Search found 62 matches
- Sat Mar 17, 2018 1:53 pm
- Forum: *Enzyme Kinetics
- Topic: Final-Collision Theory
- Replies: 3
- Views: 916
Re: Final-Collision Theory
Can someone please explain the difference between the collision theory and the transition state theory? Collision theory states that molecules need to collide with the correct orientation and with enough kinetic energy for a reaction to occur. Transition state theory builds on the collision theory....
- Sat Mar 17, 2018 1:41 pm
- 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: S and delta S
- Replies: 4
- Views: 595
Re: S and delta S
You can find the absolute entropy by using the Boltzmann equation S=kB*lnW. kB is Boltzmann's constant (1.38x10-23J/K).
- Sat Mar 17, 2018 11:27 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Catalysts
- Replies: 2
- Views: 385
Re: Catalysts
Catalysts are not included in the overall balanced reaction because they are present in the beginning and regenerated at the end. If you think about it, the catalyst would cancel out. Also, catalysts participate in the reaction by lowering the activation energy, but they are not necessarily reactant...
- Fri Mar 16, 2018 12:11 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Negative Ea
- Replies: 3
- Views: 565
Re: Negative Ea
The activation energy cannot be negative, but it can be decreased with the addition of a catalyst as it provides an alternative pathway between reactants and products.
- Fri Mar 16, 2018 12:08 am
- Forum: *Organic Reaction Mechanisms in General
- Topic: Functional Groups Overlapping
- Replies: 3
- Views: 2718
Functional Groups Overlapping
Is it possible for functional groups to overlap? For example, can an aldehyde contain part of an amine?
- Sun Mar 11, 2018 12:43 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.63
- Replies: 2
- Views: 383
Re: 15.63
For the problem, you would just plug in the values given into the equation ln(k2/k1)=Ea/R((1/T1)-(1-T2)). The activation energy isn't negative, so it would just be ln(k2/(1.5x1010)=38/.008314((1/298)-(1/310)).
- Sun Mar 11, 2018 9:25 am
- Forum: *Organic Reaction Mechanisms in General
- Topic: Activation Energy and Temperature
- Replies: 3
- Views: 1212
Re: Activation Energy and Temperature
Yes, these conditions are possible. Having a high temperature just means you're more likely to reach the activation energy, though it may be higher. The effect these two conditions would have on the rate constant, in terms of increasing or decreasing it, would depend on the magnitude of each condit...
- Sun Mar 11, 2018 9:21 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: reversible/irreversible
- Replies: 2
- Views: 375
Re: reversible/irreversible
There are a couple formulas for work that differ based on if a reaction is undergoing reversible or irreversible expansion. If the reaction is reversible, then use w=-nRTln(V final /V initial ), where T is the absolute temperature. When a reaction is irreversible and the external pressure is constan...
- Sun Mar 11, 2018 9:10 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: figure 15.27 interpretations
- Replies: 1
- Views: 243
Re: figure 15.27 interpretations
For figure 15.26, the reason that there is a peak followed by a sudden drop is that at low temperatures, there are not many molecules that have a high enough kinetic energy to collide and react. Therefore, instead, there is a higher fraction of molecules towards the beginning of the plot of kinetic ...
- Mon Mar 05, 2018 9:59 pm
- Forum: General Rate Laws
- Topic: rate constants
- Replies: 7
- Views: 876
Re: rate constants
Kayla Tchorz-Dis 1F wrote:I'm also a little confused on the concept of K. In our notes it says K is dependent on temperature and activation energy. So if it's dependent, how is it a constant?
The rate constant is constant in that it won't change under those specified conditions.
- Mon Mar 05, 2018 9:57 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Determining overall rate order
- Replies: 3
- Views: 543
Re: Determining overall rate order
Sometimes the presence of products does affect the reaction order, but I believe that we aren't really going to deal too much with that in this class. An example of this can be found on page 621 of the textbook in case you wanted to know more!
- Wed Feb 28, 2018 8:19 pm
- Forum: Second Order Reactions
- Topic: Half Life
- Replies: 2
- Views: 466
Half Life
I have a general question about the half life of reactions. Why does the half life of a zero order and second order reaction depend on the initial concentration of the reactant, but the concentration of a first order reaction does not? Can someone explain this conceptually?
- Wed Feb 28, 2018 8:27 am
- Forum: First Order Reactions
- Topic: Order of a reaction
- Replies: 3
- Views: 478
Re: Order of a reaction
To add on, the order of a reaction reveals to us what power of the concentration the rate is proportional to. For example, in a first-order reaction, its rate is proportional to the first power of the concentration. Also changing the concentration of different reactants has different effects on the ...
