Chem 401 Unit 1 (Kinetics & Thermo) Review

Transcription

1 KINETICS 1. For the equation 2 H 2(g) + O 2(g) 2 H 2 O (g) How is the rate of formation of H 2 O mathematically related to the rate of disappearance of O 2? 1 Δ [H2O] Δ[O 2] = 2 Δt Δt 2. Determine the relative reaction rates of the four substances involved in the following chemical reaction. Place the appropriate numbers in the boxes. 3. For the following reaction, the rate of disappearance of A is equal to M/s at the start of the reaction. What are the rates of change for B, C, and D at this time? 2A + B 3C + D Rate of Change of B = M/s Ans: Rate of Change of C = M/s Ans: Rate of Change of D = M/s Ans: Based on the graph below, determine the instantaneous rate of change of [x] at 10 seconds. Instantaneous Rate of Change of X = Ans: rise = (0.11M-0.07M) = 0.04M Run = (15s 5s) = 10s Slope = 0.04/10 = 0.004M/s 1

2 5. Based on the data below, what are the average rates of change of [O 2 ] and [NO 2 ] over the interval 0 to 660 seconds? Time (s) [NO] M [O 2 ] M [NO 2 ] M Average Rate of Change of O 2 = = ( )/(660-0) = -1.5x10-6 M/s Average Rate of Change of NO 2 = = ( )/(660-0) = 3.0x10-6 M/s 6. If the rate constant, k=350 s -1 for a certain reaction, what is the overall order for that reaction? First Order 8. The following reaction occurs in the gaseous state. 2 NO (g) + 2 H 2(g) N 2(g) + 2H 2 O (g) The following rate data were obtained. Trial Initial [NO] Initial [H 2 ] Initial rate(m/s) M 0.20 M M 0.30 M M 0.20 M (a) Determine the rate law for this reaction (including all exponents & rate constant). Trial 2/Trail 1 conc ratio = 1.5 rate ration = 1.5 H 2 coefficient = 1 Trial 3/Trial 1 conc ration = 2.0 rate ratio = 4 NO coefficient = 2 Using trial 3 data to get k, = k(0.40) 2 (0.20) k = = 7.5 M -2 s -1 (0.40) 2 (0.20) Rate Law: rate = 7.5 M -2 s -1 [NO] 2 [H 2 ] (b) What is the order or the reaction with respect to: NO? 2 H 2? 1 Overall? 3 (c) What would the initial rate of the reaction be if the initial concentrations were: [NO] = 0.10M [H 2 ] = 0.30 M Rate = 7.5 M -2 s -1 (0.10M) 2 (0.30M) = M/s 2

3 9. Determine the rate constant of the following reaction at 298K, where A (the Arrhenius pre-exponent factor)=1.2x10 12, and Ea=11.6kJ/mol k = Ae Ea/RT NO (g) + O 3(g) NO 2(g) + O 2(g) k = (1.2x10 12 )e [(-11600J)/(8.314J/K mol)(298k)] = 1.1 x Complete the data table below for the reaction A + B C. Then calculate the slope of the plot of ln k vs 1/T (Arrhenius Plot) and the activation energy of the reaction. Be sure to use unrounded values in your calculations. K (M -1 s -1 ) T (K) Ln k 1/T (K -1 ) Slope = ( ) = = K ( K K -1 ) Ea = (-R)(slope) = ( J/K mol)(-917.2k) =7.63kJ/mol 11. From the following graph, determine the activation energy and the enthalpy of the reaction. Ea = 25 kj/mol ΔH = -150 kj/mol 12. According to the graph below, about what percentage of the collisions will have sufficient kinetic energy to form products? % (50%) 13. In order for a reaction to occur reactant molecules must: A) collide B) with sufficient energy C) in the correct orientation 3

4 14. The mechanism for the reaction 3 ClO - ClO 3 - ClO - + ClO - ClO Cl - ClO - + ClO 2 - ClO Cl - (slow) (fast) Derive the rate law for this reaction: Rate = k[clo - ] Cl - is 15. Identify the catalyst(s) and/or reaction intermediate(s) in the following reaction mechanism. catalyst(s): NO (g) reaction intermediate(s): NO 2(g), O (g) 16. At a certain temperature, the reaction 2B C + D obeys the rate law: rate = 1.14 x 10-3M -1 s -1 [B] 2 If 5.00 mol of B is initially present in a 1.00 L container at that temperature, how much B is left after 117 seconds? We need to start by recognizing the fact that this reaction follows second order kinetics. From our equation sheet: 1/[A] t = kt + 1/[A] o 1-3 [B] t = (1.14 x 10 )(117) [B] t = 1/0.333 = 3.00M = 3.00 mol/l 3.00 mol x 1.00L = 3.00 mol B L for 2 nd order = = Write a rate equation for the following elementary step: BeO H + H 2 BrO 3 + Rate = k [BeO 3 - ][ H + ] 2 Coefficients match exponents for elementary steps and elementary reactions. 18. For a given reaction, the concentration of reactant A was doubled and the rate of reaction remained the same. We can conclude that the reaction is what order with respect to A?. zero order 4

5 THERMODYNAMICS 1. Predict the sign for the following reaction and then calculate the value of ΔS using the information below and compare to your prediction for ΔS. O 2(g) + 2 H 2(g) 2 H 2 O (g) Predicted sign: (-) fewer moles of substance compound S J/mol K O 2(g) H 2(g) H 2 O (g) ΔS = 2(188.7) [(205.0)+2(130.6)] = = J/mol K 2. Predict the value of ΔG under the following conditions: A) ΔH negative and ΔS positive X ΔG < 0 ΔG > 0 Cannot predict the sign of ΔG B) ΔH positive and ΔS positive ΔG < 0 ΔG > 0 X Cannot predict the sign of ΔG C) ΔH positive and ΔS negative ΔG < 0 X ΔG > 0 Cannot predict the sign of ΔG D) ΔH negative and ΔS negative ΔG < 0 ΔG > 0 X Cannot predict the sign of ΔG 3. Determine the value of ΔG in kj at 25 C for the following reaction given that ΔS =326.4J/K and ΔH =571.6kJ. 2 H 2 O (l) 2 H 2(g) + O 2(g) 4. Calculate the change in entropy for the following reaction. (Hint use your appendix data / green sheet.) Ca 2+ (aq) + 2 OH - (aq) Ca(OH) 2(s) ΔS = 83.4 [ (-10.90] = 83.4 (-74.9) = J/K mol 5

6 5. Will the decomposition of potassium chlorate be spontaneous at low temperatures, high temperatures, or all temperatures? Use values from you appendix/handout. Reaction: 2 KClO 3(s) 2 KCl (s) + 3 O 2(g) ΔH f = ΔH rxn = 2(-436.5) 2(-397.7) = kj/mol S = ΔS rxn = [3(205.2) + 2(82.6)]-2(143.1)=+494.6J/mol K Since ΔH rxn is (-), it supports spontaneity. Since ΔS rxn is (+), it supports spontaneity. The reaction is spontaneous at all temperatures! 6. True or False F A) Gases have less entropy than liquids. T B) When solids dissolve they tend to increase in entropy. F C) The greater the moles of solid, the greater the entropy F D) CH 4 (g) has greater entropy than CH 3 -CH 2 -CH 2 -CH 3(g) T E) An endergonic reaction is always non-spontaneous. 6