Answers to Assigned Problems from Chapter 2

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Lewis Fletcher
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1 Answers to Assigned Problems from Chapter mol of ice has a volume of g/ g cm 3 = cm 3 1 mol of water has a volume of g/ g cm 3 = cm 3 V(ice water) = 1.63 cm 3 mol 1 (PV) = atm dm 3 = J mol 1 H = 6025 J = U + (PV) = U (0.165 J mol 1 ) U = 6025 J mol 1 = kj mol 1 (Difference between H and U is only J mol 1.) Work done on the system = J mol mol of water at 100 C has a volume of g g cm 3 = cm g Volume of steam = g cm 3 = cm3 Volume increase = cm 3 mol 1 = dm 3 mol 1 (PV) = atm dm 3 mol 1 = J mol 1 = 3.06 kj mol 1 H = 4063 J mol 1 = U + ( J mol 1 ) U = 37.6 kj mol 1 Work done by the system = w = 3.06 kj 2.4. Heat the water from 10 C to 0 C: (1) q 1 = C P dt = C P (T 2 T 1 ) = 753 J mol 1 Freeze the water at 0 C: (2) q 2 = 6025 J mol 1 Cool the ice from 0 C to 10 C: (3) q 3 = 377 J mol 1 Net q = = 5649 J mol 1 = H = 5.65 kj mol Heat evolved = = J Molar mass of CH 3 COCH 3 = = g mol 1 Heat evolved in the combustion of 1 mol

2 Chapter 2 2 = J = kj a. U = kj mol 1 b. CH 3 COCH 3 (l) + 4O 2 (g) 3CO 2 (g) + 3H 2 O(l) n(gases) = 1 H = J mol 1 = kj mol a. Heat capacity of man = J K J Temperature rise = J K 1 = 35.7 C Final temperature = = 72.7 C b. H = J mol 1 /18.0 g mol 1 = 2411 J g The reaction is Mass of water required = Zn + H 2 SO 4 ZnSO 4 + H 2 (g) J 2411 J g 1 = 4340 g = 4.34 kg Thus, 1 mol of gas is liberated by each mole of Zn, i.e., by g. One hundred grams therefore liberates (100/65.37)mol = 1.53 mol of H 2. The work done by the system is P V: w = P V = n H2 RT = 1.53 mol J K 1 mol K = 3790 J = 3.79 kj The work in a sealed vessel ( V = 0) is zero Enthalpy for this reaction at 298 K is the enthalpy of formation of two moles of water, i.e., H (298 K) = 2 mol ( kj mol 1 ) = kj. At 800 K, using Eq. (2.52), we obtain H (800 K)/J = H (298 K) + d( ) e( ) f ( ), where d = 2d H2 O (2d H + d 2 O2 ), and e and f are defined similarly. d/(j K 1 ) = ( ) = e/(j K 2 ) = ( ) = f/(j K) = 2 0 ( ) = Therefore, H (800 K)/J = ( ) ( ) ( ) = or kj.

3 Chapter Molar mass of benzene = = g mol 1 Heat evolved in the combustion of 1 mol = kj g mol g 1 = kj a. U = kj mol 1 b. C 6 H 6 (l) O 2 (g) 6CO 2 (g) + 3H 2 O(l) v(gases) = 1.5 H = J mol 1 = kj mol H = H (C 2 H 6 ) + H (H 2 ) 2[ H (CH 4 )] H = ( 74.81) = kj mol Molar mass of CH 3 OH = = g mol 1 Amount of methanol = 5.27 g/32.04 g mol 1 = mol Heat evolved = kj g mol 527. g 1 = kj mol 1 = c U a. c U = kj mol 1 CH 3 OH(l) O 2 (g) CO 2 (g) + 2H 2 O(l) v(gases) = 0.5 c H = J mol 1 = kj mol 1 b. CH 3 OH(l) O 2 (g) CO 2 (g) + 2H 2O(l) (1) H = kj mol 1 H 2 (g) O 2 (g) H 2O(l) (2) f H = kj mol 1 C(s) + O 2 (g) CO 2 (g) (3) f H = kj mol 1 2 (2) + (3) (1) gives C(s) + 2H 2 (g) O 2 (g) CH 3OH(l) (4) f H = 2( ) = kj mol 1

