Electric force magnitude of NH3

In summary, the conversation discusses the structure of NH3 and the calculation of the magnitude of the electric force acting on each ion. The distance between the ions is determined to be "a" and the calculations are simplified using the variable C. The force on H1 is found to be <(3^(3/2)-1-.5, 1.5-sqrt(3)/2, 3(.8164965809)> with both the x and y components being zero. However, there is an error in the calculation and it is unclear where it occurred.
  • #1
EV33
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Homework Statement



The structure of NH3 the molecule is approximately that of an equilateral tetrahedron, with three H+ ions forming the base and an N3− ion at the apex of the tetrahedron. The length of each side is 1.64 10-10 m. Calculate the magnitude of the electric force that acts on each ion. (Let the H+ ions be in the x-y plane with H1 at (0, 0, 0), H2 at (a, 0, 0), and H3 (a/2,(a*sqrt(3))/2,0), and the N is at (a/2,a/(sqrt(3)*2),a*sqrt(2/3))

where a = 1.64 10-10 m. To simplify our calculations we'll set (ke^2)/a^2 = C = 8.56 10^-9 N. Use the following variable as necessary : C.)

Find the magnitude of force on H1

Homework Equations


(K* lq1*q2l)/r^2=F(E)

The Attempt at a Solution



First of all after sketching a little diagram I concluded that the force of h2 on h1 was < - ,0,0>, h3 on h1 <-,-,0), and N on h1 <+,+,+>. Made these conclusions by looking at the direction the force would be. In the end I used these signs to give my magnitudes direction in order to add the vectors.

I found the distance between them all from h1 by using the distance formula. They have the same distance which is "a".

Then I set up triangles for them all except the first one because it is pushed straight down the x axis.

So for h2 on h1 I got (Ke^(2))/a^(2)<-1,0,0>

For h3 on h1 I got a triangle with a side in the x direction with a length of a/2 and the side in the y was (a*sqrt(3))/2. So I used tan(o/a)=30 degrees.

So I took the sin and cos of 30, and figured I needed to multiply my force by 1/2 to get the x comp, and root3 over 2 to get my y comp.
(Ke^(2))/a^(2)<-.5,-sqrt(3)/2,0>


Then for my N on h1 I did the same process and got

(3ke^2)/a^2<sqrt(3)/2,.5,.8164965809>

Then I added all my vectors together.

<(3^(3/2)-1-.5, 1.5-sqrt(3)/2, 3(.8164965809)>


The x and y component are apparently supposed to both be zero but I don't get them cancelling out.

Does anyone see where my error occured? Or at least can anyone tell me if I am taking the right steps.
 
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  • #2
I cleaned up my post a little bit. so hopefully it is understandable now.
 
  • #3




It looks like you are on the right track with your calculations. However, it's important to keep in mind that the magnitude of the electric force is a scalar quantity and does not have a direction. The direction of the force is represented by the vector you obtained, but the magnitude is simply a numerical value.

Also, when adding vectors, it's important to consider the direction and magnitude of each vector. In this case, the x and y components of the force from H2 and H3 on H1 should cancel out, leaving only the force from N on H1. So the final magnitude of the electric force on H1 would be:

(3ke^2)/a^2 <0, 0, 3(sqrt(3)/2)>

I hope this helps clarify your calculations and understanding of electric forces in molecules. Keep up the good work!
 

1. What is the electric force magnitude of NH3?

The electric force magnitude of NH3, or ammonia, depends on the amount of charge present on the molecule and the distance between it and other charged particles. It can be calculated using Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

2. How does the electric force magnitude of NH3 affect its chemical properties?

The electric force magnitude of NH3 plays a crucial role in its chemical properties. It is a polar molecule, meaning it has a slightly positive and a slightly negative end, due to the unequal distribution of electrons. This allows it to participate in hydrogen bonding, which influences its boiling point, solubility, and other physical and chemical properties.

3. Can the electric force magnitude of NH3 be manipulated?

Yes, the electric force magnitude of NH3 can be manipulated by changing the amount of charge on the molecule or the distance between it and other charged particles. This can be achieved through chemical reactions or external forces, such as an electric field.

4. How is the electric force magnitude of NH3 related to its molecular structure?

The electric force magnitude of NH3 is related to its molecular structure in that the unequal distribution of electrons in the molecule creates a dipole moment, resulting in a net electric force. The geometry of the molecule also plays a role, as the placement of the atoms affects the overall charge distribution and thus the magnitude of the electric force.

5. What are some real-world applications of understanding the electric force magnitude of NH3?

Understanding the electric force magnitude of NH3 is important in various fields, including chemistry, biology, and environmental science. It helps explain the properties and behavior of ammonia in chemical reactions, as well as its role in biological processes, such as nitrogen fixation. It also plays a role in air pollution and acid rain, as ammonia can react with other pollutants in the atmosphere, affecting air quality and ecosystems.

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