Equilibrium Separation of HCl with Covalent Bonding: Expert Assistance Needed

  • Thread starter Thread starter rushil01
  • Start date Start date
AI Thread Summary
The discussion revolves around calculating the equilibrium separation of HCl, given its electric dipole moment of 3.60 x 10^-30 C.m and a covalent bonding percentage of 82.5%. A formula involving the dipole moment and covalent character is mentioned, yielding a separation of approximately 27.3 picometers. Participants seek clarification on the formulas used, particularly the origin of the constant 6.25 x 10^18. The original poster plans to share the complete working out after consulting their lecturer.
rushil01
Messages
2
Reaction score
0
Hey guys, I am currently studying Engieering in Australia and there is one question which was given to us which i have no idea on how to complete and i was wondering if you guys could give me a hand with it;

What is the equilibrium separation of HCl? ITs measured electric dipole moment is 3.60 x 10^-30C.m and the percentage of covalent bonding in HCl is 82.5%

ANy help would be greatly appreciated. Thanx
 
Physics news on Phys.org
rushil01 said:
Hey guys, I am currently studying Engieering in Australia and there is one question which was given to us which i have no idea on how to complete and i was wondering if you guys could give me a hand with it;

What is the equilibrium separation of HCl? ITs measured electric dipole moment is 3.60 x 10^-30C.m and the percentage of covalent bonding in HCl is 82.5%

ANy help would be greatly appreciated. Thanx

r =\ \mu_D/q\cdot P\ =\ 6.25\cdot10^{18}\ \mu_D/P\ =\ <br /> <br /> 6.25\cdot10^{18}\cdot3.6\cdot10^{-30}/0.825\ =\ 2.73\cdot10^{-11}\ m\ =\ 27.3\ pm
 
What are those formulas you came up with and where did 6.25*10^18 come from.
Thanks
 
i got my lecturer to do it and he used the formula p=qd. I'll post the full working out when i have a little more time
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top