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AUGUST 27TH 2010

Chem 251 TUT + SOLUTIONS

1.5� A constant Volume perfect gas thermometer indicates a pressure of 6.69 kPa at the triple point

temperature of water (273.16 K). (a) what change of pressure indicates a change of 1.00 K at this temperature? (b) What pressure indicates a temperature of 100.00C? (C) what change of pressure indicates a change of 1.00 K at the latter temperature

SolutionsSolutionsSolutionsSolutions

a)a)a)a) This is a perfect gas, thus can use PThis is a perfect gas, thus can use PThis is a perfect gas, thus can use PThis is a perfect gas, thus can use P1111VVVV1111/T/T/T/T1111 = P= P= P= P2222VVVV2222/T/T/T/T2222Since this is at constant volume; equation simplifies to: PSince this is at constant volume; equation simplifies to: PSince this is at constant volume; equation simplifies to: PSince this is at constant volume; equation simplifies to: P1111/T/T/T/T1111 = P= P= P= P2222/T/T/T/T2222TTTT2222 = the change + original temperature = 1.00 k + 273.16 = 274.16 k= the change + original temperature = 1.00 k + 273.16 = 274.16 k= the change + original temperature = 1.00 k + 273.16 = 274.16 k= the change + original temperature = 1.00 k + 273.16 = 274.16 k

PPPP2222 = (6.69 = (6.69 = (6.69 = (6.69 kPakPakPakPa / 273.16 k) x 274.16 k / 273.16 k) x 274.16 k / 273.16 k) x 274.16 k / 273.16 k) x 274.16 k = = = = 6.71449… 6.71449… 6.71449… 6.71449… kPakPakPakPa

Change in Pressure = 6.71449… Change in Pressure = 6.71449… Change in Pressure = 6.71449… Change in Pressure = 6.71449… ---- 6.69 6.69 6.69 6.69 = = = = 0.0244911…0.0244911…0.0244911…0.0244911…

Answer = A change in 0.0245 Answer = A change in 0.0245 Answer = A change in 0.0245 Answer = A change in 0.0245 kPakPakPakPa, would indicate a change of 1.0 K., would indicate a change of 1.0 K., would indicate a change of 1.0 K., would indicate a change of 1.0 K.

b)b)b)b) Perfect gas, therefore can use same equation from part (a);Perfect gas, therefore can use same equation from part (a);Perfect gas, therefore can use same equation from part (a);Perfect gas, therefore can use same equation from part (a); PPPP2222 = (6.69 = (6.69 = (6.69 = (6.69 kPakPakPakPa / 273.16 k) x / 273.16 k) x / 273.16 k) x / 273.16 k) x 373.16 373.16 373.16 373.16 k k k k = = = = 9.139114072… 9.139114072… 9.139114072… 9.139114072… kPakPakPakPa

Answer = 9.14 Answer = 9.14 Answer = 9.14 Answer = 9.14 kPakPakPakPa

c)c)c)c) Same Same Same Same steps as in part steps as in part steps as in part steps as in part (a)(a)(a)(a)

PPPP2222 = (9.14 = (9.14 = (9.14 = (9.14 kPakPakPakPa / 373.16 k) x 374.16 k / 373.16 k) x 374.16 k / 373.16 k) x 374.16 k / 373.16 k) x 374.16 k = = = = 9.1644935… 9.1644935… 9.1644935… 9.1644935… kPakPakPakPa

Change Change Change Change in pressure = 9.164493… in pressure = 9.164493… in pressure = 9.164493… in pressure = 9.164493… ---- 9.14 = 9.14 = 9.14 = 9.14 = 0.024493… 0.024493… 0.024493… 0.024493… kPakPakPakPa

Answer = 0.0245 Answer = 0.0245 Answer = 0.0245 Answer = 0.0245 kPakPakPakPa

2.4(A)2.4(A)2.4(A)2.4(A)

� A sample consisting of 1.00 mol of perfect gas atoms, for which CV,m = 3/2R, initially at p1 = 1.00 atm and T1 = 300 K, is heated reversibly to 400 K at constant volume. Calculate the final pressure, ∆U, q, and w.

Solution:

This This This This is a perfect gas, thus can use Pis a perfect gas, thus can use Pis a perfect gas, thus can use Pis a perfect gas, thus can use P1111VVVV1111/T/T/T/T1111 = P= P= P= P2222VVVV2222/T/T/T/T2222Since Since Since Since this is at constant volume; equation this is at constant volume; equation this is at constant volume; equation this is at constant volume; equation simplifies tosimplifies tosimplifies tosimplifies to: P: P: P: P1111/T/T/T/T1111 = P= P= P= P2222/T/T/T/T2222PPPP2222 = (1.00 / 300) x 400 = = (1.00 / 300) x 400 = = (1.00 / 300) x 400 = = (1.00 / 300) x 400 = 1.333333… 1.333333… 1.333333… 1.333333… atmatmatmatm

Answer = Final pressure = 1.33 Answer = Final pressure = 1.33 Answer = Final pressure = 1.33 Answer = Final pressure = 1.33 atmatmatmatm