4 Chapter 2 4 This value is slightly different from the value listed in Appendix D. Experimental data is constantly being evaluated causing changes in listed values. This problem may have used older data. c. CH 3 OH(l) CH 3 OH(g) v H = kj mol 1 (5) (4) + (5) gives C(s) + 2H 2 (g) O 2 (g) CH 3 OH(g) f H = kj mol C(graphite) + 3H 2 (g) C 2 H 6 (g) (1) f H = kj mol 1 C(graphite) + O 2 (g) CO 2 (g) (2) f H = kj mol 1 H 2 (g) O 2 (g) H 2O(l) (3) f H = kj mol 1 3 (3) + 2 (2) (1) gives C 2 H 6 (g) O 2 (g) 2CO 2 (g) + 3H 2 O(l) c H = kj mol First, perform a multiple regression on z = a + bx + cy using the definitions z = C P,m ; x = T and y = 1/T 2. The result is z = x 871.4y. In other words, we find that d = J K 1 mol 1 ; e = J K 2 mol 1 ; f = J K mol 1. Below, we present two plots of this function, one in the range 15 T 275, and another, in the range 10 T 25. It can be seen that the function becomes negative at T 16.1 K. A negative heat capacity is obviously unphysical. This is an indication that the temperature dependence of heat capacities of solids at low temperature cannot be expressed using the model used here (see Chapter 16, Section 6) Function Plot 4 Function Plot z z x x

5 Chapter C 3 H 6 (g) O 2 (g) 3CO 2 (g) + 3H 2O(l) (1) c H = kj mol 1 C(graphite) + O 2 (g) CO 2 (g) (2) f H = kj mol 1 H 2 (g) O 2 (g) H 2O(l) (3) f H = kj mol 1 3 (2) + 3 (3) (1) gives 3C(graphite) + 3H 2 (g) C 3 H 6 (g) f H = 53.1 kj mol Perform a multiple regression on z = a + bx + cy using the definitions z = C P,m ; x = T and y = 1/T 2. The result is z = x y. In other words, we find that d = J K 1 mol 1 ; e = J K 2 mol 1 ; f = J K mol Function Plot 1.00 CP,m T 500 A plot of the fit is 2.23 H = = 44.1 kj mol C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O (1) c H = kj mol 1 CH 3 CHO O 2 2CO 2 + 2H 2 O (2) c H = kj mol 1

6 Chapter 2 6 CH 3 COOH + 2O 2 2CO 2 + 2H 2 O (3) c H = kj mol 1 Reaction (a) is (1) (2); H = = kj mol 1 Reaction (b) is (2) (3); H = = kj mol Let us list all the reactions involved: 1. CH 2 CHCN O 2 (g) 3 CO 2 (g) H 2 O(g) + (1/2)N 2 (g); ch = kj mol 1 2. C(graphite) + O 2 (g) CO 2 (g); f H = kj mol 1 3. H 2 (g) O 2 (g) H 2 O(g); fh = kj mol C(graphite) + H 2 (g) C 2 H 2 (g); f H = kj mol 1 5. C(graphite) H 2 (g) N 2 (g) HCN(g); fh = kj mol 1 The desired reaction is HCN(g) + C 2 H 2 (g) CH 2 CHCN. To generate this equation from the five given above, we need to manipulate them to get 3(Eq. 2) + (3/2)(Eq. 3) (Eq. 1) (Eq. 4) (Eq. 5). Performing the same manipulations with the enthalpies, we obtain From Appendix D H = 3( ) + (3/2)( ) ( ) = kj mol 1. 2C + 3H O 2 C 2 H 5OH (1) f H = kj mol 1 2C + 2H 2 + O 2 CH 3 COOH (2) f H = kj mol 1 H O 2 H 2 O (3) f H = kj mol 1 Reaction is (2) + (3) (1); H = kj mol 1 This differs slightly from the result of Problem 2.19, namely = kj mol From Appendix D, f H for ethanol and acetic acid are and kj mol 1, respectively. Therefore, the enthalpy change for the reaction C 2 H 5 OH(l) + O 2 (g) CH 3 COOH(l) is ( ) = kj mol 1. Since ethanol is fed in at the rate of kg h 1, and only 42 mole % of ethanol is converted, the actual heat released in the reaction per hour is (MW of ethanol = g mol 1 ) kg h kg mol 1 1 ( kj mol ) 0.42 = kj h Therefore, heat will have to be removed at the rate of kj h 1.

