This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 215 – Exam 1 – sutcli ff e – (51630) 1 This printout should have 28 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Notes: Assume ideal behaviour. The two vapor pressures given are for the pure liquids. Calculate the vapor pressure at 25 ◦ C of a mixture of benzene and toluene in which the mole fraction of benzene is 0.650. The vapor pressure at 25 ◦ C of benzene is 94.6 torr and that of toluene is 29.1 torr. 1. 71.7 torr correct 2. 51.3 torr 3. 61.5 torr 4. 84.4 torr 5. 124 torr Explanation: 002 10.0 points If a reaction is endothermic, and the temper ature is increased, the equilibrium constant will 1. increase. correct 2. increase if Δ G is negative. 3. remain the same. 4. decrease. 5. increase if Δ G is positive. Explanation: Endothermic reactions absorb heat, favor ing formation of products and so K , which reflects the ratio of reactants to products, in creases. 003 10.0 points Consider the reaction: C graphite ( s ) + O 2 ( g ) ↔ CO 2 ( g ) Δ G ◦ = 400 kJ · mol 1 · K 1 Which of the following is a possible value of K for this reaction? 1.0.56 2. 10 70 3. 10 70 correct 4. 0.56 Explanation: Δ G ◦ = 400 kJ · mol 1 · K 1 is very large and negative, requiring that K be a very large number. 004 10.0 points The specific heat for liquid argon Ar( ) is 25 . 0 J mol · ◦ C and gaseous Ar(g) is 20 . 8 J mol · ◦ C . The enthalpy of evaporation is 6506 J/mol. How much energy is required to convert 1 mole of liquid Ar from 5 ◦ C below its boiling point to 1 mole of gaseous Ar 5 ◦ C above its boiling point? 1. 6290 J 2. 6555 J 3. 6495 J 4. 6735 J correct 5. 6960 J Explanation: SH Ar( ) = 25 . 0 J mol · ◦ C SH Ar(g) = 20 . 8 J mol · ◦ C Δ H = 6506 J/mol Δ T = 5 ◦ C In this example, argon liquid is warmed 5 degrees, then undergoes a phase change from liquid to gas, and the resulting gas is warmed 5 degrees. For each one of the steps in this process, an equation must be written, and the total amount of energy required is the sum of the energies from the three individual steps. For the process of heating liquid argon five degrees, we use the equation Version 215 – Exam 1 – sutcli ff e – (51630) 2 q = (SH) (m) Δ T , where SH = specific heat, m = mass or moles, and Δ T = temperature change. Substituting the appropriate values for liq uid argon, q = (1 mol) 25 . 0 J mol · ◦ C (5 ◦ C) = 125 J For the phase change, a di ff erent equation is used: q = Δ H vap or fus (m) where, again, m = mass or moles . Using the values we have in this example, q = (6506 J / mol)(1 mol) = 6506 J and finally, for the process of heating gaseous argon five degrees, we use the same equation as before, but with the specific heat appropri ate to gaseous argon: q = (SH) (m) Δ T = (1 mol) 20 . 8 J mol · ◦ C (5 ◦ C) = 104 J The sum of these three steps is the total amount of energy required: q = 6735 J 005 10.0 points The Tyndall e...
View
Full
Document
This note was uploaded on 10/06/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.
 Spring '07
 Holcombe

Click to edit the document details