CHEM 36 General Chemistry EXAM #2 March 13, 2002

Transcription

1 CHEM 36 General Chemistry EXAM #2 March 13, 2002 Name: Key INSTRUCTIONS: Read through the entire exam before you begin. Answer all of the questions. For questions involving calculations, show all of your work -- HOW you arrived at a particular answer is MORE important than the answer itself! Circle your final answer to numerical questions. The entire exam is worth a total of 150 points. Attached are a periodic table and a formula sheet jam-packed with useful stuff. Good Luck! Page Possible Points Points Earned TOTAL:

2 1. For the dissociation of acetic acid in water: at 25.0 o C, K a = 1.76 x CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) a. [10 pts] Calculate G o (kj/mol) for this reaction. DG o = -RTLnK = -( J/mol-K)( K)Ln(1.76 x 10-5 ) = x 10 4 J/mol x 10-3 kj/j = kj/mol b. [5 pts] At equilibrium, what is the value of G? For any system at equilibrium: DG = 0 2. [10 pts] Calculate the molarity of an aqueous acetic acid solution that is 36.0% acetic acid (by mass). The density of the solution is g/ml and the molecular weight of acetic acid is g/mol g HAc x 1 mol HAc x g soln x 1000 ml = mol HAc/L g soln g HAc ml soln L = 6.26 M HAc 2

3 3. [15 pts] Given equilibrium constants for reactions 1 and 2, compute K 3 : (1) H 2 O (l) + H 2 O(l) H 3 O + (aq) + OH - (aq) K 1 = 1.0 x (2) NH + 4 (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) K 2 = 5.7 x (3) NH 3 (aq) + H 2 O (l) NH + 4 (aq) + OH - (aq) K 3 =??? Reaction (1) Reaction (2) = Reaction (3) H 2 O (l) + H 2 O(l) H 3 O + (aq) + OH - (aq) K 1 NH 3 (aq) + H 3 O + (aq) NH + 4 (aq) + H 2 O (l) 1/K 2 NH 3 (aq) + H 2 O (l) NH + 4 (aq) + OH - (aq) K 3 So: K 3 = K 1 (1/K 2 ) = 1.0 x /5.7 x = x 10-5 = 1.8 x Chloroacetic acid dissociates in aqueous solution to produce the chloroacetate and hydronium ions: CH 2 ClCOOH (aq) + H 2 O (l) CH 2 ClCOO - (aq) + H 3 O + (aq) a. [10 pts] Compute K at 25.0 o C. given the following set of equilibrium concentrations: [CH 2 ClCOOH] = M [CH 2 ClCOO - ] = M [H 3 O + ] = M K = [CH 2 ClCOO - ][H 3 O + ] = (0.0112)(0.0112) = x 10-3 = 1.41 x 10-3 [CH 2 ClCOOH] (0.0888) b. [5 pts] Calculate the ph of this solution. ph = -Log[H 3 O + ] = -Log(0.112) = =

4 5. Consider the following endothermic ( H o = kj/mol) reaction: At 25.0 o C, K = 3.2 x CO 2 (g) + H 2 (g) CO (g) + H 2 O (l) If we start with 1.0 atm of each of the gas phase species in a closed vessel containing some water: a. [10 pts] Is the system at equilibrium? If not, will the reaction proceed to the left or to the right in order to reach equilibrium? Q = P CO = (1.0) = 1.0 >> K P CO2 P H2 (1.0)(1.0) If Q > K, then there is an excess of CO and reaction will shift to the: Left Right No Change (circle one) b. [15 pts] Calculate the partial pressure of H 2 (g) for the system at equilibrium. CO 2 (g) + H 2 (g) CO (g) + H 2 O (l) Initial Change +x +x -x Equilibrium 1.00+x 1.00+x 1.00-x 3.2 x 10-4 = (1.00 x) (1.00+x)(1.00+x) A quadratic: 3.2 x 10-4 x x x 10-1 = 0 x = P H2 = x = = 2.0 atm 4

5 c. [5 pts each] How will the equilibrium partial pressure of H 2 (g) be affected by the following: i. CO 2 (g) is added to the system (at constant volume). ii. The volume of the reaction vessel is halved. iii. Argon gas is added to the system (at constant volume). iv. The temperature of the system is increased. 6a. [10 pts] Arrange the following acids in order of increasing strength (briefly explain your reasoning): CH 3 CH 2 OH, HClO 4, CH 3 COOH, H 2 O CH 3 CH 2 OH < H 2 O < CH 3 COOH < HClO 4 strong acid (most # O) only one EN O to accommodate negative charge OH - can stablize charge with one O and a hydrogen two EN O to accommodate negative charge b. [10 pts] Identify the corresponding conjugate bases for these acids: Acid Conjugate Base CH 3 CH 2 OH CH 3 CH 2 O - HClO 4 - ClO 4 CH 3 COOH CH 3 COO - H 3 O + H 2 O c. [5 pts] Arrange the conjugate bases in order of increasing strength. ClO 4 - < H 2 O < CH 3 COO - < CH 3 CH 2 O - 5

6 7. [5 pts each] Given the K a values for the following acids: CH 2 ClCOOH (Chloroacetic Acid): K a = 1.4 x 10-3 C 6 H 5 COOH (Benzoic Acid) K a = 6.5 x 10-5 HCN (Hydrocyanic Acid) K a = 6.2 x a. Which acid is the weakest? HCN has the smallest K a, so it is the weakest acid b. Write the acid dissociation equilibrium reaction for the acid you identified in part a. HCN (aq) + H 2 O (l) CN - (aq) + H 3 O + (aq) c. Which of the conjugate bases for these acids is the weakest? The conjugate base of the strongest acid will be the weakest: CH 2 ClCOOH is the strongest acid, so CH 2 ClCOO - is the weakest conjugate base d. Calculate the value of K b for the base you identified in part c. For a conjugate acid/base pair: K a K b = K w So: K b = K w /K a = 1.0 x /1.4 x 10-3 = 7.14 x = 7.1 x e. Write the base dissociation (K b ) equilibrium reaction for the base you identified in part c. CH 2 ClCOO - (aq) + H 2 O (l) CH 2 ClCOOH (aq) + OH - (aq) 6

7 Extra Credit! pts The triple bond in the N 2 molecule is very strong, but at high enough temperatures even it breaks down. At 5000 K, when the total pressure exerted by a sample of nitrogen is 1.00 atm, N 2 (g) is 0.65% dissociated at equilibrium: N 2 (g) 2 N (g) At 6000 K, with the same total pressure, the proportion of N 2 (g) dissociated at equilibrium rises to 11.6%. Calculate the H of this reaction. We need K at 5000 K (T 1 ) and at 6000 K (T 2 ): At 5000 K: N 2 2N I C (0.0065) E So: K 1 = (P N ) 2 = (0.0130) 2 = 1.70 x 10-4 P N At 6000 K: N 2 2N I C (0.116) E So: K 2 = (P N ) 2 = (0.232) 2 = 6.09 x 10-2 P N Now, plug into Van t Hoff equation and solve for DH: Ln(K 2 /K 1 ) = -(DH o /R)[(1/T 2 ) (1/T 1 )] = -DH o ( x 10-6 ) DH o = x 10 6 J fi 1.5 x 10 3 kj 7