A is the frequency factor (related to the number of collisions)

Transcription

1 Chemistry Week 10 Worksheet Notes Oregon State University Ea RT 1. Discuss k e k is the rate is the frequency factor (related to the number of collisions) Ea is the activation energy R is the gas constant (8.314 J/mol K) T is the temperature Rate = k[] x [B] y If you want to increase the rate you can: increase the concentrations increase k To increase k you can: increase increase T decrease Ea (use a catalyst). The following are initial rate data for: + B C + D Experiment Initial [] Initial [B] Initial Rate Determine x and y for the rate law is Rate = k[] x [B] y When [B] is held constant and [] is doubled trails 1 and, the rate stays constant. This implies that y = 0. When [] is held constant and [B] is doubled trails 1 and 3, the rate quadruples. This implies that y =. Rate = k[] 0 [B]

2 3. Based on the thermodynamic data plotted below, determine the activation energy (E a ) and the change in energy (ΔE) for the reaction + B C + D. E a = 700 kj/mol 600 kj/mol = +100 kj/mol ΔE = 00 kj/mol 600 kj/mol = -400 kj/mol (exothermic) 4. Based on the thermodynamic data plotted below, determine the activation energy (E a ) and the change in energy (ΔE) for the reaction + B C + D. E a = 700 kj/mol 400 kj/mol = +300 kj/mol ΔE = 600 kj/mol 400 kj/mol = +00 kj/mol (endothermic)

3 5. What could be responsible for the lowering of the activation energy (as pictured below)? Catalyst. 6. Consider the reaction H O (l) H O (l) + O (g) with the rate law Rate = k[h O ]. How can the rate be increased? Hint: k = e -Ea/RT. 1. Increase the hydrogen peroxide concentration.. Increase k (this can be accomplished by increasing, decreasing E a, increasing T). 7. student obtains.00 M CH 3 COOH (aq) [Ka = 1.8 x 10-5 ]. Determine [CH 3 COOH (aq)], [CH 3 COO - (aq)], and [H + (aq)]. 1. Balanced equation: CH 3 COOH (aq) CH 3 COO - (aq) + H + (aq). Equilibrium expression: K = products/reactants K = [CH 3 COO - (aq)] [H + (aq)] / [CH 3 COOH (aq)] 3. Create an ICE Table: CH 3 COOH (aq) CH 3 COO - (aq) H + (aq) I (initial).00 M 0 0 C (change) -x +x +x E (final; equilibrium).00 x x x K = 1.8 x 10-5 = [x][x] / [.00 x] Lose the x in the denominator:

4 1.8 x 10-5 = [x][x] / [.00 x] 1.8 x 10-5 = [x][x] / [.00] x = [CH 3 COOH (aq)] =.00 x = = M [CH 3 COO - (aq)] = x = M [H + (aq)] = x = M 8. Write the equilibrium law expression for the reaction CO (g) CO (g) + O (g) [ O ][ CO] K c = [ CO ] 9. The following reaction was allowed to come to equilibrium at 755 K. SO (g) + NO (g) NO(g) + SO 3 (g) The equilibrium concentrations were analyzed and found to be: [SO ] = 3.30 M [NO ] = 1.03 M [NO] = 4.44 M [SO 3 ] = 16. M What is the value of K c? K c = [ NO][ SO [ SO 3 ][ NO ] ] = (4.44)(16.) (3.30)(1.03) = For the reaction equation below, K c = 5.0 x 10-0 NO(g) + Cl (g) NOCl(g) Which predominate at equilibrium? () (B) (C) (D) PRODUCTS GRETLY predominate. PRODUCTS moderately predominate. RECTNTS moderately predominate. RECTNTS GRETLY predominate. Because K c is small, the ratio of product over reactants is small, meaning there is little product and much reactant. (E) There are EQUL amounts of PRODUCTS and RECTNTS. 11. For the following reaction NO (g) + I (g ) NOI (g) The partial pressures at equilibrium were: P(NO) = atm P(I ) = 0.8 atm P(NOI) = 1.9 atm What is the value of K p?

