Pv nrt p 0328 mol 00821 l atm mol1 k 1 378 k 0969

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Unformatted text preview: e container? PV = nRT P= 0.328 mol × 0.0821 L atm mol−1 K −1 × 378 K = 0.969 atm 10.5 L One point is earned for the correct substitution. One point is earned for the correct pressure. At 105°C, AsF5(g) decomposes into AsF3(g) and F2(g) according to the following chemical equation. → AsF5(g) ← AsF3(g) + F2(g) (b) In terms of molar concentrations, write the equilibrium-constant expression for the decomposition of AsF5(g). [AsF3 ] [F2 ] [AsF5 ] K= One point is earned for the correct equation. (c) When equilibrium is established, 27.7 percent of the original number of moles of AsF5(g) has decomposed. (i) Calculate the molar concentration of AsF5(g) at equilibrium. 100.0% − 27.7% = 72.3% [AsF5] = 0.723 × 0.0313 M = 0.0226 M One point is earned for the correct concentration. © 2008 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® CHEMISTRY 2008 SCORING GUIDELINES (Form B) Question 1 (continued) (ii) Using molar concentrations, calculate the value of the equili...
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This note was uploaded on 05/05/2011 for the course CHEM 504 taught by Professor John during the Fall '11 term at American College of Computer & Information Sciences.

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