1. Kinetics 2000D By: Charles Liao

2. 2000D a.Referring to the data in the table below, calculate the standard enthalpy change, for the reaction at 25˚C. O 3(g) + NO (g)  O 2(g) + NO 2(g) ∆H =33 ( kJ mol-1) -(143+90) ( kJ mol-1) =-200 ( kJ mol-1) O 3 (g) NO (g) NO 2 (g) Standard enthalpy of formation,  H at 25  C (kJ mol -1 ) 143 kJ mol -1 90 kJ mol -1 33 kJ mol -1

3. 2000D b. Make a qualitative prediction about the magnitude of the standard entropy change, ∆S˚, for the reaction at 25˚C. Justify your answer. A very small change is present because both sides of the equation contain 2 moles of gas. Because Entropy is the measure of disorder and molecules in their gas state contain the least amount of order, very little change is observed from 2 moles to 2 moles.

4. 2000D c. On the basis of your answers to parts (a) and (b), predict the sign of the standard free-energy change, for the reaction at 25˚C. Explain your reasoning. Because the reaction’s ∆S˚ is very little and the equation to determine free energy change is ∆G˚= ∆H˚-T ∆S˚, it can be assumed that with a negative ∆H˚ and at 25˚C or 298˚K, that the reaction is spontaneous. By having a spontaneous reaction, ∆G is inherently Negative

5. 2000D d. Use the information in the table below to write the rate-law expression for the reaction, and explain how you obtained your answer. k= [O 3 ][NO] Based on Experiments 1 and 2, by doubling the [NO] concentration, the rate doubles. Therefore the reaction is first order with respect for [NO]. The doubling of the [O3] concentration also doubles the rate therefore making the reaction a first order with respect to [O3] as well. Experiment Number Initial [O 3 ] (mol L -1 ) Initial [NO] (mol L -1 ) Initial Rate of Formation of [NO 2 ] (mol L -1 s -1 ) 10.001 x 2 0.0022x 30.0020.0012x 40.002 4x

6. 2000D e. The following three-step mechanism is proposed for the reaction. Identify the step that must be the slowest in order for this mechanism to be consistent with the rate-law expression derived in part (d).  Step I:O3 + NO  O + NO3  Step II:O+ O3  2 O2  Step III:NO3 + NO  2 NO2 In Part D, it was found that the Rate-law expression was k= [O 3 ][NO]. Therefore the slow step of the reaction must be Step 1. The Slow step is the rate-determining step in equations.