Search found 17 matches
- Fri Feb 27, 2015 12:33 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: When to use steady-state vs. pre-equilibrium
- Replies: 4
- Views: 1368
Re: When to use steady-state vs. pre-equilibrium
Justin's right, for this class we use the pre-equilibrium approach to find the overall rate law!
- Tue Feb 24, 2015 10:31 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Drawing proposed structures for an activated complex
- Replies: 3
- Views: 3188
Re: Drawing proposed structures for an activated complex
Hi Glenda! This is explained on page 72 of the course reader! But basically for part a) you know that CH3CHO ---> CH3 + CHO so you know that the CH3CHO bond is breaking into CH3 and CHO and you would represent this by [CH3........CHO] to depict where the bond is breaking. and you would do the same t...
- Tue Feb 24, 2015 10:14 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Decrease in Activation Energy
- Replies: 2
- Views: 787
Re: Decrease in Activation Energy
It's not reaction 2 because the energy of the products is higher than the reactants so the reaction is endothermic and non spontaneous, whereas reaction 1 is exothermic and spontaneous so lowering the activation energy would benefit reaction 1 more because it would favor the products.
- Tue Jan 20, 2015 8:51 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Equipartition Theorem
- Replies: 1
- Views: 464
Re: Equipartition Theorem
I would make sure to understand the homework problems because often times they will be repeated on the midterm/final. As long as you understand the concepts in the homework problems well, I believe you should be fine!
- Wed Jan 14, 2015 10:57 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 7.15 clarifications
- Replies: 5
- Views: 1098
Re: 7.15 clarifications
When a solid turns into a liquid you know that heat is absorbed and q is positive and since the change is occurring at a constant temperature you know that ΔU = 0. From here you can use the formula: ΔU = q+w = 0 to solve for w. If w is positive you know that work was done on the system and if w is n...
- Wed Jan 14, 2015 8:23 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Does a monatomic atom have rotational motion?
- Replies: 3
- Views: 2556
Re: Does a monatomic atom have rotational motion?
I'm not entirely positive but I believe you're right because molecules have rotational and translational energy while atoms only have translational energy!
- Wed Jan 14, 2015 8:16 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess's Law, Multiplying Coefficients
- Replies: 2
- Views: 8653
Re: Hess's Law, Multiplying Coefficients
We have to write the reaction like this in the beginning: H2(g) + 1/2 O2(g) ----> H20 (l) ΔHformation: -286 kJ because the standard enthalpy of formation of water for one mol is: -286 kJ so in the reaction you want one mol of water which means 02 has to have 1/2 as the coefficient for it to be balan...
- Wed Jan 14, 2015 12:11 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: HW 7.33
- Replies: 1
- Views: 1078
Re: HW 7.33
If you know the molar heat capacity and the specific heat capacity you can calculate the molar mass of the element and then find it on the periodic table: Molar Heat Capacity: 3R = 3 (8.314 J/k/mol) Specific Heat Capacity: 0.392 J/k/g We know that the molar heat capacity = specific heat capacity * m...
- Tue Jan 13, 2015 11:58 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Cs and Cm
- Replies: 1
- Views: 1536
Re: Cs and Cm
I think it's fine as long as you know to use grams when using specific heat capacity Cs and use moles when using molar heat capacity Cm because the units would cancel out in each case!
- Tue Jan 13, 2015 11:56 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: HW #109b
- Replies: 1
- Views: 831
Re: HW #109b
The solutions manual depicts the reaction as 60C(gr)----->60C(g) bond enthalpy: 60(+717kJ/mol) = +43030 kJ/mol and then forming 60 mol of C-C single bonds and 30 mol of C=C double bonds. We know that when a bond is formed it's an exothermic process and releases energy so you would multiply 60 and 30...
- Tue Jan 13, 2015 11:33 pm
- Forum: Phase Changes & Related Calculations
- Topic: Hw 7.37
- Replies: 2
- Views: 7570
Re: Hw 7.37
There are two parts to this problem since you need to melt the 80.0 grams of ice at 0.00 degrees C and then heat the liquid water to 20 degrees C. So you need to find the heat needed in order to melt the ice and add it to the heat needed to heat the water to 20 degrees C. 1.) In order to find ΔH of ...
- Mon Jan 12, 2015 8:55 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat Capacity of ideal gas
- Replies: 4
- Views: 979
Re: Heat Capacity of ideal gas
Problems 7.27 and 7.29 are directly related to section 7.10!
- Sun Jan 11, 2015 10:58 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Self-test 7.4 A
- Replies: 3
- Views: 1489
Re: Self-test 7.4 A
The formula for change in internal energy is: Δu= q+pΔV This question uses information given in example 7.4. We know from example 7.4 that it's placed in a constant-volume calorimeter so the change in volume=0 which means that pΔV=0 so Δu=q+0. We also know from the given information in 7.4 that the ...
- Sun Jan 11, 2015 9:52 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Understanding reversible, isothermic expansion
- Replies: 2
- Views: 3276
Re: Understanding reversible, isothermic expansion
In the text they say: "a reversible process is one that can be reversed by an infinitely small change in a variable (an “infinitesimal” change). For example, if the external pressure exactly matches the pressure of the gas in the system, then the piston moves in neither direction. If the extern...
- Sat Jan 10, 2015 1:29 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Molar heat capacity of an ideal monatomic gas
- Replies: 1
- Views: 834
Re: Molar heat capacity of an ideal monatomic gas
The book gets 5/2 R from the chart: IMG_4189.jpg Since Kr is an atom and you're trying to find heat released at constant pressure you use the Cp,m for an atom on the chart which is 5/2R. They get 20.8 J*mol^/1 which is Cp,m when they multiply 5/2 by the Gas Constant R! From there they use the equati...
- Sat Jan 10, 2015 1:20 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Contribution to the heat capacity
- Replies: 3
- Views: 3026
Re: Contribution to the heat capacity
7.29 Predict the contribution to the heat capacity CV,m made by molecular motions for each of the following atoms and molecules: (a) HCN; (b) C2H6; (c) Ar; (d) HBr. You have to use the chart in the textbook on page 255: IMG_4189.jpg Since Ar is an atom and you're trying to find the Cv,m which means ...
- Mon Jan 05, 2015 7:34 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Why E=0 for isothermal processes?
- Replies: 6
- Views: 2522
Re: Why E=0 for isothermal processes?
Change in U (internal energy) = 0 for an ideal gas(pg.248). Internal energy is a state function because it only depends on the current state but heat(q) and work(w) are not state functions. I think if the system does work it loses some energy so some heat would flow back in to make up for the lost e...