- Sun Feb 25, 2018 10:55 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Order
- Replies: 5
- Views: 617
Re: Order
The textbook briefly talks about fractional order reactions (page 621). I believe that if the fractional order is in the denominator, it means that the reaction slows down as the concentration of it increases. On the other hand, if the fractional order is in the numerator, the reaction rate increase...
- Sun Feb 25, 2018 10:48 pm
- Forum: General Rate Laws
- Topic: rate constants
- Replies: 7
- Views: 876
Re: rate constants
In addition to what is stated above, the rate constant k is independent of the concentrations of the reactants. You would multiply k by the concentrations to find the rate, but the concentrations themselves do not impact k.
- Sat Feb 24, 2018 9:17 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Order
- Replies: 5
- Views: 617
Re: Order
To add on, the order also reveals something about the rate of reaction. For example, a first-order reaction is one in which the rate of it is directly proportional to the concentration of the reactant concentration. A second-order reaction's rate is proportional to the second power of the concentrat...
- Wed Feb 21, 2018 11:44 pm
- Forum: General Rate Laws
- Topic: NO2 Example from Lecture
- Replies: 2
- Views: 454
Re: NO2 Example from Lecture
The equation is -d[NO2]/dt = d[NO]/dt = 2 * d[O2]/dt because as you stated, the rate at which NO2 concentration decreases is equal to the rate at which NO concentration increases. In addition, I believe that because O2 concentration increases half as fast, 2 x the rate of oxygen concentration increa...
- Wed Feb 21, 2018 11:23 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: mistake in Ch 14. #35.b?
- Replies: 1
- Views: 262
Re: mistake in Ch 14. #35.b?
Yes, this error is included in the Errors in Solutions Manual document posted on Lavelle's website.
- Tue Feb 20, 2018 6:44 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: HW 14.9
- Replies: 2
- Views: 239
Re: HW 14.9
To get the number of mols of electrons, you write a half reaction for both the oxidation and reduction. In 14.9a, for example, 2Ce 4+ (aq) is reduced while 3I - (aq) is oxidized. The two half reactions: 2Ce 4+ (aq) + 2e - --> 2Ce 3+ (aq) 3I - (aq) --> I 3 - (aq) + 2e- From this, you can see that 2 e...
- Mon Feb 19, 2018 3:36 pm
- Forum: Balancing Redox Reactions
- Topic: 14.1 chromium
- Replies: 1
- Views: 255
Re: 14.1 chromium
Each O is usually 2-, and there are 7 of them. 7 x 2- = 14-. The overall charge of the molecule is 2-. The difference between the overall charge and the oxygens is 12+. Each Cr is therefore 6+.
- Mon Feb 19, 2018 3:31 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.15b [ENDORSED]
- Replies: 1
- Views: 252
14.15b [ENDORSED]
This problem asks for the half-reactions and the cell diagram of H+(aq) + OH-(aq) --> H2O(l), the Bronsted neutralization reaction. Where did the O2 in both the half reactions come from (in the solutions manual)?
- Sun Feb 18, 2018 8:48 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Nernst Equation
- Replies: 4
- Views: 763
Re: Nernst Equation
In addition to what was stated above, the Nernst equation allows us to estimate the potentials of cells under nonstandard conditions.
- Sun Feb 18, 2018 8:40 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: inert metals
- Replies: 1
- Views: 259
Re: inert metals
I believe that the inert metal used is usually Pt(s) or C(gr). Inert metals are only necessary if the reactants do not act as electrodes.
- Sat Feb 17, 2018 7:12 pm
- Forum: Balancing Redox Reactions
- Topic: Oxidizing & Reducing Agents
- Replies: 6
- Views: 1152
Re: Oxidizing & Reducing Agents
The species that causes oxidation is the oxidizing agent. It accepts electrons released by the species being oxidized. The oxidation agent in a redox reaction gets reduced. The species that brings about reduction is the reducing agent. It supplies electrons to the species being reduced. The reducing...
- Sat Feb 17, 2018 8:54 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Electrolytes
- Replies: 1
- Views: 238
Electrolytes
In a galvanic cell, what exactly constitutes an electrolyte? Is it the solution the electrodes are in?
- Tue Feb 13, 2018 7:17 pm
- 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: S=klnW
- Replies: 7
- Views: 3472
Re: S=klnW
You would use this equation if the question asked you to calculate absolute entropy (instead of the change in entropy) of a system.
