Asparagine
IUAPC name: 2-amino-3-carbamolypropanoic acid
C4H8N2O3
Molecular Weight: 132.12 g
Physical Properties
- Melting Point: 235 °C
- Boiling Point: 438.029 °C
- Flashpoint: 218.712 °C
- Density: 1.405 g/cm3
- Solubility: soluble in acid and alkali solution; slightly soluble in water; hardly soluble in ethanol, ethyl ether, methanol, and benzene
Intermolecular Forces
Dispersion forces are present in all atoms or molecules that have electrons, so asparagine exhibits dispersion forces. The compound will have higher dispersion forces, because dispersion forces increase with increasing molar mass. Greater dispersion forces result in increased boiling points, this is also reflected in asparagine as it has a boiling point of 438.029 °C. In addition, dipole-dipole forces exist in asparagine because it is a polar molecule. Hydrogen bonding also takes place. For example, the NH2 atom has one unshared pair of electrons and can form one hydrogen bond, the oxygen atoms of C=O groups have two unshared pairs of electrons and can form two hydrogen bonds with hydrogen atoms.
Brittany Lask
ReplyDeleteVictoria Decker
Question for Blog 3
This question relates to calculating E̊cell along with choosing one of the following formulas to relate the E̊cell and the K value.
Calculate the E̊cell for each of the following balanced redox reactions and determine whether each one is spontaneous. Choose one of the three reactions to calculate the K value using the following formula, E̊cell =(0.0592 V)/n log〖(k)〗(Located on page 836). Use the following webpage to calculate the values dealing with the E ̊cell and the half reductions; https://sites.google.com/site/chempendix/potentials.
Find E̊cell for each of the following reactions, along with determining which is spontaneous.
A) O2(g) + 2 H2O(l) + 4Ag(s) yields 4OH-(aq)+ 4Ag+(aq)
B) Cu(s) + 2H+(aq) yields Cu2+(aq)+ H2(g)
C)Br2(l) + 2I-(aq) yields 2Br –(aq) + I2 (s)
In all the above formulas, the number before the element is the coefficient and the number within the element is to be noted as subscript.
Use the provided equation again located on page 836 to determine K in relation with the information that was found using the E̊cell from one of the above reactions.
The question asked is of significance in relation to the topic in class we have been discussing. If you know the standard electrode potential you can then determine if the cell is spontaneous or nonspontaneous. I feel like the question wasn't exactly at the level of difficulty that you asked, but it was still an appropriate question to ask.
ReplyDeleteA) Anode: Ag(s)--> Ag(aq) + 2e- E= 0.80 V
Cathode: O2(g)+2H2O(l)+4e- --> 2OH-(aq) E= 0.40
Ecell= Ecathode- Eanode
= 0.40-0.80
=-0.40 V overall it's negative so it's nonspontaneous
B) Anode: Cu(s) --> Cu^2+(aq)+2e- E= 0.34
Cathode: 2H+ + 2e- --> H2(g) E= 0.00
Ecell= 0.00-0.34
= -0.34 V overall it's negative so it's nonspontaneous
C) Anode: 2I-(aq) --> I2(s)+ 2e- E=1.07
Cathode: Br2(l)+2e- --> 2Br- E=0.54
Ecell=1.07-0.54
=0.53 V overall it's positive so it's sponatneous
Determine K: for equation C
Ecell= 0.0592/n *log(k)
0.53 = 0.0592/2 *log(k)
17.91=log(k)
10^17.91=k
8.13*10^17=k