# ZUMDAHL Marathon Problem: Chemistry Experts Only!!

<p>Hi everyone!! For you geniuses out there... :P I'm having problems with question 95 of Chatper 6 in Zumdahl Fifth Edition. I've been pondering and thinking about it for like two hours now... argh!! </p>

<p>Here's the question:</p>

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<p>A sample consisting of 22.7 g of a nongaseous, unstable compound X is placed inside a metal cylinder with a radius of 8.00 cm, and a piston is carefully placed on the surface of the compound so that, for all practical purposes, the distance between the bottom of the cylinder and the piston is zero. (A hole in the piston allows trapped air to escape as the piston is placed on the compound; then this hole is plugged so that nothing inside the cylinder can escape.) The piston-and-cylinder apparatus is carefully placed in 10.00 kg of water at 20.00C. The barometric pressure is 778 torr.</p>

<p>When the compound spontaneously decomposes, the piston moves up, the temperature of the water reaches a maximum of 29.52C, and then it gradually decreases as the water loses heat to the surrounding air. The distance between the piston and the bottom of the cylinder, at the maximum temperature, is 59.8 cm. Chemical analysis shows that the cylinder contains 0.300 mol carbon dioxide, 0.250 mol liquid water, 0.025 mol oxygen gas, and an undetermined amount of a gaseous element A.</p>

<p>It is know that the enthalpy change for the decomposition of X, according to the reaction described above, is -1893 kJ/mol X. The standard enthalpies of formation for gaseous carbon dioxide and liquid water are -393.5 kJ/mol and -286 kJ/mol, respectively. The heat capacity for water is 4.184 J/C x g. The conversion factor between L x atm and J can be determined from the two values for the gas constant R, namely, 0.08206 L x atm/mol x K and 8.3145 J/mol x K. The vapor pressure of water at 29.5C is 31 torr. Assume that the heat capacity of the piston-and-cylinder apparatus is negligible and that the piston has negligible mass.</p>

<p>Given the preceding information, determine</p>

<p>a. The formula for X.</p>

<p>b. The pressure-volume work (in kJ) for the decomposition of the 22.7g sample of X.</p>

<p>c. The molar change in internal energy for the decomposition of X and the approximate standard enthalpy of formation for X.</p>

<p>Any ideas anyone? This is very involved.</p>

<p>Since it's an odd problem, I recommend you get your handy dandy partial solutions guide and take a look at it. :)</p>

<p>The solution guide only lists answers up to the challenge problems :P</p>

<p>The marathon problem answers are reserved for teachers only :(</p>

<p>Thanks for the reply anyways <3</p>

<p>Darnit. I knew I shoulda gotten the complete solutions guide I found on ebay a few months back...it was the guide and the book for \$25...what a deal.</p>

<p>Anyway...I'm actually studying for a buffer/common ion test tomorrow so...yeah. We finished chap 1-4, 13, 14, and now we're on 15...</p>

<p>I know man! I would kill for a complete solutions guide :P</p>

<p>Our school finished chapters 1-5. We started chapter 7, but we're going to have the chapter 6 and 16 (Thermochemistry 1 and 2) exam tommorow :)</p>

<p>Good luck to you man.</p>

<p>Volume of cylinder @ max temp:</p>

<p>(pi * .00800 m^2 *.0598 m) = 1.20235 x 10^-5 L^3</p>

<hr>

<p>Mass of the gaseous products:</p>

<p>22.7 g X - (.250 mol H2O)(18.0154 g H2O/1 mol H2O) = 18.19615 g gas</p>

<hr>

<p>10000 g H2O * 9.52 C * 4.184 J/(C x g) = 398316.8 J = 398.3168 kJ</p>

<p>398.3168 kJ * (1 mol X / 1893 kJ) = .2104156 mol X</p>

<p>--</p>

<p>22.7 g X - (.300 mol CO2)(44.0098 g CO2 / 1 mol CO2) - ( .250 mol H2O) (18.0154 g H2O/1 mol H2O) - (.025 mol O2)(31.9988 g O2 / 1 mol O2) = </p>

<p>4.19324 g A</p>

<hr>

<p>sum of enathalpy = formation of CO2 + H2O + O2 + A</p>

<p>398.3168 kJ = (.300 mol CO2)(-393.5 kJ / 1 mol CO2) + ( .250 mol H2O) (-286 H2O/1 mol H2O) + 0 + A</p>

<p>587.8668 kJ = A</p>

<hr>

<p>The above is info I thought may be helpful</p>

<p>I think that that you need to find the pressure (use given and subtract out the water vapor?) Then use ideal gas law (PV = nRT) to get moles of gas.</p>

<p>I suppose that you could subtract out the volume of the traces amount of liquid water, then use ideal gas law.</p>

<p>From there, you have total moles of gas, subtract the knowns to get moles of A.</p>

<p>Divide the mass of A by the moles of A to get A's molar mass.</p>

<p>I would hazard a guess that A is probably a familar molar mass.</p>

<p>Then empirical formula X, now that you have the composition and moles of all it's products, then find the real formula.</p>

<p>That'd be my approach, but checking to see if it works will take too long :P</p>

<p>I'm probably wrong though XD</p>

<p>Anyway, my take on (a) lol</p>

<p>Geez that's above an AP level question isn't it? My class is using Silberberg and we finished Thermochemistry and now we're on Quantum Theory.</p>

<p>adidasty, without a doubt. I can't see this coming up an AP exam.</p>

<p>You did thermochem before quantum, weird. Quantum was one of the first things we did last year. We use Masterton/Hurley.</p>