Rates of Reaction Important to know how fast a reaction goes. This information is only comes from experimental data. Some reactions are very fast ( µ seconds) others very slow ( years ). What factors affect the reaction rates ? 1) Concentration of reactants and products. 2) Surface area of solid or liquid reactants 3) Temperature 4) pH 5) Catalysis 6) Presence of inhibitors. Lets take a simple reaction A B Reaction rate = change in [A] with time = - ∆ [A] ∆ t note: Greek letter delta, ∆ means “change in” -ve sign because [A] is decreasing units of reaction rate = mol l –1 s–1 = M s–1 Instead of following the change in [reactant] we could follow change in [product] as :- - ∆ [A] = ∆ [B] ∆ t ∆ t Rate equation Rate of reaction is related to concentration by the rate equation: - ∆ [A] = k [A] n ∆ t Where k is termed the rate constant and n is the order of the reaction. k and n can only be determined by experiment the order of a reaction (n) is normally 0, 1 or 2 Some reactions follow 1st order kinetics i.e. the rate depends on the concentration of a single reactant ∴ - ∆ [A] = k [A]1 ∆ t what are the units of the rate constant ? units of rate = (units of rate constant) (units of conc.) ∴ units of rate constant = units of rate / units of conc. = M s-1 / M = s-1 for 1st order reactions k has units of time –1 ( e.g. s –1 or min. –1 ) Determination of k Difficult to obtain an accurate measurement of k as the graph is in the form of a hyperbola (curve). ∴ need to re-arrange the rate equation to give a straight line graph. - ∆ [A] = k [A] ∆ t This equation can transformed into an equation that relates concentration of A at the start of a reaction, [A]0, to its concentration at any other time t, [A]t ln [A]t - ln [A]o = -kt ln [A]t = -kt + ln [A]o ∴ = an equation for a straight line ( y = mx + c ) Plot the natural log ( ln ) for concentration against time and obtain a negative sloped straight line. We don’t have to use concentration [A] in the equation we could use pressure, UV absorbance, radioactive decay etc. to determine a reaction rate. This form of the equation also allows us to calculate the fraction of reactant remaining after a given time. e.g. What fraction of H2O2 remains after 25 mins ? ( k = 2.8 x 10-2 min –1 ) ln [A]0 / [A]t = k t = 2.8 x 10-2 . 25 = 0.7 ∴ [A]0 / [A]t = 2.02 ( not 5.01 remember ln not log10 ) ≈ so [A]t ½ [ A]0 Half – Life Reactions may be characterised by their half-life, i.e. the time taken for the concentration to drop to one half of the initial concentration. ≈ [A]t ½ [ A]0 or k t½ = 0.693 = ln 2 for a 1st order reaction t½ = 0.693 / k it is independent of the initial concentration of reactants. 2nd Order Reactions the rate depends on the reactant concentration squared or on the concentration of two reactants. 2 A C the reaction rate = k [A]2 i.e. if the concentration of A is doubled, the rate is x 4 units of rate = (units of rate constant) (units of conc.)2 units of rate constant = units of rate / (units of conc.)2 = M s-1 / M2 = M-1 s-1 and A + B D reaction rate = k [A].[B] in the case 2 A C To obtain a straight line graph we need to integrate the rate equation between t and t = 0 - ∆ [A] = k [A] 2 ∆ t 1 / [A]t = kt + 1/[A]0 ∴ Plot 1 / [A]t vs time to get a positive sloped graph where k = gradient. What is the half-life of a 2nd order reaction? t½ = 1 / k [A]0 nd remember the units for t½ (for a 2 order reaction) are M-1 s-1 You cannot predict the reaction order (n) from the reaction stoichiometry !!! e.g. NO2 + CO NO + CO2 ≠ Reaction rate k [NO2] . [CO] 2 Experimentally determined rate = k [NO2] Catalysts Homogeneous:- normally occurring in solution Heterogeneous:- taking place at the interface between two phases. Biological (enzyme) catalysts :- part homogeneous, part heterogeneous in character. Enzymes are very large organic molecules that present very complicated kinetic patterns as they catalyse a variety of vital life functions. - extremely sensitive (easily denatured) - efficient (can be 10,000 times faster than chemical catalysts) - highly selective : ability of an enzyme to select certain functional groups out of many possible. - highly specific : this is a property of the reaction catalysed by the enzyme Enzyme Kinetics This can provide information about the way enzymes catalyse reactions. Velocity of reactions is measured by the amount of substrate [S] reacting or product [P] formed under specific conditions. Enzyme activity is usually measured in µ moles of substrate transformed in one 1 minute. The Michaelis-Menton model for enzyme kinteics assumes the following steps are involved in an enzyme reaction: E + S ES E + P reversible formation of the enzyme-substrate complex ES, followed by the conversion to product. we assume : 1. Enzyme catalyst is unchanged by the reaction 2. The total concentration of enzyme is constant throughout the reaction. 3. production formation is irreversible Unfortunately the reaction order can be complicated and vary with conditions e.g. pH, temperature and [S]. Derivation of the rate equation uses a kinetic criterion known as the “steady state approximation”. That is in the enzymatic reaction the concentration of intermediate, ES, remains constant. i.e. rate of formation of ES = rate of ES breakdown ∴ K1 [E] [S] = K2 [ES] + K –1 [ES] total amount of enzyme [E]o = [E] + [ES] K1 [E] [S] = K2 [ES] + K –1 [ES] ∴ [ES] = K1 [E]o [S] / K1 [S] + K2 + K –1 rate of production of P = K2 [ES] v = K2 [E]o [S] ( K2 + K –1 / K1 ) + [S] v = Kcat [E]o [S] = vmax [S] Km + [S] Km + [S] Kcat is the turnover number. The number of molecules turned over by one molecule of enzyme per second. - good indicator of how quick the enzyme is. Km is the Michaelis constant for the enzyme and is the concentration of substrate at which half maximal rate is observed. - rough indicator of how tightly an enzyme binds its substrate Values of Kcat and Km can be measured for an enzyme by measuring the rate while varying the substrate conc. At high substrate conc. [S] >> Km ∴ The rate equation reduces to v = Kcat [E]o = vmax vmax = the maximum enzyme rate As with the 1st and 2nd order rate equations we need to represent the rate equation as a straight line. The most common form is the Lineweaver-Burk Plot 1 / v = Km /[S] . 1 / vmax + 1/ vmax A plot of 1 / v against 1/[S] gives a straight line of slope Km/Vmax and intercepts on the x and y axis of – 1/Km and 1/Vmax respectively.