Consider the equilibrium system described by the chemical reactionbelow, which has a value of Kc equal to 1.79 at a certain temperature. Ifa reaction mixture initially contains 0.050 M H2S and 0.050 M SO2, whatwill the equilibrium concentration of H2O be?2 H2S(g) + SO2(g)=3 S(s) + 2 H20(g)1NEXT >Based on the given values, set up ICE table in order to determine the unknown.2 H2S(g)SO2(g)3 S(s)2 H20(g)++Initial (M)|Change| (M)Equilibrium| (M)5 RESET0.050+x+2x+3x-x-2x-3x0.050 + x0.050 + 2x0.050 + 3x0.050 - x0.050 - 2x0.050 -3x0.025 + x0.025 + 2x0.025 + 3x0.025 - x0.025 - 2x0.025 -3x

Consider the equilibrium system described by the chemical reaction below, which has a value of Kc equal to 1.79 at a certain temperature. If a reaction mixture initially contains 0.050 M H2S and 0.050 M SO2, what will the equilibrium concentration of H2O be? 2 H2S(g) + SO2(g) = 3 S(s) + 2 H;O(g)

Consider the equilibrium system described by the chemical reaction below, which has a value of Kc equal to 1.79 at a certain temperature. If a reaction mixture initially contains 0.050 M H2S and 0.050 M SO2, what will the equilibrium concentration of H2O be? 2 H2S(g) + SO2(g) = 3 S(s) + 2 H;O(g)

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter14: Chemical Equilibrium

Section: Chapter Questions

Problem 89AP

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Question

100%

Consider the equilibrium system described by the chemical reaction
below, which has a value of Kc equal to 1.79 at a certain temperature. If
a reaction mixture initially contains 0.050 M H2S and 0.050 M SO2, what
will the equilibrium concentration of H2O be?
2 H2S(g) + SO2(g)=3 S(s) + 2 H20(g)
1
NEXT >
Based on the given values, set up ICE table in order to determine the unknown.
2 H2S(g)
SO2(g)
3 S(s)
2 H20(g)
+
+
Initial (M)
|Change
| (M)
Equilibrium
| (M)
5 RESET
0.050
+x
+2x
+3x
-x
-2x
-3x
0.050 + x
0.050 + 2x
0.050 + 3x
0.050 - x
0.050 - 2x
0.050 -3x
0.025 + x
0.025 + 2x
0.025 + 3x
0.025 - x
0.025 - 2x
0.025 -3x

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