever reaction or space we're looking at. The two ways that internal energy can change is through heat...
- Thu Feb 07, 2019 10:37 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Change in entropy and its relation with temperature
- Replies: 2
- Views: 325
Re: Change in entropy and its relation with temperature
Change in entropy can be calculated using the formula: deltaS = q / T. In this case, delta S is change in entropy, q is the heat of a reversible reaction, and T is absolute temperature. Mathematically speaking, the lower your temperature is, the greater delta S will be, thereby allowing a greater in...
- Mon Feb 04, 2019 10:47 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: ΔU=q+w
- Replies: 1
- Views: 214
Re: ΔU=q+w
Internal energy is the total energy that is contained in a particular system, and this would essentially be the kinetic energy plus the potential energy of all the particles involved. Now, realistically speaking, it is essentially impossible to calculate the KE and PE of every single little atom, it...
- Mon Feb 04, 2019 10:36 pm
- Forum: Calculating Work of Expansion
- Topic: Converting between units?
- Replies: 2
- Views: 240
Re: Converting between units?
In the case of converting L * atm to joules, we multiply by 101.325 because 1 joule is equal to 1 kg*m^2*s^-2. This means that we have to convert liters into meters, where 1 m^3 = 1000 L, and we have to convert atm to pascals, which is 1 atm = 101,325 Pa. We convert to pascals because 1 Pa = 1 kg * ...
- Mon Feb 04, 2019 10:25 pm
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
- Replies: 179
- Views: 22929
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
For question 6 on worksheet 4, how exactly are we supposed to determine which volume we use to find the pressure for the work component of the internal energy change? To my understanding, the pressure was calculated using PV = nRT, with V = 998L, n = 31.9 mol, R = 0.08206 L * atm / mol * K, T = 311K...
- Thu Jan 31, 2019 8:30 pm
- Forum: Calculating Work of Expansion
- Topic: Problem 8.27
- Replies: 2
- Views: 238
Re: Problem 8.27
In this particular problem, you're given the initial conditions of the gas sample prior to any change in volume. This means you can calculate the number of moles in the gas sample, which can then be used in the reversible work equation to find the total work in both cases. PV = nRT -> n = PV/RT. Hop...
- Wed Jan 30, 2019 9:59 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: HW Problem 4A.9
- Replies: 3
- Views: 266
Re: HW Problem 4A.9
For this problem, think about where the heat is going. Because there is a temperature difference, energy will be transferred from the copper and into the water. Since heat released = - heat absorbed, we can use q = mC \Delta T for both substances, and we solve for T final since we should know all ot...
- Wed Jan 30, 2019 9:53 pm
- Forum: Phase Changes & Related Calculations
- Topic: Phase change equations
- Replies: 1
- Views: 246
Re: Phase change equations
The difference really is just a matter of definition. Standard enthalpy of phase change (of formation, bond enthalpy, etc.) or any enthalpy really, is the measurement of energy required to do something to one mole of something, be it breaking a bond, forming the substance from its purest constituent...
- Wed Jan 30, 2019 9:47 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Internal Energy
- Replies: 1
- Views: 156
Re: Internal Energy
Internal energy is defined as the "total store of energy in a system," which means that to measure it would require us to know the energy of every single individual particle in a system (KE/PE). This is impossible since we would have to individually calculate out these energies for every a...
- Wed Jan 23, 2019 8:58 pm
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
- Replies: 179
- Views: 22929
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
Can someone explain how to do number 4b on worksheet 1. I do not seem to understand the steps to take. The Haber process is used to synthesize ammonia gas (NH ) from nitrogen gas ( ) 3 N2 and hydrogen gas ( H ). A system at equilibrium contains 1.85M H2 1.36M N2 and 2.91 x 10 ^−3 NH3 at a constant ...
- Tue Jan 22, 2019 5:46 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: autoprotolysis
- Replies: 1
- Views: 532
Re: autoprotolysis
Autoprotolysis is the process by which a molecule donates a hydrogen ion to an identical molecule. While we did learn about water in lecture, by this definition, it would also mean that polyprotic acids could potentially undergo autoprotolysis. For example, the hydrogen sulfate ion, or conjugate bas...
- Tue Jan 22, 2019 5:35 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Hmwrk 11.77 6th edition
- Replies: 2
- Views: 273
Re: Hmwrk 11.77 6th edition
More generally, if delta H is positive, it means that the internal energy of the reaction has increased (endothermic reaction). If delta H is negative, the opposite is true (exothermic reaction). From there, it's simply a matter of seeing how a temperature increase can drive endothermic reaction dir...
- Mon Jan 21, 2019 9:55 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Chem Eq. Module #4 Question 15 Conflicting Answers
- Replies: 2
- Views: 321
Chem Eq. Module #4 Question 15 Conflicting Answers
15. The photosynthesis reaction, 6 CO2(g) + 6 H2O(l) ⇌ C6H12O6(aq) + 6 O2(g), is endothermic. What effect will the following changes have on the equilibrium composition. a) Water is added. b) The partial pressure of CO2 is decreased. A. a) Decrease [C6H12O6] and [O2], b) Decrease [O2] B. a) Decrease...