5 [ NOI] K p = [ NO] [ I ] = (1.9) (0.448) (0.8) =.3 1. How does a catalyst effect equilibrium? Equilibrium will be reached in less time, but a catalyst does not shift equilibrium. 13. What is meant by dynamic equilibrium? Not static a forward rate indicates reactants are being converted to products and a reverse rate indicates products are being converted to reactants. The rates are equal so concentrations are constant. 14. Which of the following reactions would have the same value for K p and K c? () (B) (C) NO + Br (g) NOBr(g) SO (g) + O (g) SO 3 (g) N (g) + O (g) NO(g) moles of reactants and moles of products. (D) (E) PCl 3 (g) + Cl (g) PCl 5 (g) NH 3 (g) N (g) + 3 H (g) 15. Write the equilibrium law expression for the reaction CrSe(s) + 3 O (g) CrO(s) + SeO (g) [ SeO ] K c = 3 [ O ] (No solids!) 16. The following reaction is at equilibrium: H (g) + Br (g) HBr (g) ΔH o = -7 kj Does the concentration of HBr (g) increase or decrease when the system is heated? Explain. Because heat is given off when the forward reaction occurs, an increase in temperature will shift equilibrium to the left. [HBr] 17. The following reaction is at equilibrium: H (g) + Br (g) HBr (g) ΔH o = -7 kj Does the concentration of Br (g) increase or decrease when H (g) is added? Explain. When H (g) is added, Br (g) will be consumed to produce HBr (g). [Br ]

6 18. The following reactants were combined in a 1.0 L reaction vessel at 337 K: NO (g) + O (g) NO (g) K c = 1.35 fter a short period, the concentrations of reactants and products were found to be as follows: [NO] = 1.31 M, [O ] = 1.0 M, [NO ] = 1.54 M. Is the system at equilibrium? Explain. Q = [ NO ] [ NO] [ O ] = (1.54) (1.31) (1.0) = 1.35 Q = K, so the system is at equilibrium. 19. student obtains M C 6 H 5 COOH (aq) [Ka = 6.6 x 10-5 ]. Determine [C 6 H 5 COOH (aq)], [C 6 H 5 COO - (aq)], and [H + (aq)]. 1. Balanced equation: C 6 H 5 COOH (aq) C 6 H 5 COO - (aq) + H + (aq). Equilibrium expression: K = products/reactants K = [C 6 H 5 COO - (aq)] [H + (aq)] / [C 6 H 5 COOH (aq)] 3. Create an ICE Table: C 6 H 5 COOH (aq) C 6 H 5 COO - (aq) H + (aq) I (initial) M 0 0 C (change) -x +x +x E (final; equilibrium) x x x K = 6.6 x 10-5 = [x][x] / [0.500 x] Lose the x in the denominator: 6.6 x 10-5 = [x][x] / [0.500 x] 6.6 x 10-5 = [x][x] / [0.500] x =

7 [C 6 H 5 COOH (aq)] = x = = M [C 6 H 5 COO - (aq)] = x = M [H + (aq)] = x = M 0. The following data were obtained for the reaction: N O 5 4 NO + O Time [N O 5 ] [NO ] s mol/l mol/l Rate mol/l s = *( ) = M/10 s = 0.00 M/s k 1/s ssume the above data have been obtained by measuring N O 5 concentration at 10 s intervals, during the reaction.. What is the concentration of NO after 10 seconds? 0.0 M (see table for calculation). Why is the concentration 0.0 M and not 0.101M? Stoichiometry! B. What is the average RTE that NO forms during the FIRST 10 s, in mol/l s? 0.0 M / 10 s = 0.00 M/s C. s the reaction proceeds, the RECTION RTE (choose one of the following): () (B) (C) Increases Decreases s the reaction proceeds the rate decreases because less reactant is present. Remains constant 1. For the hypothetical reaction + B C + D The rate law equation is experimentally found to be: Rate = k[] [B] What is the order with respect to reactant?. What is the overall order for the above reaction? 3.. For the hypothetical reaction + B C + D The rate law equation is experimentally found to be: Rate = k[] 3 [B] 0

