Chemistry I -- Final Exam

Transcription

1 Chemistry I -- Final Exam 01/1/13 Periodic Table of Elements Constants R = J / mol K 1 atm = 760 Torr = 1.01x10 5 Pa = L atm / K mol c = m/s = L kpa / K mol h = Js Mass of e, m e = kg k = J/K 1

2 1. (10%) If.00 mol of an idea gas at 300K and 3.00 atm expands from 6.00 L to L and has a final pressure of 1.00 atm, isothermally and reversibly, calculate the w, and ΔU?. Benzoic acid has a known internal energy of combustion (-351 kj/mol). When a bomb calorimeter was calibrated by burning 0.85 g of benzoic acid (C 6 H 5 COOH) in oxygen, the temperature rose 1.94 C. A 0.77 g sample of the sugar D-ribose (C 5 H 10 O 5 ) was then burned in excess oxygen in the same calorimeter, forming gaseous CO and liuid H O. The temperature of the calorimeter rose from 1.81 C to.7 C. (a) (5%) Write a balanced euation for the combustion. (b) (5%) Calculate heat capacity of bomb calorimeter. (c) (5%) Calculate the internal energy change, ΔU, for the combustion of 1.00 mol ribose molecules. 3. (5%) The standard molar heat of combustion of the isomers m-xylene and p-xylene are kj/mol and kj/mol, respectively. Use the data, together with Hess s Law, to calculate the value of ΔH r for the isomerization reaction m-xylene p-xylene. 4. (10%) An idea gas mixture consists of N and N. At room temperature, only translational and rotational motion contribute to heat capacity of the mixture, C v, is found to be 9R (translation) and 4R (rotational), respectively. Find the mole fraction of N. 5. (10%) Given the following standard reaction enthalpies: O (g) O(g) ΔH = kjmol NO(g) + O 3 (g) NO (g) + O (g) ΔH =00.0 kjmol And the standard enthalpy of formation of ozone, kjmol, calculate the standard reaction enthalpy for the reaction NO(g) + O(g) NO (g) 6. (8%) (a) Define statistical entropy and explain it; (b) Which would you expect to have a higher molar entropy at 0 K, single crystals of BF 3 or of COF? Why?

3 7. (8%) List the following substances in order of increasing molar entropy at 98 K: H O(l), H O(g); H O(s), C(s, diamond). Explain your reasoning. 8. (8%) Show that if two metal blocks with different temperatures, T 1, T and T 1 < T, are placed in contact, then the direction of spontaneous change is toward the eualization of temperatures. Assume that the temperatures of the blocks remain constant individually. 9. (10%) (a) Verify the Clausius ineuality, S. (b) Use the change in free T energy to explain how an endothermic reaction can be spontaneous. 10. (6%) Calculate the standard Gibbs free energy of formation of HI(g) at 5 C. Given: H f (HI, g) = 6.48 kj mol -1 ; S m (in J K -1 mol -1 ): H (g) = 130.7, I (s) = 116.1, HI(g)= (10%) Show that the change in Gibbs free energy for a process is eual to the maximum nonexpansion work that the system can do at constant temperature and pressure, i.e. dg = dw e,max or G = w e,max. 3

4 Answers: 1. isothermally ΔT=0 ΔU = C v ΔT = C v 0 = 0. (%) w = nrtln(v f /v i ) = ln(18/6) = 5480J = 5.48 kj. (4%) =ΔU-w = kj. (4%).. (a) C H O 5O 5CO 5H (5%) (g) O(l) (b) benzoic acid C 6 H 5 COOH, 1.1 g.mol g 1.1 g.mol mol Heat released on combustion of 0.85g: ( kj 3 mol)( 351kJ.mol 1 ) heat capacity of bomb calorimeter, C = T 1.96 kj = = 11.3 kj (5%) (c) D-ribose, C H 5 10O5, molecular mass = g mol g g.mol mol Bomb calorimeter is a fixed volume. U = + w = Heat released by reaction, (released) = heat gained by calorimeter, (gained) (released) (gained) 0 (released) (gained) CT (11.3 kj. C kj 1 )(.7C 1.81C) On combustion of 0.77g (or mol ) of D-ribose U ( released) kj Per mole of D-ribose, 10.30kJ U mol 17.5 kj.mol 1 (5%) 4