7 Chapter a. Assume that all the ice melts. The process would absorb heat from the 1000 g water. The water produced by the melting ice (100 g of it) will be initially at 0 C. It will warm up by absorbing heat from the 1000 g water present. Suppose that the final temperature is t C; then the total heat gained by the 100 g water is 100 g Heat gained = g mol 1 [6 025 J mol J K 1 mol 1 (t 0) K] = J J K 1 t K. Finally, all of the water is at the temperature t. Using the relationship heat lost = heat gained, we get J J K 1 t K = [75.3 J K 1 mol g g mol 1 (t 20) K]. Solving for t, we get t = 10.9 C. Since this is not below 0 C, it follows that all of the ice in fact melts. The final temperature is 10.9 C. b. It is obvious that now not all of the ice will melt. (If we assumed it all melted, we would find the final temperature to be below 0 C.) The final temperature of the water is now 0 C, and if we suppose that x g of the ice melts, the heat balance equation is x g 6025 J mol 1 = 75.3 J K 1 mol g 20 K x = 250 g a. Since a bomb calorimeter is a constant-volume instrument, the heat evolved is U. The heat evolved per gram is thus J/ g = J g 1 The molecular weight of sucrose is , and therefore c U m = J g g mol 1 = J mol 1 = 5635 kj mol 1 The equation for the combustion reaction is C 12 H 22 O 11 (s) + 12O 2 (g) 12CO 2 (g) + 11H 2 O(l) The change n is therefore zero, and by Eq. 2.41, H = U; thus c H m = 5635 kj mol 1 b. From the equation, with the use of Hess s law, it can be deduced that f H m = 12 f H m (CO 2,g) + 11 f H m (H 2 O,l) c H m (sucrose) = ( ) + ( ) kj mol 1 = = 2231 kj mol Note that we need to find the intersection of the isotherm that passes through the initial state and the adiabat that passes through the final state. Let this point be (P 0,V 0 ) at the temperature of the isotherm T i. For adiabatic processes, (Eq. 2.90), T f / T i = (V f / V 0 ) γ 1, where 3 3 γ = R + R / R = 5/

8 Chapter 2 8 Therefore, γ 1 = 2/3. The final volume is V f = RT f /P f = ( dm 3 bar K 1 mol K)/2.0 bar = dm 3. Therefore, V 0 = V f (T f /T i ) 3/2 = (253.2/298.0) 3/2 = dm a. 1 mol in 22.7 dm 3 at 273 K exerts a pressure of 1 bar; 2 mol in dm 3 exert a pressure of 4 bar. b. PV = = bar dm 3 1 bar dm 3 = 100 J bar dm 3 = kj c. C V,m = C P,m R = = 21.1 J K 1 mol a. Zero b = 4220 J = 4.22 kj c J = 4.22 kj d. P = 4 373/273 = 5.47 bar = 547 kpa e. P 2 V 2 = = bar dm 3 = 6208 J = kj f. H = = 5888 J = 5.89 kj a. V = /273 = 15.5 dm 3 b. w = P V = 4 ( ) = 16.6 bar dm 3 c. q = = 5880 J = 5.88 kj d. H = 5880 J = 5.88 kj = 1660 J = 1.66 kj e. U = H P V = = 4220 J = 4.22 kj (= 2C V,m T = ) a. Zero b. P 2 = 8 bar = 800 kpa c. w = ln 2 = 3150 J = 3.15 kj d. q = 3147 J = 3.15 kj e. Zero Initial volume of gas = nrt =.. = dm P 3 10 Final volume = dm 3 V = dm 3

9 Chapter 2 9 a. Work done by the gas = = bar dm 3 = 3632 J = energy transferred to surroundings. b. U = H = 0 c. q = 3632 J a. Work done by gas = nrt ln V final V initial = ln 5 = 7310 J b. U = H = 0 c. q = 7310 J = 7.31 kj C V,m = = J K 1 mol 1 γ = = a. P 2 = P 1 V γ = = bar T 2 = T 1 V γ 1 = = K b. U m = C V,m (T 2 T 1 ) = 20.49( ) = 2361 J mol 1 Amount of H 2, n = = mol U = 430 J for mol H m = 28.80( ) = 3319 J mol 1 H = 604 J for mol a. Initial volume, = nrt P P 1 1/γ P 2 V 2 = = (10) = dm 3 b. T 2 = T 1 P 2 V 2 P 1 = = = dm = K

