{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

winter midterm 2005

# winter midterm 2005 - CHEM 14B YOUR NAME Al/SWﬁ(J‘...

This preview shows pages 1–6. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHEM 14B YOUR NAME Al/SWﬁ/(J‘ ..... Instructor: Dr. Laurence Lavelle WINTER 2005 MIDTERM (Total number of pages = 8) (Total points = 50) (Total time = 50 mins) YOUR DISCUSSION SECTION ...................................... YOUR TA ................................. Write in pen. Show all your work. Check your units and significant figures. Think Clearly. Good Luck. Q1. A sample of gas in a cylinder of volume 3.42 L at 298 K and 2.57 atm expands to 7.39 L by two different pathways. (lOpt) Path A is an isothermal, reversible expansion. Calculate the energy, work and heat for path A. Because the process is isothermal, E = 0 {/ﬂ) E = q + w therefore q = —w. V For an isothermal, reversible expansion w = —nRT 1n 72 0%) 1 . n is obtained from the ideal gas law: n 2 ﬂ = (2.57 atm)t13.42 L1 = 0359 m0] 071') RT (0.082 06 L - atm - K -mol )(298 K) w = —(0.359 mol)(8.314 J ~ K"1 -mol’1)(298 K) lnﬂ = —685 J QM/IL) q=+685J [/7117 3'42 Path B involves two steps. In the first, the gas is cooled at constant volume to 1.19 atm. In the second step, the gas is heated and allowed to expand against a constant external pressure of 1.19 atm until the final volume is 7.39 L. Calculate the work for path B. A/ﬂ/I‘) For step 1, because the volume is constant, w = 0 and E = q. In step 2, there is an irreversible expansion against a constant opposing pressure, which is calculated from w = —I:XAV Q W The constant opposing pressure is given, and AV can be obtained from V ﬁnal ~V initial 2 7.39 L — 3.42 L w = —(l.l9 atm)(7.39 L — 3.42 L) 2—. ‘atm= —. -atm M =— / (217742;) ( 472L )[1L-atm] 4791 (7419/ The total work for partBis 0J+ (—479 J)=—479J 6W (WM) . . Gut/”1M Q2. The autoprotolysis constant of water, Kw, is 9.9197 x 10"5 at 25°C and 1.4689 x 10'14 at 30°C. Assuming AHr° for this reaction to be independent of temperature, calculate ASrO for the autoprotolysis reaction. Suggest an interpretation of the sign. (lOpt) AG° = —RT1n K = AH°—TAS° 67%? mm = —<8-31451K"mol'1)(298.15K)1n(9.9197 x 10-15) 67"?) : 7.9933><10“Jmor1 (14¢) “”303 = “(8-31451K"mol")(303.15K)1n(1.4689xio-‘4>QM} = 8.0283x1041mor‘ (Wt) AG°298 —AG°303 = AH°—(298.15K)AS°—[AH°—(303.15K)AS°] mt) —3.5097><102Jmol'1 = (5.00K)AS° (Wt) AS° = —70.194JK'lmol“ Q 1"?) The sign is negative indicating that the product ions create more order in the system probably because of their interaction with solvating water molecules. (4,74%) d/MM) ”@zﬁr W W MW W. Q3A. A piece of metal of mass 20.0 g at 100.0°C is placed in a calorimeter containing 50.7 g of water at 220°C. The final temperature of the mixture is 257°C. What is the specific heat capacity of the metal? Assume that there is no energy lost to the surroundings. (5pt) M W 7‘ M W I: 0 heat lost by metal = ~heat gained by water (W (20.0 g)(25.7°C —100.0°C)(CS) @7127 = —(50.7 g)(4.18 J - (°C)" .g" )(25.7°C — 220°C) C5 = 0.53 J - (Ocy1 -g‘1 g If) B. In a working electrochemical cell (+ cell voltage), the cations in the salt bridge move toward the cathode. True or False (Circle your answer.) (2pt) true C. When equilibrium is reached in an electrochemical cell, the voltage reaches its maximum value. True or False (Circle your answer.) (2pt) false D. The standard potential of the Cu2+/Cu electrode is +0.34 V and the standard potential of the cell IH/ 095 C 4 77/406 Ag<s> l Agcxs) lc1-<aq> || Cu2+(aq) l Cu<s> is +0.12 V. What is the standard potential of the AgCl/Ag,Cl— electrode? (5pt) 5:“ ’2 Eérnwog “gram; (OJ) 0./7,1/ : 0.?gcl/ “5,240.96 0W Q4A. For a second-order reaction, a straight line is obtained from a plot of B. C. (2pt) l/ [A] vs t The rate law for the following mechanism is . (2pt) N02(g) + F2(g) —> N02F(g) + F(g) k1, 810W F(g) + N02(g) ~—> N02F(g) k2, fast rate = k1[N02][F2] The decomposition of hydrogen peroxide, 2 H202(aq) —> 2 H20(l) + 02(g), follows first—order kinetics with respect to H202 and has k = 0.0410 min"1 for the rate law for the decomposition of H202. (a) If the initial concentration of H202 is 0.35 mol-L", what is its concentration after 10 min ? (3pt) [A],=[A]oe‘k’ Q71? 5V ﬁg] 2 “M fgﬂ], [H202], = [Hzozioe‘k’ [H202], = 0.35 molL‘1 x 6(‘0‘0410mn4xl0mim (71/17 = 0.23 moI-L‘1 (Wt) (b) How much time (in minutes) will it take for the H202 concentration to decrease from 0.50 mol-L‘] to 0.10 mol-L‘l ? (3pt) In [A] = —k t + 1n [A10 Inllillekt CW 1n[9—59]=0.0410min‘1xr (Mt/t) 0.10 r = ———1‘3—5— = 39 min Z 0.0410 mjn'1 (c) How much time (in minutes) is needed for the H202 concentration to decrease by one—fourth ? (3pt) . 1 3 . A reduction of —, means — remams. 4 4 “13122010 1 4 [H.201 “ E t=—~——-— =70min 2.00410min1 (M (M 01/ ((1) Calculate the time needed for the H202 concentration to decrease by 75%. (3pt) A reduction of 75%, means 25% or 1 remains. 4 rim] “IF-l [H202] 1 1.39 k k 0.0410 min 1 (W M W - —34 min ...
View Full Document

{[ snackBarMessage ]}