(a) M.W. of H2 = 2, H2O = 18 root mean speed of H2 = 640 x (18/2)1/2 = 1920 (m/s) (3 points)

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Roland Newton
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1 Chemistry(I) Final Exam 1/11/2008 ANSWER Total Score: 102 points Time: 2 hours (18:00-20:00) You are not allowed to ask TAs questions regarding the content of the test. However, using non-programmable calculators is allowed. 1. (a) Give a brief statement about Charles s law. (b) Why can we have the idea about the lowest possible temperature (absolute zero) according to this law? (c) A gas occupies 506 inch 3 at 10 psi/-4 F, and 313 inch 3 at 20 psi/104 F. Use Boyle s law and Charles s law to find the absolute zero in degrees Fahrenheit ( F). (4+4+4 points) (a) For a fixed amount of gas under constant pressure (2 points), the volume varies linearly with the temperature, V=constant x T. (2 points) (b) According to Charles s law, lower temperature causes smaller volume of a gas. The smallest volume of an ideal gas can have is 0. So the temperature at which an ideal gas reach zero volume is the absolute zero. (4 points) (c) 313 inch 3 at 20 psi/104 F 626 inch 3 at 10 psi/104 F (Boyle s law) let x F is equal to absolute zero under constant pressure, V 1 /T 1 = V 2 /T 2 (Charles s law) 506 / (-4 - x) = 626 / (104-x) x = ( F) (4 points) 2. Through a series of enzymatic steps, glucose and oxygen are consumed in our bodies to provide energy. C 6 H 12 O 6(s) + 6 O 2(g) 6 CO 2(g) + 6 H 2 O (l) Given that the partial pressure of oxygen in the air is 0.2 atm (pressure of air = 1.0 atm) and the temperature is 300 K. (a) Calculate the volume of air needed to consume 18 g of glucose. (b) Actually, in one breathing, 500 ml of air is taken into the lung and only 1/4 of oxygen can diffuse into the blood. Calculate the number of breathing needed. (R = L atm/mol K) (3+3 points) (a) M.W. of glucose = 180 ; 18 g glucose = 0.1 mol 0.6 mol O 2 is needed PV = nrt, 0.2xV = 0.6x0.082x300 V = 73.8 (L) (3 points) (b) 73.8/0.5/(1/4) = (3 points)

2 3. Consider the properties of H 2(g) and H 2 O (g). (a) Assume they are ideal gases. The root mean spead of H 2 O (g) is 640 m/s at 25 C. Calculate the root mean spead of H 2(g)? (b) The van der Waals a parameters of these two gases are and L 2 atm/mol 2. Use your knowledge of the factors governing the magnitudes of a to assign each of these two gases to it s a value. ( (P+an 2 /V 2 )(V-nb) = nrt) (3+4 points) (a) M.W. of H 2 = 2, H 2 O = 18 root mean speed of H 2 = 640 x (18/2) 1/2 = 1920 (m/s) (3 points) (b) parameter a represents the attraction between gas molucules, L 2 atm/mol 2 H 2 O L 2 atm/mol 2 H 2 (4 points) 4. Compute the fraction of the space occupied by atoms in: a) cubic closed-packed structure, b) cubic body-centered structure, c) cubic primitive structure. (15 points) a) volume of a single atom = 4/3 π r 3 number of atoms in a unit cell = 4 total volume occupied by atoms = 16/3 π r 3 length of the diagonal of the wall of the unit cell = 4 r unit cell constant = a = 2 2 r volume of the unit cell = 16 2 r 3 fraction of space occupied by atoms = π/(3 2) = 74.05% b) c) volume of a single atom = 4/3 π r 3 number of atoms in a unit cell = 2 total volume occupied by atoms = 8/3 π r 3 length of the diagonal of the unit cell = 4 r unit cell constant = a = 4/ 3 r volume of the unit cell = 64/ 27 r 3 fraction of space occupied by atoms = π 27/(24) = 68.02%

$3 fraction of space occupied by$ atoms = π/6 = 52.36% 5.

