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

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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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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