"This is a favored process, but with a high energy barrier."
Search found 21 matches
- Sat Mar 18, 2017 9:29 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3725311
- Sun Mar 12, 2017 1:52 pm
- Forum: *Alcohols
- Topic: Determining phenol functional groups
- Replies: 3
- Views: 1439
Re: Determining phenol functional groups
I also don't think so. Alcohol functional group is a hydroxyl group that attaches to a saturated carbon. However, because the OH of phenol functional group is attached to a benzene ring, the hydroxyl group of phenol attaches to an unsaturated carbon. Therefore the phenol functional group is differen...
- Sun Mar 05, 2017 11:58 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Ratio of Rate Constants (Energy Barriers)
- Replies: 2
- Views: 573
Re: Ratio of Rate Constants (Energy Barriers)
This can be explained with the Arrhenius equation: k=A*Exp(-EA/(R*T)), which is determined through experiments. A higher energy level means reactants must collide with higher energy, which is harder to achieve. Therefore the rate is lower and the rate constant is smaller.
- Sun Feb 26, 2017 10:08 pm
- Forum: General Science Questions
- Topic: Gases in Atmosphere
- Replies: 1
- Views: 529
Re: Gases in Atmosphere
Because ozone is constantly produced in the upper atmosphere, where there is intense ultraviolet light. UV light has shortwave length and high frequency(higher energy), and can break O2 into two oxygen atoms. The single oxygen atom combines with O2 to form O3. So even though ozone tends to sink or b...
- Sun Feb 19, 2017 10:02 am
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3725311
- Sun Feb 12, 2017 5:21 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Practice Midterm Winter 2013: Q5B
- Replies: 3
- Views: 796
Re: Practice Midterm Winter 2013: Q5B
I think here "configuration" infers electron configuration rather than geometry orientation. Clearly F has a more complex electron configuration than H. A more complex electron configuration means more possible microstates. Therefore the more F the molecule has, the larger the molar entropy.
- Sun Jan 29, 2017 10:25 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Quiz question #7
- Replies: 1
- Views: 499
Re: Quiz question #7
Actually this process leads to an decrease in entropy. For an isothermal process, ΔG=q(reversible)/T
Clearly, since the system (the glass of water) is losing heat, q is negative Thus ΔG is also negative, which means decrease in entropy.
Clearly, since the system (the glass of water) is losing heat, q is negative Thus ΔG is also negative, which means decrease in entropy.
- Sun Jan 22, 2017 3:55 pm
- Forum: Calculating Work of Expansion
- Topic: Example of reversible expansion?
- Replies: 1
- Views: 556
Example of reversible expansion?
For irreversible expansions, an example would be a piston with atmosphere as external environment. But can some one give me a real-life situation as an example of a reversible expansion, if there's any? I understand in reversible expansion process, the pressures of the system and exterior are always...
- Sat Jan 14, 2017 9:38 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Using Reaction Enthalpies
- Replies: 1
- Views: 398
Re: Using Reaction Enthalpies
You can decide which method to use by looking at what information you are given in the question. When you know the reaction enthalpies of reactions related to your given reactions, you use Hess's law. For example, if you are given N2+ O2= 2NO ΔH1°=180.5 kJ and N2+2O2=2NO2 ΔH2°=66.4 kJ, and want to ...
- Sat Dec 03, 2016 10:16 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Why aqua is monodentate?
- Replies: 2
- Views: 769
Re: Why aqua is monodentate?
Thanks so much!
- Fri Dec 02, 2016 10:02 am
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Why aqua is monodentate?
- Replies: 2
- Views: 769
Why aqua is monodentate?
According to its Lewis structure, aqua(H2O) has two lone pairs. This means that aqua can possibly bind to two molecules, right? But why it is monodentate?
- Tue Nov 29, 2016 6:02 pm
- Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
- Topic: Calculating Bond Order
- Replies: 2
- Views: 3464
Re: Calculating Bond Order
Basically you can calculate the bond order by counting the bonds, with the help of the lewis structure. For example, O2 has a double bond and has a bond order of 2. If we draw the lewis structure for NO3-, we may find out that it's a resonance structure with one double bond and two single bonds. So ...
