CH 104: CHEMICAL KINETICS Chemical kinetics is the study of the rates of reactions. The rate of a reaction is the change in concentration per unit of time. Some reactions are very fast. For example, H3O+(aq) + OH–(aq) → 2H2O(l) is completed in about 0.0000001 seconds. Some reactions are very slow. For example, 2H2(g) + O2(g) → 2H2O(l) is completed in about 1,000,000,000 years. Which is these reactions is faster, (a) Na(s) and Br2(l), or (b) the rusting of Fe(s)? (a) Na(s) and Br2(l)

CHEMICAL KINETICS Factors affecting the rates of chemical reactions: Nature of reactants Presence or absence of catalysts Solvent Concentration of reactants Temperature In Part A of today’s experiment you will measure the affect of the concentration of reactants on rate. In Part B of today’s experiment you will measure the affect of temperature on rate.

aA + bB + cC + … → dD + eE + fF + … CHEMICAL KINETICS Given the following general reaction: aA + bB + cC + … → dD + eE + fF + … The rate equation equals: This rate has been arbitrarily defined as the disappearance of A (–Δ[A]/Δt). However, it could have been defined as the disappearance of any reactant, or the appearance of any product. m need not equal a, n need not equal b, etc. m is “the order in A”, n is “the order in B”, etc. m + n + p + … is “the overall order” m, n, p, etc. usually equals 0, 1, or 2; however, they may also equal 1/2, 3/2, etc. k is the specific rate constant. It is a constant for any given reaction in a specific solvent and at a specific temperature. What does k equal when all the concentrations are 1 M? Rate = k[1]m[1]n[1]p Rate = k

CHEMICAL KINETICS AND CONCENTRATION S2O82–(aq) + 3I–(aq) → 2SO42–(aq) + I3–(aq) The method of initial rates is used to measure the orders of a reaction. For example, the order in S2O82–(aq) is measured as follows. Step #1: To find the order in S2O82–(aq), select the experiments with different initial concentrations of S2O82–(aq) and equal concentrations of I–(aq). What are these experiments? Experiments 1 and 2. In Part A of today’s experiment you must assign the initial concentrations 3 different of reactants (CH3COCH3, I2, and H+). How will you do this so that you can measure the order of each reactant? Step #2: Use the ratio of these rate equations to solve for the order in S2O82–(aq). 2 = 2m m = 1 Therefore, the order in S2O82–(aq) is 1. Experiment Initial Concentrations, M Initial Rate, mol L–1 s–1 S2O82–(aq) I–(aq) 1 2 3 0.038 0.076 0.060 0.030 R1 = 1.4 x 10–5 R2 = 2.8 x 10–5 R3 = 1.4 x 10–5

CHEMICAL KINETICS AND CONCENTRATION S2O82–(aq) + 3I–(aq) → 2SO42–(aq) + I3–(aq) What is the order in I–(aq)? Step #1: To find the order in I–(aq), select the experiments with different initial concentrations of I–(aq) and equal concentrations of S2O82–(aq). What are these experiments? Experiments 2 and 3. Step #2: Use the ratio of these rate equations to solve for the order in I–(aq). 2 = 2n n = 1 Therefore, the order in I–(aq) is also 1. Experiment Initial Concentrations, M Initial Rate, mol L–1 s–1 S2O82–(aq) I–(aq) 1 2 3 0.038 0.076 0.060 0.030 R1 = 1.4 x 10–5 R2 = 2.8 x 10–5 R3 = 1.4 x 10–5

CHEMICAL KINETICS AND CONCENTRATION S2O82–(aq) + 3I–(aq) → 2SO42–(aq) + I3–(aq) What is the overall order? The Order in S2O82–(aq) + The Order in I–(aq) = 1 + 1 = 2 Therefore, the overall order is 2. Experiment Initial Concentrations, M Initial Rate, mol L–1 s–1 S2O82–(aq) I–(aq) 1 2 3 0.038 0.076 0.060 0.030 R1 = 1.4 x 10–5 R2 = 2.8 x 10–5 R3 = 1.4 x 10–5