- Mon Feb 12, 2018 9:41 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: General Question [ENDORSED]
- Replies: 1
- Views: 350
Re: General Question [ENDORSED]
Larger molecules tend to have greater entropy than smaller ones because heavier molecules have more vibrational energy states. In this way, the molecules have a greater number of possible positions they can take up. Using the formula S=k*lnW with W being the number of ways the atoms or molecules in ...
- Fri Feb 09, 2018 11:52 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Homework Question 9.13
- Replies: 4
- Views: 523
Re: Homework Question 9.13
But how do you know that "for a change in temperature, the energy supplied as heat is related to the increase in temperature by the heat capacity, C". Is that always true? The definition of heat capacity is the ratio of the heat supplied to the rise of temperature produced: C= q/ \Delta T...
- Fri Feb 09, 2018 11:14 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: work
- Replies: 2
- Views: 292
Re: work
There are two types of work a system can do. There is nonexpansion work and expansion work. Expansion work is the work that concerns a change in the volume of a system. The equation relating work to volume is w=-P ex \Delta V, but this only applies when external pressure is constant. When external p...
- Fri Feb 09, 2018 11:02 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Homework Question 9.13
- Replies: 4
- Views: 523
Re: Homework Question 9.13
I think you actually mean R is used when volume changes and C v when temperature changes. The equations in which these values are plugged in are both derived from others. On pages 321 and 323, the derivations are explained for the equation for entropy change of a system with temperature change and o...
- Fri Feb 09, 2018 11:51 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Question 9.21
- Replies: 4
- Views: 540
Re: Question 9.21
This question wants you to use your knowledge about the possible states/number of ways the molecules can be arranged to calculate the overall entropy. This can be done using the equation S=k*lnW, where k is Boltmann's constant: 1.381 x 10 -23 J/K and W is the number of ways that the molecules can be...
- Fri Feb 09, 2018 8:25 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.35 Explained
- Replies: 2
- Views: 334
Re: 9.35 Explained
This problem is also on the solutions manual error sheet, where there is an explanation. The actual answer is that the change in entropy of B is less than that of C. However, The change in entropy of C is equal to the change in entropy of A. (B<C=A)
- Thu Feb 08, 2018 10:10 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Temperature and Spontaneity
- Replies: 2
- Views: 389
Temperature and Spontaneity
Is it true that when H and S are both negative that decreasing the temperature increases spontaneity? If so, could someone explain why this is the case? I know it has to do with the equation G=H-TS, but beyond the math, what is the explanation?
- Tue Feb 06, 2018 5:17 pm
- 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: 9.21
- Replies: 3
- Views: 412
Re: 9.21
Kb is Boltzmann's constant, so it is 1.381 x 10-23 J/K. It doesn't change from problem to problem. Hope this helps!
- Mon Feb 05, 2018 5:54 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.17
- Replies: 10
- Views: 1058
Re: 11.17
The problem is: Calculate the reaction Gibbs free energy of N 2 (g) + 3H 2 (g) \rightarrow 2NH 3 (g) when the partial pressures of N 2 ,H 2 , and NH 3 are 4.2 bar, 1.8 bar, and 21 bar, respectively. For this reaction, K=41 at 400. K. I used the equation \Delta G r = \Delta G r (standard) + RTlnQ to ...
- Mon Feb 05, 2018 10:55 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.17
- Replies: 10
- Views: 1058
11.17
For this problem, my answer was -2736.87 J, which is -2.7 kJ/mol when accounting for significant figures. The solution manual has -27 kJ/mol as the final answer. Is there something wrong with my math that is giving me the incorrect answer? My math looks exactly like what is shown in the manual.
- Thu Feb 01, 2018 9:48 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Total Entropy Change
- Replies: 1
- Views: 250
Total Entropy Change
The equation for total entropy change is \Delta S total = \Delta S system + \Delta S surr . My question is why doesn't the entropy change of the system always equal the entropy change of the surroundings? I know that only when the total entropy change is positive will the process be spontaneous, but...
- Wed Jan 31, 2018 4:22 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Phase change question demonstrated in Lecture
- Replies: 4
- Views: 645
Re: Phase change question demonstrated in Lecture
I believe that this just means that when delta H is positive, the reaction is endothermic. This doesn't favor the forward process because you have to add heat to the reactants; it isn't favorable or spontaneous in this fashion. However, it can become favorable when the temperature is high enough.
