Predict the equilibrium concentration of H₂O in the reaction described below (for which Kc = 1.79 at the reaction temperature) by constructing an ICE table, writing the equilibrium constant expression, and solving for the equilibrium concentration. Complete Parts 1-3 before submitting your answer. 2 H₂S(g) + SO₂(g) = 3 S(s) + 2 H₂O(g) Initial (M) Change (M) Equilibrium (M) -3.x 0.025 + 2x The reaction mixture initially contains 0.050 M H₂S and 0.050 M SO₂. Fill in the ICE table with the appropriate value for each involved species to determine the concentrations of all reactants and products. 0 1 2 H₂S(g) + 0.050 + x 0.025 + 3x 0.050 0.050 + 2x 0.025 - x +X 2 SO₂(g) 0.050 + 3x 0.025 - 2x +2x 0.050- x 0.025 -3x = 3 3 S(s) +3x 0.050 - 2x -X + 0.050-3x NEXT > 2 H₂O(g) RESET -2x 0.025 + x

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ISBN:9781305957404
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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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before submitting your answer.
[x]
[3.x]³
[0.050- 3x1²
PREV
NEXT
Using the data from your ICE Table (Part 1), construct the expression for the equilibrium constant, Kc. Each
reaction participant must be represented by one tile. Do not combine terms.
0
[x]²
[0.050 -x]
0.043
[0.050-3x]³
1
0.0019
2 H₂S(g) + SO₂(g) = 3 S(s) + 2 H₂O(g)
before submitting your answer.
1
0.0034
Kc
[x]³
[0.050 -x]²
[2x]
0.011
[0.050 -x]³
0.067
2
[H₂O]eq
=
2 H₂S(g) + SO₂(g) = 3 S(s) + 2 H₂O(g)
0.0053
[2x]²
2
[0.050 - 2x]
= 1.79
PREV
ased on the information from your ICE Table (Part 1) and the Kc expression (Part 2), solve for the
quilibrium concentration of H₂O.
0.33
[2x]³
M
[0.050 - 2x]²
3
0.0038
[3.x]
[0.050 - 2x]¹
3
RESET
0.65
[3x]2
[0.050-3x]
RESET
0.022
Transcribed Image Text:before submitting your answer. [x] [3.x]³ [0.050- 3x1² PREV NEXT Using the data from your ICE Table (Part 1), construct the expression for the equilibrium constant, Kc. Each reaction participant must be represented by one tile. Do not combine terms. 0 [x]² [0.050 -x] 0.043 [0.050-3x]³ 1 0.0019 2 H₂S(g) + SO₂(g) = 3 S(s) + 2 H₂O(g) before submitting your answer. 1 0.0034 Kc [x]³ [0.050 -x]² [2x] 0.011 [0.050 -x]³ 0.067 2 [H₂O]eq = 2 H₂S(g) + SO₂(g) = 3 S(s) + 2 H₂O(g) 0.0053 [2x]² 2 [0.050 - 2x] = 1.79 PREV ased on the information from your ICE Table (Part 1) and the Kc expression (Part 2), solve for the quilibrium concentration of H₂O. 0.33 [2x]³ M [0.050 - 2x]² 3 0.0038 [3.x] [0.050 - 2x]¹ 3 RESET 0.65 [3x]2 [0.050-3x] RESET 0.022
Predict the equilibrium concentration of H₂O in the reaction described below (for which Kc
= 1.79 at the reaction temperature) by constructing an ICE table, writing the equilibrium
constant expression, and solving for the equilibrium concentration. Complete Parts 1-3
before submitting your answer.
2 H₂S(g) + SO₂(g) = 3 S(s) + 2 H₂O(g)
Initial (M)
Change (M)
Equilibrium (M)
-3.x
0.025 + 2x
The reaction mixture initially contains 0.050 M H₂S and 0.050 M SO₂. Fill in the ICE table with the
appropriate value for each involved species to determine the concentrations of all reactants and products.
0
1
2 H₂S(g) +
0.050 + x
0.025 + 3x
0.050
0.050 + 2x
0.025 - Xx
+X
2
SO₂(g)
0.050 + 3x
0.025 - 2x
+2x
0.050 - x
0.025-3x
3
3 S(s)
+3x
0.050 - 2x
--X
+
0.050-3x
NEXT >
2 H₂O(g)
RESET
-2x
0.025 + x
Transcribed Image Text:Predict the equilibrium concentration of H₂O in the reaction described below (for which Kc = 1.79 at the reaction temperature) by constructing an ICE table, writing the equilibrium constant expression, and solving for the equilibrium concentration. Complete Parts 1-3 before submitting your answer. 2 H₂S(g) + SO₂(g) = 3 S(s) + 2 H₂O(g) Initial (M) Change (M) Equilibrium (M) -3.x 0.025 + 2x The reaction mixture initially contains 0.050 M H₂S and 0.050 M SO₂. Fill in the ICE table with the appropriate value for each involved species to determine the concentrations of all reactants and products. 0 1 2 H₂S(g) + 0.050 + x 0.025 + 3x 0.050 0.050 + 2x 0.025 - Xx +X 2 SO₂(g) 0.050 + 3x 0.025 - 2x +2x 0.050 - x 0.025-3x 3 3 S(s) +3x 0.050 - 2x --X + 0.050-3x NEXT > 2 H₂O(g) RESET -2x 0.025 + x
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