CHEMICAL KINETICS AND CONCENTRATION S2O82–(aq) + 3I–(aq) → 2SO42–(aq) + I3–(aq) What is the rate constant (k) for this reaction? Rate = k[S2O82–]1[I–]1 1.4 x 10–5 = k[0.038][0.060] Experiment Initial Concentrations, M Initial Rate, mol L–1 s–1 S2O82–(aq) I–(aq) 1 2 3 0.038 0.076 0.060 0.030 R1 = 1.4 x 10–5 R2 = 2.8 x 10–5 R3 = 1.4 x 10–5

CHEMICAL KINETICS AND CONCENTRATION S2O82–(aq) + 3I–(aq) → 2SO42–(aq) + I3–(aq) What is the rate of this reaction when [S2O82–] = 0.050 M and [I–] = 0.025 M? Rate = (6.1 x 10–3 L mol–1 s–1)[S2O82–]1[I–]1 Rate = (6.1 x 10–3 L mol–1 s–1)[0.050][0.025] Rate = 7.7 x 10–6 mol L–1 s–1 Experiment Initial Concentrations, M Initial Rate, mol L–1 s–1 S2O82–(aq) I–(aq) 1 2 3 0.038 0.076 0.060 0.030 R1 = 1.4 x 10–5 R2 = 2.8 x 10–5 R3 = 1.4 x 10–5

CHEMICAL KINETICS AND TEMPERATURE For example, fuels such as gasoline, oil, and coal are relatively inert at room temperature; however, they rapidly burn at elevated temperatures. In addition, many foods last almost indefinitely in a freezer; however, they spoil quickly at room temperature. In Part B of today’s experiment you will measure the affect of temperature on rate. Experience tells us that the rates of reactions increase with temperature.

CHEMICAL KINETICS AND TEMPERATURE The activation energy (Ea) is the minimum energy that is needed for molecules to react. In other words, Ea is the height of the energy barrier between reactants and products.

CHEMICAL KINETICS AND TEMPERATURE Svante Arrhenius noted that the temperature dependence of the specific rate constant is mathematically similar to the Boltzmann distribution of energies.

CHEMICAL KINETICS AND TEMPERATURE The Arrhenius equation describes the relationship between the specific rate constant (k), the activation energy (Ea), and the absolute temperature (T). A graph of ln k versus 1/T is called an Arrhenius plot. It is a straight line with slope of m = –Ea/R and a y-intercept of b = ln A. k is the specific rate constant. Ea is the activation energy. R is the gas constant, 8.314 J mol–1 K–1. T is the temperature in Kelvin. A is a constant for a given reaction.

CHEMICAL KINETICS AND TEMPERATURE Calculate the Ea for this reaction. 2HI(g) → H2(g) + I2(g) Step #1: Complete this table. k (M–1 s–1) ln k t (°C) T (K) 1/T (K–1) 3.52 x 10–7 3.02 x 10–5 2.19 x 10–4 1.16 x 10–3 3.95 x 10–2 283 356 393 427 508 –14.860 556 0.00180 –10.408 629 0.00159 –8.426 666 0.00150 –6.759 700 0.00143 –3.231 781 0.00128

CHEMICAL KINETICS AND TEMPERATURE Step #2: Use Excel to plot ln k versus 1/T. Then calculate the slope (–Ea/R) of this Arrhenius plot.

CHEMICAL KINETICS AND TEMPERATURE Step #3: Calculate Ea. Slope = (–Ea/R) –Ea = (Slope)R Ea = –(Slope)R Ea = –(–2.24 x 104 K)( 8.314 J mol–1 K–1) Ea = 1.86 x 105 J mol–1 Ea = 186 kJ mol–1

SAFETY Give at least 1 safety concern for the following procedure. Using acetone (CH3COCH3), hydrochloric acid (HCl), and iodine (I2). These are irritants. Wear your goggles at all times. Immediately clean all spills. If you do get either of these in your eye, immediately flush with water. Acetone is extremely flammable. Never use it near a flame or spark. Your laboratory manual has an extensive list of safety procedures. Read and understand this section. Ask your instructor if you ever have any questions about safety.

SOURCES Barnes, D.S., J.A. Chandler. 1982. Chemistry 111-112 Workbook and Laboratory Manual. Amherst, MA: University of Massachusetts. McMurry, J., R.C. Fay. 2004. Chemistry, 4th ed. Upper Saddle River, NJ: Prentice Hall. Petrucci, R.H. 1985. General Chemistry Principles and Modern Applications, 4th ed. New York, NY: Macmillan Publishing Company.