[x] PREV NEXT Using the data from your ICE Table (Part 1) construct the expression for the equilibrium constant. Each reaction participant must be represented by one tile. Do not combine terms. [0.20 - 2x] [0.40-2x] [x]² your answer. [0.20 + 2x] [0.40 + 2x] Kc = 0 [2x] 2 NOBr(g) = 2 NO(g) + Br₂(g) 1 [0.20-21]² Delore submitting your answei. [0.40-2x]² 0.013 [2x]² [0.20 + 2x]² [0.40 + 2x]² 2 0.17 [0.20 -x] [0.40 -x] 2 NOBr(g) = 2 NO(g) + Br₂(g) 2 = 3.07 x 10-4 0.0054 [0.20 + x] [0.40 + x] PREV Based on the information from your ICE Table (Part 1) and Kc expression (Part 2), solve for concentration of NO at equilibrium. [NO]eq M 3 0.028 [0.20-x]² [0.40-x]² 3 RESET RESET 0.0108 [0.20 + x)² [0.40 + x)²
[x] PREV NEXT Using the data from your ICE Table (Part 1) construct the expression for the equilibrium constant. Each reaction participant must be represented by one tile. Do not combine terms. [0.20 - 2x] [0.40-2x] [x]² your answer. [0.20 + 2x] [0.40 + 2x] Kc = 0 [2x] 2 NOBr(g) = 2 NO(g) + Br₂(g) 1 [0.20-21]² Delore submitting your answei. [0.40-2x]² 0.013 [2x]² [0.20 + 2x]² [0.40 + 2x]² 2 0.17 [0.20 -x] [0.40 -x] 2 NOBr(g) = 2 NO(g) + Br₂(g) 2 = 3.07 x 10-4 0.0054 [0.20 + x] [0.40 + x] PREV Based on the information from your ICE Table (Part 1) and Kc expression (Part 2), solve for concentration of NO at equilibrium. [NO]eq M 3 0.028 [0.20-x]² [0.40-x]² 3 RESET RESET 0.0108 [0.20 + x)² [0.40 + x)²
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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Question
Kc= ?
[NO]eq= ?
![before submitting your answer.
[0.20-2x]
PREV
NEXT
Using the data from your ICE Table (Part 1) construct the expression for the equilibrium constant. Each
reaction participant must be represented by one tile. Do not combine terms.
[0.40-2x]
[x]²
[0.20 + 2x]
[0.40 + 2x]
1
Kc =
0
[2x]
1
2 NOBr(g) = 2 NO(g) + Br₂(g)
Delore submitting your answei.
[0.20-2x]²
[0.40-2x]²
0.013
[2x]²
[0.20 + 2x]²
[0.40 + 2x]²
[NO]eq
2
=
0.17
[0.20-x]
[0.40 -x]
2 NOBr(g) = 2 NO(g) + Br₂(g)
2
0.0054
= 3.07 x 10-4
[0.20+ x]
[0.40 + x]
M
Complete Parts 1-3
PREV
Based on the information from your ICE Table (Part 1) and Kc expression (Part 2), solve for concentration of
NO at equilibrium.
3
0.028
[0.20-x]²
[0.40-x]²
3
RESET
RESET
0.0108
[0.20+ x)²
[0.40 + x)²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb0140d99-b557-46d4-a2e7-675ca4b314d8%2F3e5a471a-5381-4938-bfa2-aa90c9050ba2%2Fktyahum_processed.jpeg&w=3840&q=75)
Transcribed Image Text:before submitting your answer.
[0.20-2x]
PREV
NEXT
Using the data from your ICE Table (Part 1) construct the expression for the equilibrium constant. Each
reaction participant must be represented by one tile. Do not combine terms.
[0.40-2x]
[x]²
[0.20 + 2x]
[0.40 + 2x]
1
Kc =
0
[2x]
1
2 NOBr(g) = 2 NO(g) + Br₂(g)
Delore submitting your answei.
[0.20-2x]²
[0.40-2x]²
0.013
[2x]²
[0.20 + 2x]²
[0.40 + 2x]²
[NO]eq
2
=
0.17
[0.20-x]
[0.40 -x]
2 NOBr(g) = 2 NO(g) + Br₂(g)
2
0.0054
= 3.07 x 10-4
[0.20+ x]
[0.40 + x]
M
Complete Parts 1-3
PREV
Based on the information from your ICE Table (Part 1) and Kc expression (Part 2), solve for concentration of
NO at equilibrium.
3
0.028
[0.20-x]²
[0.40-x]²
3
RESET
RESET
0.0108
[0.20+ x)²
[0.40 + x)²
Transcribed Image Text:An initial gas sample of NOBr decomposes according to the reaction below. The
decomposition of NOBr has a Kc equal to 3.07 x 10-4 at 297 K. Predict the concentration
of NO at equilibrium by constructing an ICE table, writing the equilibrium constant
expression, and solving for the concentration of NO at equilibrium. Complete Parts 1-3
before submitting your answer.
2 NOBr(g) 2 NO(g) + Br₂(g)
Initial (M)
Change (M)
Equilibrium (M)
-X
0.20 + 2x
If an initial solution of 0.20 M NOBr decomposes, fill in the ICE table with the appropriate value for each
involved species.
0
-2x
1
0.40 - 2x
2 NOBr(g)
0.20
0.20
0.20 - 2x
-2x²
-2x
0.40 + 2x
0.40
2
0.20-x
+X
0.20 + x
2 NO(g)
0
+2x
+2x
+.x²
3
0.40 - x
+
+2x
0.40 + x
Br₂(g)
0
+x
NEXT
+x
RESET
+21²
0.20 - 2x
>
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