Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 5 - Problems: 5.156

Answer

Pfinal = 0.323 atm and PI[sub]2[/sub] = 0.0332 atm

Work Step by Step

I2(s) + 7F2(g) --> 2IF7(g) Initially, nF[sub]2[/sub] = PV/RT = (350 x 1/760)(2.50)/(0.0821)(250) = 0.0561 mol F2(g) nI[sub]2[/sub] = 2.50/126.9*2 = 0.0098 mol I2(s) The F2 gas is the limiting reactant. Therefore, 0.0561 mol F2 x (1 mol I2 / 7 mol F2) = 8.01x10-3 mol I2 consumed in reaction. 0.0561 mol F2 x (2 mol IF7 / 7 mol F2) = 0.0160 mol IF7 produced. nI[sub]2[/sub] remaining = 0.0098 - 8.01x10-3 = 0.0018 mol I2(g) remaining. Pfinal = ntotalRT/V = (nI[sub]2 remaining[/sub] + nIF[sub]7[/sub])RT/V = (0.0117 + 0.0160)(0.0821)(550)/2.50 = 0.323 atm. PI[sub]2[/sub] = nI[sub]2[/sub]RT/V = (0.0018)(0.0821)(550)/2.50 = 0.0332 atm.
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