5 3. m-xylene combustion products ΔH r = kj + combustion products p-xylene ΔH r = kj => m-xylene p-xylene ΔH r = +.9 kj 4. N is atom, C v = 3/ R. (%) N is diatomic molecule, C v = 5/ R. (%) Assume x moles of N, y moles of N Translational: 3/ R (x+y) = 9R (%) Rotational: yr = 4R (%) y = 4, and x = mole fraction of N = 1/3, mole fraction of N = /3 (%) 5. NO(g) + O 3 (g) NO (g) + O (g) ΔH =00.0 kjmol (%) 3/ O (g) O 3 (g) ΔH = kjmol (%) O(g) ½ O (g) ΔH = kjmol (%) NO(g) + O(g) NO (g) ΔH = kjmol (4%) 6. (a) S = k lnw, (%) k: Boltzmann constant 對 於 解 釋 W 以 下 3 種 說 法, 任 一 種 說 法 皆 正 確 (%) W 等 於 在 相 同 的 能 量 下, 不 同 分 子 排 列 組 合 的 總 數 W: the number of ways that the atoms or molecules in the sample can be arranged and yet still give rise to the same total energy W: the number of different microstates that correspond to the same energy (b) COF. (%) 以 下 兩 種 說 法, 任 一 說 法 皆 正 確 (%) COF and BF 3 are both trigonal planar molecules, but it would be possible for the molecule to be disordered with the fluorine and oxygen atoms occupying the same locations. Because all the groups attached to boron are identical, such disorder is not possible. 5

6 W(COF )=3 and W(BF 3 )=1, W(COF ) > W(BF 3 ) COF has a higher molar entropy. 7. C(s, diamond) < H O(s) < H O(l) < H O(g) (4%) It is easy to order HO in its various phases because entropy will increase when going from a solid to a liuid to a gas. ( 或 根 據 物 質 三 態 所 佔 領 的 空 間 及 運 動 的 自 由 度, 可 以 直 接 說 明 H O(s) < H O(l) < H O(g)) (%) 以 下 兩 種 說 法, 任 一 說 法 皆 正 確 (%) The main uestion concerns where to place C(s, diamond) in this order, and that will essentially become a uestion of whether C(s, diamond) should have more or less entropy than HO(s), because we would automatically expect C(s, diamond) to have less entropy than any liuid. Because water is a molecular substance held together in the solid phase by weak hydrogen bonds, and in C(s, diamond) the carbon is more rigidly held in place and will have less entropy. Diamond 結 晶 的 晶 格 中 只 有 C 原 子, 而 水 結 晶 的 晶 格 中 有 兩 的 H 原 子 及 O 原 子 因 此 W(H O, solid) > W(C, diamond) 8. 以 下 兩 種 說 法, 任 一 說 法 皆 正 確 Assume the transfer of energy as heat is reversible. The change in entropy for each block (block 1 and block ) are: S and S (%) 1 1 T1 T Energy is transferred, then the change in entropy for the system is: 1 S S1 S (%) T1 T Only when energy is transferred from hot to cold block, 6

7 1 > 0, < 0 and 1 = -, (%) S = 1 /T 1 1 /T > 0, (%) the process is spontaneous. Assume the transfer of energy as heat is reversible and the heat is assigned as, where >0. If the energy is transferred from hot (T ) to cold block (T 1 ), the change in entropy for each block (block 1 and block ) are: 1 = T = T (4%) Then, the change in entropy for the total system is: S S 1 S - T T 1 T T T1T 1 (%) T > T 1 and >0 T T1 S > 0, (%) T1T The process for the energy transferred from hot (T ) to cold block (T 1 ) is spontaneous. 9. (a) (5%) Clausius ineuality S /T. For reversible processes, S = /T. For irreversible processes, rev > irrev. S = rev /T > irrev /T. (b) (5%) If G r < 0, ( G r is negative); the process is spontaneous. Gr Hr TS r If a reaction is endothermic H r is positive, then the reaction will be spontaneous only if TS r is larger than H r. So, effectively, a reaction that is endothermic can be spontaneous only if the entropy change in the system outweighs the enthalpy change s effect on the entropy of the surroundings. 10. ½ H (g) + ½ I (s) HI(g) H = 1 mol H f (HI, g) = 1 mol (+6.48) kj mol -1 ; = kj (%) S = S m (HI, g) {0.5 S m (H, g) +0.5 S m (I, s)} 7

8 = 1 mol 06.6 J K -1 mol -1 { 0.5 mol mol 116.1} J K -1 mol -1 = +83. J K -1. (%) G = H -TS = *83./1000 = kj (%) 11. 批 改 此 題 同 學 請 注 意 : 以 下 的 w e 寫 成 w non (non 代 表 non-expansion work) 也 可 以 給 分! 如 果 使 用 Δ 而 不 使 用 d 也 算 正 確! dg dh TdS( at constant temperature) dh du PdV( at constant pressure) dg du PdV TdS ( at constant temperature and pressure) (%) dg d dw PdV TdS For a reverdible process Insert d and dw into dg dg dw d d e Because dw e rev dw dw rev dw PdV dw dg TdS PdV dw TdS dw expansion e e dw PdV TdS e is the maximun nonexpansion work that thesystemcan do, therefore dg = dw e, max at constant T and P 8