8 What is the order with respect to reactant? 3. What is the overall order for the above reaction? For the hypothetical reaction + B C + D The rate law equation is experimentally found to be: Rate = k[] 0.5 [B] 3 What is the order with respect to reactant? 0.5. What is the overall order for the above reaction? Consider the following rate law: Rate = k[] [B] 0 What are the units on k for the above rate law? Rate = k[] [B] 0 M/s = k (M) (M) 0 M/s = k M (1) k = (M/s)/M k = 1/M٠s or L/mol٠s 5. For a hypothetical reaction: + B C + D The rate law is experimentally found to be: Rate = k[] 0 [B] What is the reaction rate if k = s -1 and [] = 1.11 M, and [B] = 1.91 M? Rate = ( s -1 )(1.11 M) 0 (1.91 M) = M/s 6. The following are initial rate data for: + B C + D Experiment Initial [] Initial [B] Initial Rate First examine the concentration data to determine for which three experiments [B] was held constant so that the effect of [] could be determined.

9 Then determine for which three experiments [] was held constant so that the effect of [B] could be determined. From these experimental data, what is the order with respect to? Consider trials 1 and 4: [B] was held constant and [] was doubled. The rate quadrupled, revealing []. What is the order with respect to B? Consider trials 1 and : [] was held constant and [B] was doubled. The rate doubled, revealing [B] 1. What is the overall order of this reaction? 3. What is the rate law equation? (Circle the best choice below.) Rate = k[] Rate = k[b] Rate = k[][b] Rate = k[] [B] Rate = k[][b] What is the value of k for these data? Experiment 1 data: Rate = k[] [B] M/s = k[0.0 M] [0.1 M] k = /M ٠s or L /mol ٠s 7. ssume that the following first order reaction FClO FClO + O has a rate constant k = /min. Given initial FClO conc. = 0.73 M What will be the FClO concentration after 16 min? ln o = -kt ln e [ln 0.73 M 0.73 M = -( /min)(16 min) ] [-( /min)(16 min)] = e

10 0.73 M 0.73 M = 0.58 M [-( /min)(16 min)] = e = ssume that the following first order reaction CH 3 CHO CH 4 + CO has a half life of 10 min. If the initial CH 3 CHO conc. is M, what will be the CH 3 CHO concentration after 5 min? ln o = -kt Step 1: Calculate k 1 ln = -k(10 min) = -k(10 min) k = /min ln e [ln M M = -( /min)(5 min) ] [-( /min)(5 min)] = e M M = 0.15 M [-( /min)(5 min)] = e = If a certain reaction had a half-life of 105 s, what UNITS would you assign to the rate constant? 1/s

11 30. Based on the thermodynamic data plotted below, determine the activation energy (E a ) and the change in energy (ΔE) for the reaction + B C + D. E a = 700 kj/mol 600 kj/mol = +100 kj/mol ΔE = 00 kj/mol 600 kj/mol = -400 kj/mol (exothermic) 31. Consider the reaction H O (l) H O (l) + O (g) with the rate law Rate = k[h O ]. How can the rate be increased? Hint: k = e -Ea/RT. 1. Increase the hydrogen peroxide concentration.. Increase k (this can be accomplished by increasing, decreasing E a, increasing Temperature). 3. student obtains.00 M CH 3 COOH (aq) [Ka = 1.8 x 10-5 ]. Determine [CH 3 COOH (aq)], [CH 3 COO - (aq)], and [H + (aq)]. 1. Balanced equation: CH 3 COOH (aq) CH 3 COO - (aq) + H + (aq). Equilibrium expression: K = products/reactants K = [CH 3 COO - (aq)] [H + (aq)] / [CH 3 COOH (aq)] 3. Create an ICE Table: CH 3 COOH (aq) CH 3 COO - (aq) H + (aq) I (initial).00 M 0 0 C (change) -x +x +x E (final; equilibrium).00 x x x

12 K = 1.8 x 10-5 = [x][x] / [.00 x] Lose the x in the denominator: 1.8 x 10-5 = [x][x] / [.00 x] 1.8 x 10-5 = [x][x] / [.00] x = [CH 3 COOH (aq)] =.00 x = = M [CH 3 COO - (aq)] = x = M [H + (aq)] = x = M