2] = 1.00 M, [H2] = 1.00 M, [NH3] = 4.00 M 5a.) What is Qc at this point? Which way will the reaction go? 5b.) if x is the concentration of N2 that reacts, fill in the rest of the ice chart at left. Now, suppose that the N2 concentration is increased to 300 M. The ICE setup would look like this: N2(g) + 3 H2(g)  2 NH3 (g) I 300 1 4 C E 5c) It is too hard to solve this problem for x! So I am going to give you the answer. The value of x turns out to be 0.279. Notice that x cannot ever be greater than 0.33 (or 1/3) – why is that? [N2] = 1.00 M, [H2] = 1.00 M, [NH3] = 4.00 M 5a.) What is Qc at this point? Which way will the reaction go? 5b.) if x is the concentration of N2 that reacts, fill in the rest of the ice chart at left. Now, suppose that the N2 concentration is increased to 300 M. The ICE setup would look like this: N2(g) + 3 H2(g)  2 NH3 (g) I 300 1 4 C E 5c) It is too hard to solve this problem for x! So I am going to give you the answer. The value of x turns out to be 0.279. Notice that x cannot e

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter12: Chemical Equilibrium

Section: Chapter Questions

Problem 117IP: In a solution with carbon tetrachloride as the solvent, the compound VCl4. undergoes dimerization:...

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[N₂] = 1.00 M, [H₂] = 1.00 M, [NH₃] = 4.00 M

5a.) What is Qc at this point? Which way will the reaction go?

5b.) if x is the concentration of N2 that reacts, fill in the rest of the ice chart at left.

Now, suppose that the N₂ concentration is increased to 300 M. The ICE setup would look like this:

N₂(g) + 3 H₂(g)  2 NH₃ (g)

I 300 1 4

5c) It is too hard to solve this problem for x! So I am going to give you the answer. The value of x turns out to be 0.279. Notice that x cannot ever be greater than 0.33 (or 1/3) – why is that?

[N₂] = 1.00 M, [H₂] = 1.00 M, [NH₃] = 4.00 M

5a.) What is Qc at this point? Which way will the reaction go?

5b.) if x is the concentration of N2 that reacts, fill in the rest of the ice chart at left.

Now, suppose that the N₂ concentration is increased to 300 M. The ICE setup would look like this:

N₂(g) + 3 H₂(g)  2 NH₃ (g)

I 300 1 4

5c) It is too hard to solve this problem for x! So I am going to give you the answer. The value of x turns out to be 0.279. Notice that x cannot ever be greater than 0.33 (or 1/3) – why is that?

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