10 Chapter 2 10 c. C P,m C V,m = J K 1 mol 1 C P,m C P,m = 1.31; C V,m C V,m 1 = 0.31; C P,m C V,m = 0.31 C V,m 0.31 C V,m = J K 1 mol 1 C V,m = J K 1 mol 1 C P,m = J K 1 mol 1 U m = 26.82( ) = 3984 J K 1 mol 1 U = J K 1 for 0.1 mol H m = ( ) = 5219 J K 1 mol 1 H = 522 J K 1 for 0.1 mol a. C P,m C V,m = R = J K 1 mol 1 C P,m = (T/K) b. T = 0, U = H = 0. It could be made to occur adiabatically by allowing free expansion, with w = q = C V,m = 5 2 R and C P,m = 7 2 R γ = 7 5 T 2 γ 1 V 2 T 1 = (Eq. 2.90) Therefore, T 2 = T 1 V 2 γ 1 = K 1 2 2/5 = K U m = C V,m (T 2 T 1 ) = 5 R( ) K 2 = 1501 J mol 1 = 1.5 kj mol 1 H m = C P,m (T 2 T 1 ) = 7 R( ) K 2 = 2100 J mol 1 = 2.1 kj mol For 1 mol of a van der Waals gas P + a V 2 m Therefore (V m b) = RT (Eq )

11 Chapter 2 11 T = P(V m b) R a + V 2 m R (V m b) and T P V = V m b R a. Work done on the system = V 2 PdV = RT ln V 2 = ln 10 = 9757 J = 9.76 kj mol b. w = RT V 2 dv V b Put V b = x; dv = dx = RT V 2 dx x = RT ln(v b)v 2 = RT ln b V 2 b = ln = J mol 1 = kj mol 1 More work is done in (b) because of the greater ratio of free volumes a. w = RT ln = ln 20 V 2 5 = 1153 J mol 1 = 1.15 kj mol 1 b. P = RT V a V 2 w = V 2 PdV = RT V 2 dv V + a 2 dv V V 2 = +RT ln V 2 a 1 1 V 2 = = 1211 J = 1.21 kj mol 1 In (b), additional work has to be done against the molecular repulsions The reversible work done on the system is w rev = V 2 PdV = nrt V 2 dv V nb = nrt ln V 2 nb nb = (J) ln = J ln(0.92/9.92) = = J = 11.9 kj

12 Chapter 2 12 From Eq , U is zero since a is zero. To obtain H we must calculate (PV): P 1 = 2RT/( 0.08) = /9.92 = 5.03 bar P 2 = 2RT/(V ) = /0.92 = 54.2 bar P 1 = = 50.3 bar dm 3 = Pa m 3 = J P 2 V 2 = = 54.2 bar dm 3 = J (PV) = P 2 V 2 P 1 = 390 J The reversible work is H = U + (PV) = = 390 J w rev = V 2 PdV = V 2 = nrt ln V 2 nrt V n 2 a n2 a 1 1 V 2 V 2 dv = (J) ln (1/20) (3.00) (Pa m 6 ) = = J = 17.7 kj The change in internal energy is obtained from the relationship: du = U V T dv = n2 a V 2 dv U = V 2 n 2 a V 2 dv = n 2 a 1 V V 2 = n 2 a 1 1 V 2 = (3.00) (Pa m 6 ) = 4703 J = 4.70 kj To obtain H we calculate P 1 and P 2 V 2 : = 20 dm 3 P 1 = (nrt/ ) n 2 a/ m m m m 3 = [ (bar)/20] (10 5 Pa/bar) [ (Pa m 6 )]/( m 3 ) 2 = Pa Pa = kpa V 2 = 1 dm 3

13 Chapter 2 13 P 2 = [ (bar)/1] (10 5 Pa/bar) [ (Pa m 6 )]/(10 3 m 3 ) 2 = Pa Pa = Pa = 2529 kpa P 1 P 2 V 2 = ( m 3 ) Pa = 7232 J = (10 3 m 3 ) Pa = J (PV) = = 4703 J H = = 9406 J = 9.41 kj.

U = 4.18 J if we heat 1.0 g of water through 1 C. U = 4.18 J if we cool 1.0 g of water through 1 C.

U = 4.18 J if we heat 1.0 g of water through 1 C. U = 4.18 J if we cool 1.0 g of water through 1 C. CHAPER LECURE NOES he First Law of hermodynamics: he simplest statement of the First Law is as follows: U = q + w. Here U is the internal energy of the system, q is the heat and w is the work. CONVENIONS