3 volume of a single atom = 4/3 π r 3 number of atoms in a unit cell = 1 total volume occupied by atoms = 4/3 π r 3 unit cell constant = a = 2 r volume of the unit cell = 8 r 3 fraction of space occupied by atoms = π/6 = 52.36% 5. The three most popular structures of ionic crystals are: 1) rock-salt structure, 2) cesium chloride structure, 3) zinc-blende structure. Identify these three structures among the figures below. (6 points) A B C D E F Answer: 1C, 2B, 3E 6. Rank the following liquids in order of increasing viscosity at 50º C: C 6 H 5 OH, C 6 H 5 SH, C 6 H 6. (6 points) Question 5.23 (5.25 in old edition) from the textbook. Correct answer is C 6 H 6 < C 6 H 5 SH < C 6 H 5 OH

4 7. Calculate the lattice enthalpy of silver chloride from the following data. (8 points) enthalpy of formation of Ag(g): +284 kj mol 1 first ionization energy of Ag(g): +731 kj mol 1 enthalpy of formation of Cl(g): +122 kj mol 1 electron affinity of Cl(g): +349 (ΔH = 349) kj mol 1 enthalpy of formation of AgCl(s): 127 kj mol 1 -( 127 kj mol 1 )+(122 kj mol 1 )+(731 kj mol 1 )+(284 kj mol 1 )-(349 kj mol 1 ) = 915 kj mol 1 8. Calculate the standard reaction enthalpy for the following reaction. (7 points) N 2 H 4 (l) + H 2 (g) 2NH 3 (g) Given:N 2 H 4 (l) + O 2 (g) N 2 (g) + 2H 2 O(g) ΔH = 543 kj mol 1 2H 2 (g) + O 2 (g) 2H 2 O(g) ΔH = 484 kj mol 1 N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔH = 92.2 kj mol 1 N 2 H 4 (l) + O 2 (g) N 2 (g) + 2H 2 O(g) ΔH = 543 kj mol 1 2H 2 O(g) 2H 2 (g) + O 2 (g) ΔH = +484 kj mol 1 + N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔH = 92.2 kj mol N 2 H 4 (l) + H 2 (g) 2NH 3 (g) ΔH = kj mol 1 Ans: kj mol 1 9. Use the bond enthalpies in following tables to estimate the reaction enthalpy for the reactions: (A) CH 4 (g) + Cl 2 (g) CH 3 Cl(g) + HCl(g) (B) N 2 (g) + 3F 2 (g) 2NF 3 (g) (5 points) (5 points)

5 (A) ΔH(C-H) + ΔH(Cl-Cl) - ΔH(C-Cl) - ΔH(H-Cl) = = 115 kj. mol -1 (B)ΔH(N 2 ) + 3ΔH(F 2 ) - 6ΔH(N-F) = *158 6*195 = 248 kj. mol (12%) At 0 0 C, H fusion (H 2 O)= 6.01kJmol -1 and S fusion (H 2 O)= JK -1 mol -1. Calculate the S sys for each of the following processes. (1) H 2 O (l), C H 2 O (l), 0 0 C H sys = 750 Jmol -1 (2) H 2 O (l), 0 0 C H 2 O (s), 0 0 C H sys = Jmol -1 (3) H 2 O (s), 0 0 C H 2 O (s), C H sys = -370 Jmol -1 (4) H 2 O (l), C H 2 O (s), C H sys = Jmol -1 Given: C p (s)= 37 JK -1 mol -1 & C p (l)= 75 JK -1 mol -1 for H 2 O. (1) S sys = C p (l)ln(273/263)= 2.80 JK -1 (2) S sys = - S fusion = JK -1 (3) S sys = C p (s)ln(263/273)= JK -1 (4) S sys = = (JK -1 ) 11. (6%) Calculate the change in molar entropy of the surroundings ( S sur ) and the change in molar Gibbs free energy of the system ( G sys ) when water freezes at C & 1 atm. S sur = - H sys /T= 5630/263= 21.41(JK -1 )

6 G sys = Hsys - T S sys = x (-20.59)= (J) Or G sys = -T S total = -263 x ( )= (J) 12. (7%) At low temperatures, heat capacities are proportional to T 3 (the Debye T 3 law: C= at 3 ). Show that, near T= 0K, the entropy of a substance is equal to onethird of its heat capacity at the same temperature. Note: = T 1 S(T) dq. 0 T S(T)= 0 T dq/t= 0 T CdT/T S(T)= 0 T at 3 dt/t= at 3 /3 = C/3

Chem 112 Exam 1 Version A Spring /16/ :00am/Odago, M. O.

Chem 112 Exam 1 Version A Spring 2011 02/16/2011 10:00am/Odago, M. O. 1. The pressure of a certain gas is measured to be 25.1 mmhg. What is this pressure expressed in units of pascals? (1 atm=1.0125 x10