- Sat Jan 19, 2019 11:49 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Problem 12.33 Sixth Edition
- Replies: 2
- Views: 320
Re: Problem 12.33 Sixth Edition
In order to solve for the hydroxide concentration of the original solution, we must work backwards from the dilution that took place. Using the equation M1 * V1 = M2 * V2, we can plug in the values of the original solution for M1 and V1, and the values for the diluted solution go into M2 and V2. The...
- Sat Jan 19, 2019 11:44 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Change in Pressure
- Replies: 2
- Views: 307
Re: Change in Pressure
If the volume decreases, then pressure should increase due to Boyle's Law (PV = c). An increase in pressure then means that the reaction wants to return to equilibrium by minimizing this increase. Therefore, it favors the side that has less moles of gaseous molecules. Another way to think about this...
- Fri Jan 18, 2019 7:34 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: 12.39 and 12.41
- Replies: 1
- Views: 222
Re: 12.39 and 12.41
The other two acids aren't on the table because they're the conjugate acids of certain bases that can be found on Table 12.2. By using the relationship of Kw = Ka * Kb, you can convert between the acid constant and the base constant to determine the relative strengths.
- Fri Jan 18, 2019 7:33 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Pressure
- Replies: 3
- Views: 442
Re: Pressure
In terms of the equilibrium constant, the value of K does not change unless temperature is affected. If the source of the pressure is caused by a change in volume, then the concentrations of the reactants and products change, which may cause either the forward or reverse reaction to be favored to re...
- Tue Jan 15, 2019 8:15 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Question on 6th Edition 6A.21
- Replies: 2
- Views: 220
Re: Question on 6th Edition 6A.21
Water is assumed to be electrically neutral assuming it is pure, so [H3O+] = [OH-]. Given Kw, it is then just a matter of finding the square root of both sides to find the concentration of H3O+ and OH-.
- Tue Jan 15, 2019 7:09 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Le Chatelier - Volume
- Replies: 2
- Views: 182
Re: Le Chatelier - Volume
I believe what was meant by the lecture is that anytime volume is changed, thereby changing the pressure, it's not the number of moles of gas that causes the forward or reverse reaction being favored, it's the change in concentration (which means the reaction quotient is affected). While the conclus...
- Tue Jan 15, 2019 7:04 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Same/different @ equilibrium
- Replies: 2
- Views: 139
Re: Same/different @ equilibrium
The important distinction is that the ratio with the exponents is the equilibrium constant, or some variation of it (one of them is also the reciprocal of K). This means that these specific ratios are unchanged across different samples sizes given the temperature is constant. Because the ratio witho...
- Thu Jan 10, 2019 8:22 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: 11.11 (6th Edition)
- Replies: 1
- Views: 105
Re: 11.11 (6th Edition)
The equilibrium constant will remain the same for a given reaction as long as the temperature remains constant. In this particular case, the equilibrium constant expression is as follows: Kc = [O2]^3 / [O3]^2. Any alteration to the reaction (reversing the direction, multiplying the stoichiometric co...
- Thu Jan 10, 2019 8:13 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: ICE Chart for HW problem 11.49 6th Edition
- Replies: 4
- Views: 342
Re: ICE Chart for HW problem 11.49 6th Edition
In this case, you would use the quadratic equation on what Kate pointed out to be the equilibrium constant expression, Kc = [0.2 + x][x]. Since you have one variable, x, and there is no other way to properly factor it, once you expand out the Kc expression, you use the quadratic formula x = (-b ± √(...
- Tue Jan 08, 2019 10:15 pm
- Forum: Ideal Gases
- Topic: Definition
- Replies: 2
- Views: 458
Re: Definition
An ideal gas is one that meets a few criteria (the gas particles do not attract nor repel other particles, their only interaction are in the form of elastic collision, the particles are randomly moving, and the volume of the particles are negligible). Ideal gas assumptions are fine for low pressure,...
- Tue Jan 08, 2019 10:11 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Calculations
- Replies: 2
- Views: 258
Re: Calculations
There isn't any mathematical difference in calculating K, Kc, or Kp. The subscripts c and p simply indicate how you're measuring the presence of your reactants and products. Kc is the equilibrium constant calculated from molar concentrations (of the units mol * L^-1), whereas Kp is the equilibrium c...
- Mon Jan 07, 2019 7:18 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Topic 5G.9 7th Edition HW Problem
- Replies: 2
- Views: 169
Re: Topic 5G.9 7th Edition HW Problem
The primary concept being tested in this question is the equilibrium constant, K, and what meaning the constant holds. For any given chemical reaction, there is a specific value of K that is dependent on the temperature, and since the problem states that the same temperature is maintained across bot...