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| || | + |M edugen.wileyplus.com/edugen/student/mainfr.uni-WileyPLUS Reader WileyPLUS Chemistry question | Chegg.com Google iley Jespersen, The Molecular Nature of Matter, 7e GENERAL CHEMISTRY-SCIENCE MAJORS (CHE 1101/1102) Home Read, Study& Practic Assignment Gradebook ORION Assignment> Open Assignment FULL SCREEN PRINTER VERSION BACK NEXT At 460 °C, the reaction SO2(g) NO2(g) has Kc-85.0. Suppose 0.109 mol of SO2, 0.0612 mol of NO2, 0.0839 mol of NO, and 0.125 mol S03 are placed in a 10.0 L container at that temperature, what will the concentrations of all the gases be when the system reaches equilibrium? NO(g) SO3(9) V Review Problem V Review Problem V Review Problem V Review Problem [SO2] = V Review Problem the tolerance is +/-2% V Review Problem V Review Problem V Review Problem [NO2] = | Review Problem 14.066 the tolerance is +/-2% Review Problem 14.070 Review Problem 14.074 Review Problem 14.078 Review 14.082 [NO] = | the tolerance is +/-2% Review Sco Review Results by Study Objective SO3] = License Agreement Privacy Policy © 2000-2016 John Wiley & Sons, Inc. All Rights Reserved. A Division of John Wiley & Sons, 11 Version 4.17.3.3Explanation / Answer
we knoow that
concentration = moles / volume
so
initially
[S02] = 0.109 / 10 = 0.0109
[N02] = 0.0612 / 10 = 0.00612
[NO] = 0.0839 / 10 = 0.00839
[S03] = 0.125 / 10 = 0.0125
now
S02 + N02 ---> NO + SO3
using ICE table
at equillibrium
[S02] = 0.0109 - x
[N02] = 0.00612 - x
[NO] = 0.00839 - x
[S03] = 0.0125 - x
Kc = [NO] [SO3] / [NO2] [SO2]
so
85 = [0.00839 - x] [0.0125-x] / [0.00612 -x] [0.0109-x]
solving
we get
x = 0.006084
so
at equilibrium
[S02] = 0.0109 - 0.006084 = 0.004816 M
[N02] = 0.00612 - 0.006084 = 0.000036 M
[NO] = 0.00839 - 0.006084 = 0.002306 M
[S03] = 0.0125 - 0.006084 = 0.006416 M
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