- Sun Nov 27, 2016 5:39 pm
- Forum: Calculating the pH of Salt Solutions
- Topic: classifying acidity/basicity of NaClO
- Replies: 1
- Views: 1090
Re: classifying acidity/basicity of NaClO
Na is not an acidic cation not because of its charge, but because NaOH is a strong acid and ionizes completely in water. So yes, ClO is the only ion reacting with water and the determining the pH. And NaClO is basic and your reason is correct For AlClO3, I think we need the specific ka and kb of Al ...
- Fri Nov 18, 2016 7:50 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Chapter 11 Q 89
- Replies: 4
- Views: 906
Re: Chapter 11 Q 89
So according to the graph, Change in A=ΔA=27.5-27.5=-10 (approximately), ΔB=5, ΔC=10 ΔA: ΔB: ΔC=2:1:2, which means for every two unit's of decrease in A's PP, there is one unit of increase in C's PP and two in D's. So the coefficient of A, B, and C is 2: 1: 2 And the equation will be: 2A --> B + 2C ...
- Fri Nov 11, 2016 12:42 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Chapter 11 Question 43
- Replies: 2
- Views: 662
Re: Chapter 11 Question 43
As initially the concentration for N2 is 0 and equilibrium concentration is x, the change in N2 concentration is x. The change in NO concentration is -2x, twice the change in N2 concentration. Given NO' s initial concentration (1.0) and change in concentration (-2x), we can know that the equilibrium...
- Fri Nov 04, 2016 7:03 pm
- Forum: Hybridization
- Topic: Homework 4.43
- Replies: 1
- Views: 524
Re: Homework 4.43
I don't know the exact reason, but we can make generalization by comparing sp3 and sp2 orbitals. The sp2 orbitals has less p-character and more s-character than sp3 (you may think of sp2 has 1/3 s, while sp3 has 1/4 s), and has lager bond angles. Generally, the more s-character the hybrid orbital ha...
- Sat Oct 29, 2016 10:01 am
- Forum: Hybridization
- Topic: 25.d
- Replies: 2
- Views: 666
Re: 25.d
After sulfur forms bonds with each of the fluorine, it still has a lone pair left. The side with the lone pair has a slightly negative charge, thee other side has a slightly positive charge. So SF4 is polar.
- Thu Oct 20, 2016 5:41 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3725311
- Sat Oct 15, 2016 3:38 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 3d and 4s orbitals
- Replies: 2
- Views: 1113
3d and 4s orbitals
I learnt in class that 3d orbital has lower energy level than the 4s orbital. Then why electrons first fill in 4s orbitals before they go to 3d?
- Sat Oct 08, 2016 10:50 pm
- Forum: Properties of Electrons
- Topic: Kinetic Energy of Electrons to moles? [ENDORSED]
- Replies: 1
- Views: 750
Re: Kinetic Energy of Electrons to moles? [ENDORSED]
To know the kinetic energy of the electrons, you need to know both their mass and velocity. According to the formula: E(k)=(1/2)m*v^2 The mass of one mole of electrons is 9.1*10^-31kg * 6.02*10^23/mol(Avogadro's number) * 1mol. And the unit of v^2 is (m/s)^2, the unit of the final answer is: kg*m^2/...
- Sun Oct 02, 2016 8:25 pm
- Forum: Properties of Light
- Topic: 1.57 : Wavelength of next line in series ?
- Replies: 1
- Views: 687
Re: 1.57 : Wavelength of next line in series ?
397.0nm The next wavelength is the fifth in balmer series and is for light emitted when electron transits from n=7 to n=2 by definition. First calculate the energy difference between n=7 and n=2 using the equation: ΔE=E2-E7=-hR/4-(-hr/49)=-5.008*10^(-19)J ΔE is negative because energy is emitted as ...