What Is the Net Ionic Equation for HBr(aq) and NH₃(aq)?

  • Thread starter Thread starter Barfolumu
  • Start date Start date
  • Tags Tags
    Ionic Net
AI Thread Summary
The net ionic equation for the reaction between HBr(aq) and NH₃(aq) is H⁺(aq) + NH₃(aq) → NH₄⁺(aq). The initial attempt included incorrect species and did not properly represent the reaction. It was noted that NH₃ does not equate to NH₄⁺ and OH⁻. The correct approach focuses on the proton transfer from HBr to NH₃. This clarification helps ensure accurate representation of the chemical reaction.
Barfolumu
Messages
68
Reaction score
0

Homework Statement



"Write the net ionic equation for the reaction between HBr(aq) and NH_3(aq)"


Homework Equations




The Attempt at a Solution



H^+(aq) + Br^+(aq) + NH_4^+(aq) + OH^-(aq) --> NH_4Br(aq) + H_2O(l)

[edit]
It's an online application that is a bit finicky, and I was curious if I'm at least heading the right direction, and if I'm not, then where exactly I'm messing up? I also posted this in the Chemistry forum, but then found this one. Thanks in advance.
 
Last edited:
Physics news on Phys.org
Barfolumu said:

Homework Statement



"Write the net ionic equation for the reaction between HBr(aq) and NH_3(aq)"


Homework Equations




The Attempt at a Solution



H^+(aq) + Br^+(aq) + NH_4^+(aq) + OH^-(aq) --> NH_4Br(aq) + H_2O(l)


[edit]
It's an online application that is a bit finicky, and I was curious if I'm at least heading the right direction, and if I'm not, then where exactly I'm messing up? I also posted this in the Chemistry forum, but then found this one. Thanks in advance.

Check the bolded items.
 
Thank you very much. The Barium should've remained an ion, and NH_3 is not equivelent to NH_4 + OH, so the net ionic equation is H^+(aq) + NH_3(aq) --> NH_4^+(aq).
 
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