- Wed Jan 31, 2018 4:14 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.7 vs 9.13
- Replies: 4
- Views: 370
Re: 9.7 vs 9.13
There is a mistake in the solutions manual for 9.13. So, for both problems you should use C in the equation, not R. If you look at the Solution Manual Errors page from Dr. Lavelle's website, he has it corrected on there. Hope this helps!
- Tue Jan 30, 2018 9:45 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: heat capacity and entropy
- Replies: 2
- Views: 641
Re: heat capacity and entropy
First, heat capacity is the heat required to raise the temperature of an object by 1 degree Celsius. In general, increasing molecular complexity tends to also lead to higher heat capacities, since larger molecules have more possible ways to vibrate, bend and rotate. The same is true of entropy. More...
- Sat Jan 27, 2018 8:52 pm
- Forum: Calculating Work of Expansion
- Topic: U for Reversible & Irreversible Expansion
- Replies: 2
- Views: 375
Re: U for Reversible & Irreversible Expansion
Because internal energy is a state function, it is only dependent on the current state of the system. So, the change of the internal energy is the same regardless of if the expansion was reversible or not.
- Sat Jan 27, 2018 4:42 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: equations
- Replies: 1
- Views: 189
Re: equations
For a change in volume, you would use deltaS=nRln(V2/V1), in which n=moles of gas, R=ideal gas constant, V2=the final volume, and V1=the initial volume. For a change in temperature, you would use deltaS=nCln(T2/T1), in which n=moles of gas, C=molar heat capacity, T2=final temperature, and T1=the ini...
- Tue Jan 23, 2018 8:34 pm
- 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: 1/22 Lecture (Boltzmann Equation) [ENDORSED]
- Replies: 1
- Views: 136
1/22 Lecture (Boltzmann Equation) [ENDORSED]
On the lecture slide explaining the Boltzmann equation, it says "thermodynamic property ('small' error in S)" and "statistical ('large' error in W)." I'm not exactly sure what this is saying. Could someone clarify to me their significance in this context?
- Mon Jan 22, 2018 7:17 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.41
- Replies: 1
- Views: 162
Re: 8.41
The 7.5 x 104 comes from the calculation of 400 grams water x (4.184 J/C/g) x (-45). From that I got -7.5 x 104, but because water is losing heat, you make qwater negative, turning the negative value positive (negative x negative= positive). Hope that helps.
- Fri Jan 19, 2018 7:51 pm
- Forum: Phase Changes & Related Calculations
- Topic: Phase Change Question
- Replies: 3
- Views: 401
Re: Phase Change Question
For 8.37, you consider the units that enthalpy is reported in (kJ/mol). So, you should divide the heat (measured in kJ) by moles of the substance to find delta H of vaporization.
- Fri Jan 19, 2018 6:13 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.13
- Replies: 3
- Views: 369
Re: 8.13
The value of q is negative because the problem states that "the cooling system that surrounds the cylinder absorbs 947 kJ as heat." I think this means that the surroundings absorb the heat, and the actual cylinder itself releases it.
- Fri Jan 19, 2018 6:09 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: First law potential energy
- Replies: 1
- Views: 181
Re: First law potential energy
I believe that the potential energy doesn't change because this is an isothermal expansion of an ideal gas. Even though molecules are moving apart and therefore separation is increasing, the potential energy remains the same because there are no forces between the molecules.
- Fri Jan 19, 2018 5:45 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Confusion About One of the Answers to an Example Done in Lecture
- Replies: 2
- Views: 329
Re: Confusion About One of the Answers to an Example Done in Lecture
The answer is 49 kJ/degree for both Kelvin and Celsius because both temperature scales are scaled the same. A rise of one degree Celsius is equal to a rise of one degree Kelvin. However, this would not be the case for Fahrenheit. Hope that is helpful!
- Fri Jan 19, 2018 5:34 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.23
- Replies: 5
- Views: 929
Re: 8.23
The question asks for the heat capacity(C) of the calorimeter and not the specific heat capacity because specific heat capacity is for a certain mass of a sample, which we can't necessarily assign in this problem. The definition of heat capacity is heat supplied/change in temperature. In equation fo...
- Fri Jan 19, 2018 9:28 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: HW Q.8.49 [ENDORSED]
- Replies: 2
- Views: 210
Re: HW Q.8.49 [ENDORSED]
You use PV=nRT for this problem because of the equation on page 290 of the textbook. delta H = delta U + delta n gas RT where delta n gas =the final number of moles of gas-the initial number of moles of gas. The problem gives you a delta H value you can plug in, and with that, you are able to use PV...