W = 0 because there is no change in volumeW = 0 because there is no change in volumeW = 0 because there is no change in volumeW = 0 because there is no change in volume

∆∆∆∆U = q + w; but w = 0, thus ∆U = qU = q + w; but w = 0, thus ∆U = qU = q + w; but w = 0, thus ∆U = qU = q + w; but w = 0, thus ∆U = q

CCCCV,mV,mV,mV,m = 3/2R = (3/2) * 8.31447 = = 3/2R = (3/2) * 8.31447 = = 3/2R = (3/2) * 8.31447 = = 3/2R = (3/2) * 8.31447 = 12.471705 J K12.471705 J K12.471705 J K12.471705 J K----1111 molmolmolmol----1111

CvCvCvCv = = = = 12.471705 J K12.471705 J K12.471705 J K12.471705 J K----1111 molmolmolmol----1 1 1 1 x 1 mol (from the question; 1 mol of perfect)x 1 mol (from the question; 1 mol of perfect)x 1 mol (from the question; 1 mol of perfect)x 1 mol (from the question; 1 mol of perfect)

qqqqvvvv = = = = CCCCvvvv∆∆∆∆TTTT = = = = 12.471705 J K12.471705 J K12.471705 J K12.471705 J K----1111 x (400 x (400 x (400 x (400 –––– 300) = 1247.1705 J300) = 1247.1705 J300) = 1247.1705 J300) = 1247.1705 J

Answer = Answer = Answer = Answer = ∆U = ∆U = ∆U = ∆U = q = 1.25 kJq = 1.25 kJq = 1.25 kJq = 1.25 kJ

2.152.152.152.15SilyleneSilyleneSilyleneSilylene (SiH(SiH(SiH(SiH2222) is a key intermediate in the thermal decomposition ) is a key intermediate in the thermal decomposition ) is a key intermediate in the thermal decomposition ) is a key intermediate in the thermal decomposition of of of of silicon hydrides such as silane (SiH4) and disilane (Si2H6). Moffat et al. (H.K. Moffat, K.F. Jensen, and R.W. Carr, J. Phys. Chem. 95, 145 (1991)) 95, 145 (1991)) 95, 145 (1991)) 95, 145 (1991)) report report report report ∆fH

O(SiH2) = +274 kJ mol−1. If ∆fH

O(SiH4) = +34.3 kJ mol−1 and

∆fHO(Si2H6) = +80.3 kJ mol

−1 (CRC Handbook (2004)), compute the standard enthalpies of the following reactions:

� (a) SiH4(g) →SiH2(g) + H2(g)

� (b) Si2H6(g) →SiH2(g) + SiH4(g)

Solutions:Solutions:Solutions:Solutions:

Use equation below, from the slides (see equation 2.34 on p55 in your Use equation below, from the slides (see equation 2.34 on p55 in your Use equation below, from the slides (see equation 2.34 on p55 in your Use equation below, from the slides (see equation 2.34 on p55 in your text book); text book); text book); text book);

(a)(a)(a)(a) [(+274 x 1) [(+274 x 1) [(+274 x 1) [(+274 x 1) + (+ (+ (+ (0 x1 )] 0 x1 )] 0 x1 )] 0 x1 )] –––– [[[[34.3 x 1] 34.3 x 1] 34.3 x 1] 34.3 x 1] = = = = 239.7 239.7 239.7 239.7 kJ molkJ molkJ molkJ mol−1−1−1−1

(b)(b)(b)(b) [(+274 x 1) + (34.3 x 1)] [(+274 x 1) + (34.3 x 1)] [(+274 x 1) + (34.3 x 1)] [(+274 x 1) + (34.3 x 1)] –––– [+80.3 x 1] = [+80.3 x 1] = [+80.3 x 1] = [+80.3 x 1] = 228 228 228 228 kJ molkJ molkJ molkJ mol−1−1−1−1

8/29/2010

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2.18(B)2.18(B)2.18(B)2.18(B)

� From From From From the following data, determine the following data, determine the following data, determine the following data, determine ∆∆∆∆ffffHHHHoooo for for for for

diboranediboranediboranediborane, B, B, B, B2222HHHH6(g)6(g)6(g)6(g), , , , at at at at 298 K:(1) B2H6(g) + 3O2(g) →B2O3(s) + 3H2O(g) ∆rH

O = −1941 kJ mol−1

(2) 2B(s) + 3/2O2(g) →B2O3(s) ∆rHO = −2368 kJ mol−1

(3) H2(g) + 1/2O2(g) →H2O(g) ∆rHO = −241.8 kJ mol−1

Solution:

B2O3(s) + 3H2O(g) →B2H6(g) + 3O2(g) ∆rHO = +1941 kJ mol−1

2B(s) + 3/2O2(g) →B2O3(s) ∆rHO = −2368 kJ mol−1

3H2(g) + 3/2O2(g) →3H2O(g) ∆rHO = (3 x−241.8 kJ mol−1)

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3H2(g) + 2B(s) → B2H6(g) = −1152.4 kJ mol−1