1 Exercise 3.1b pg 131

Transcription

1 In this solution set, an underline is used to show the last significant digit of numbers. For instance in x = the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit, the 9 & 3 in the example, are not significant and would be rounded off at the end of calculations. Carrying these extra digits for intermediate values in calculations reduces rounding errors and ensures we get the same answer regardless of the order of arithmetic steps. Numbers without underlines (including final answers) are shown with the proper number of sig figs. 1 Exercise 3.1b pg 131 Given 50 kj of energy is transferred reversibly and isothermally as heat to a large block of copper at temperatures (a) T = 0 C, (b) T = 70 C. In terms of given variables, this is written: q = 50 kj (a) T = 0 C, (b)t = 70 C Calculate the change in entropy S. The definition of an entropy change in given in our text book by Equation 3.2 on pg 97 as S = f i dq rev T Since the temperature is constant this simply reduces to S = q T For this expression we need absolute temperature in units of Kelvin and absolute temperatures in this exercise are found as (a) T = K, (b) T = K. Substituting these values and the value given for heat gives us the entropy changes. For part (a) S = q T = 50 kj K = J K 1 1

2 and for part (b) S = q T = 50 kj K = J K 1 (a) S = 200 J K 1 (b) S = 100 J K 1 2

3 2 Exercise 3.8b pg 131 Calculate the standard reaction entropy r S at T = 298 K of (a) Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) and (b) C 12 H 22 O 11 (s) + 12O 2 (g) 12CO 2 (g) + 11H 2 O(l) Reaction entropies are easily calculated from reactants and products standard entropies using the definition of standard reaction entropy (text book Equation 3.25a on pg 111). r S = Products ν p S m[p] Reactants ν r S m[r] The standard entropies of reactants and products can be found in Table 2.8 (pg. 919) of our text book and on substitution the standard reaction entropies are calculated as follows. r S = Sm[Zn 2+ (aq)] + Sm[Cu(s)] Sm[Zn(s)] Sm[Cu 2+ (aq)] = JK 1 mol JK 1 mol JK 1 mol JK 1 mol 1 = JK 1 mol 1 for part (a) and for part (b) r S = 12Sm[CO 2 (g)] + 11Sm[H 2 O(l)] Sm[C 12 H 22 O 11 (s)] 12Sm[O 2 (g)] = JK 1 mol JK 1 mol JK 1 mol JK 1 mol 1 = JK 1 mol 1 (a) r S = 21.0 JK 1 mol 1 (b) r S = JK 1 mol 1 3

4 3 Exercise 3.9b pg 131 Combine the reaction entropies calculated in Question 2 (Exercise 3.8(b)) with the reaction enthalpies, and calculate the standard reaction Gibbs energies at T = 298 K. To calculate the standard reaction enthalpies we ll use text book Equation 2.32 (pg. 68) r H = ν p f Hp ν r f Hr Products Reactants Substituting enthalpies of formation from Table 2.8 (pg. 919) the reaction enthalpies are calculated as r H = f H [Zn 2+ (aq)] + f H [Cu(s)] f H [Zn(s)] f H [Cu 2+ (aq)] = kjmol kjmol 1 0 kjmol kjmol 1 = kjmol 1 r H = 12 f H [CO 2 (g)] + 11 f H [H 2 O(l)] f H [C 12 H 22 O 11 (s)] 12 f H [O 2 (g)] = kjmol kjmol kjmol kjmol 1 = kjmol 1 We can combine these enthalpies with previously calculated entropies to find the Gibbs energy using the definition of Gibbs energy (Equation 3.34 on pg. 114 of your text book). For part (a) the Gibbs energy is calculated as G = H T S G = H T S = kjmol K J K 1 mol 1 1 kj 1000 J = kjmol 1 For part (b) G = H T S = kjmol K J K 1 mol 1 1 kj 1000 J = kjmol 1 4

5 (a) G = kjmol 1 (b) G = 5797 kjmol 1 5

6 4 Exercise 3.15b pg 131 Given A certain heat engine operates between T h = 1000 K and T c = 500 K. In terms of given variables, this is written: T h = 1000 K T c = 500 K (a) What is the maximum efficiency of the engine? (b) Calculate the maximum work that can be done for each q h = 1.0 kj of heat supplied by the hot source. (c) How much heat is discharged into the cold sink q c in a reversible process for each q h = 1.0 kj supplied by the hot source? The engine efficiency can be calculated using the Carnot efficency equation (Equation 3.10 on pg. 101 of our text book). η = 1 T c T h = K 1000 K = Next we can use the definition of efficency (Equation 3.8 on pg. 101) to relate the work performed w to the heat absorbed from the hot source q h. The equation is simply rearranged to solve for w. η = w q h w = ηq h = kj = kj Lastly, we can use the heat transaction expression (Equation 3.9 on pg. 101) to relate heat supplied by the hot source q h to the heat discharged into the cold source q c. η = 1 q c q h 6

7 Rearranging for q c we get the following expression. q c = q h (1 η) = 1.0 kj ( ) = kj (a) η = 0.5 (b) w = 0.5 kj (c) q c = 0.5 kj 7

8 5 Exercise 3.17b pg 132 Given The change in the Gibbs energy of a certain constant-pressure process was found to find the following expression. Calculate the value of S for the process. G J = T K We can solve this problem by analyzing the definition of Gibbs energy (Equation 3.34 pg. 114) G = H T S Here we see that the Gibbs energy is a linear function of temperature just like the fit expression. Further, the linear multiplier is S. G = H + T ( S) We can find the linear constant of the fit expression through rearrangment to G = 73.1 J + T Where the linear constant is written in parantheses. Hence, ( 42.8 J ) K and the entropy of this processes is simply S = 42.8 J K S = 42.8 J K 1 S = 42.8 J K 1 8

9 6 Exercise 3.21b pg 132 Given The pressure on a v = 1.00 dm 3 sample of water is increased from p i = 100 kpa to p f = 300 kpa. In terms of given variables, this is written: v = 1.00 dm 3 p i = 100 kpa p f = 300 kpa Estimate the change in Gibbs energy of this process. In solving this problem we ll begin with fundamental equation of chemical thermodynamics (Equation 3.52 pg. 124). dg = V dp SdT Since temperature is constant in this process, the Gibbs energy differential is given by dg = V dp which on integration gives G = pf p i For a liquid we can assume the volume remains constant for this change in pressure. V dp G = = V pf p i pf p i V dp = V (p f p i ) = 1.00 dm 3 (300 kpa 100 kpa) = dm 3 kpa Lastly we can convert this to energy units (J). dp G = dm 3 kpa (0.1 m) Pa 1 N m 2 1 J (1 dm) 3 1 kpa 1 Pa 1 N m = 200 J 9

10 G = 200 J 10