- Wed Jan 17, 2018 4:23 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.31
- Replies: 3
- Views: 241
Re: 8.31
You would use the equation for heat: q = C x delta T. Plug in the value of (5/2)R for C for part a and (3/2)R for part b. Because these are the molar heat capacities, you also have to convert the grams into moles. Hope this helps.
- Wed Jan 17, 2018 4:16 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bong Enthalpy accuracy [ENDORSED]
- Replies: 3
- Views: 340
Re: Bond Enthalpy accuracy [ENDORSED]
Bond enthalpies are the least accurate method of calculating reaction enthalpies because they are average values from many different molecules. Bond enthalpies are accurate for diatomic molecules, though. The three methods are Hess's Law, bond enthalpies, and using standard enthalpies of formation, ...
- Wed Jan 17, 2018 4:05 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Bomb Calorimeters
- Replies: 2
- Views: 527
Re: Bomb Calorimeters
I believe that bomb calorimeters can be considered isolated systems because they do not exchange matter or heat with the surroundings (what is around the calorimeter). Calorimeters are designed to measure the transfer of energy as heat. The heat of the reaction is supposed to be equal to negative he...
- Wed Jan 17, 2018 9:32 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat Transfer at Constant Volume/Pressure
- Replies: 4
- Views: 503
Heat Transfer at Constant Volume/Pressure
Can someone explain to me or clarify the following statement made in the textbook? "At constant volume, the heat transfer is interpreted as delta U; at constant pressure, it is interpreted as delta H." Why is this true? I think I partly understand that when volume doesn't change, no expans...
- Wed Jan 17, 2018 9:28 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: internal energy
- Replies: 5
- Views: 559
Re: internal energy
Internal energy still is considered a state function despite both heat and work being path-dependent. I believe that because internal energy is not path-dependent (and only state-dependent), it does not matter what combination of heat and work is present to sum to the internal energy.
- Tue Jan 16, 2018 3:38 pm
- Forum: Phase Changes & Related Calculations
- Topic: work, heat, and internal energy
- Replies: 2
- Views: 248
Re: work, heat, and internal energy
To add on to this, internal energy is the sum of heat and work, as you stated. However, each of these components does not have to be present to calculate internal energy. For example, if no work is being done by the system, then the internal energy is just set equal to heat (U=q). So, no matter what...
- Fri Jan 12, 2018 10:53 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Intensive vs extensive property
- Replies: 3
- Views: 295
Re: Intensive vs extensive property
An intensive property is one that is independent of the size of the sample. Examples include density or temperature. An extensive property, on the other hand, depends on the size of the sample. Examples of this are volume or enthalpy. Intensive properties might be helpful because they stay constant ...
- Fri Jan 12, 2018 8:50 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Enthalpy and internal energy.
- Replies: 1
- Views: 287
Re: Enthalpy and internal energy.
Enthalpy H is related to internal energy of a system. When a system and its surroundings do no work, the heat given off or absorbed by the system is equal to the change in the internal energy of the system. Also, in general, H = U + P*V, where U is internal energy; P, pressure; and V, volume of the ...
- Wed Jan 10, 2018 9:58 pm
- Forum: Phase Changes & Related Calculations
- Topic: Phase Changes
- Replies: 2
- Views: 309
Re: Phase Changes
The temperature part refers to the fact that the enthalpy of sublimation of a substance can only equal the sum of the enthalpies of fusion and vaporization when these two values are measured at the same temperature. Temperature is important to take into account because substances tend to have differ...
- Wed Jan 10, 2018 9:36 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Most Stable Form for an Element
- Replies: 4
- Views: 629
Re: Most Stable Form for an Element
I believe the most stable form of an element is, similar to what is stated above, the purest state of the element. The book also states that graphite is the most stable form of carbon at normal temperatures. This relates to the idea that the enthalpy of formation of graphite is zero, while the entha...
- Tue Jan 09, 2018 10:38 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Figure 8.26
- Replies: 1
- Views: 178
Figure 8.26
In the description that accompanies the heating curve in Figure 8.26, it states that "the heating curve for the liquid is not as steep as for the solid because the liquid has the higher heat capacity." Is this the general trend (that liquids tend to have higher heat capacities than solids ...
- Tue Jan 09, 2018 8:58 pm
- Forum: Phase Changes & Related Calculations
- Topic: Endo and Exothermic Ways to Remember
- Replies: 28
- Views: 11427
Re: Endo and Exothermic Ways to Remember
This also still sometimes confuses me, and it has since high school. In particular, I had issues with knowing what side of the reaction heat was on. I remember it now as endothermic meaning heat entering (heat is required) and exothermic as heat exiting (heat is a product).