Chemical Kinetics and Enzyme Catalysis

Integrated rate laws and linear plots

An integrated rate law expresses the reactant concentration as a function of time, so a suitable plot of the data gives a straight line whose slope or intercept yields the rate constant k. Each order has its own linear form, and the order is identified by finding which plot is linear. For a zero-order reaction, [A] = [A]_0 - kt, so a plot of [A] against t is a straight line with slope -k. For a first-order reaction, \ln([A]_0/[A]) = kt, so a plot of \ln([A]_0/[A]) against t is a straight line through the origin with slope k. For a second-order reaction with rate = k[A]^2, 1/[A] - 1/[A]_0 = kt, so a plot of 1/[A] against t is a straight line with slope k. Recognizing the order from experimental data means checking which of these plots gives a straight line.

For a second-order reaction of the type 2\mathrm{A} \rightarrow products, the rate law is -\frac{1}{2}\frac{d[A]}{dt} = k[A]^2, which integrates to 1/[A] = 1/[A]_0 + 2kt. The slope of the 1/[A] versus t plot is therefore 2k, and the rate constant must be obtained as half the slope, not from the slope directly.

For a first-order reaction the time needed to reach a fraction x of completion is t_x = \frac{1}{k}\ln\frac{1}{1-x}. Because t_{50}/t_{90} = \ln 2/\ln 10 \approx 0.30, the time to 90% completion is about 3.3 times the time to 50% completion, and this ratio is independent of the initial concentration.

• integrated rate law — an equation giving [A] as a function of t for a given order. 예) For a first-order decay, [A] = [A]_0e^{-kt}.
• linearization — rearranging the integrated rate law so that it has the form y = mx + c. 예) For first order, plot \ln[A] against t with slope -k.
• stoichiometric factor in slope — for 2\mathrm{A} consumption, the slope of 1/[A] against t equals 2k. 예) If the slope is 0.04\ \mathrm{M^{-1}s^{-1}}, then k = 0.02\ \mathrm{M^{-1}s^{-1}}.

Michaelis-Menten enzyme kinetics

The Michaelis-Menten mechanism is \mathrm{E} + \mathrm{S} \rightleftharpoons \mathrm{ES} \rightarrow \mathrm{E} + \mathrm{P}. Applying the steady-state approximation to the enzyme-substrate complex gives v = \frac{V_{max}[S]}{K_M + [S]}, where V_{max} = k_2[E]_0 and K_M = (k_{-1} + k_2)/k_1. The rate of product formation is v = k_2[\mathrm{ES}], so it is proportional to the concentration of the complex, not to the free enzyme or the substrate alone.

The maximum rate V_{max} depends on k_2 and the total enzyme concentration, and it does not depend on the substrate concentration. K_M is the substrate concentration at which the rate equals half of V_{max}. At low substrate concentration, [S] \ll K_M, so v \approx (V_{max}/K_M)[S], which is first order in substrate and also first order in enzyme because V_{max} is proportional to [E]_0. At high substrate concentration, [S] \gg K_M, the enzyme is saturated, v \approx V_{max}, and the reaction is zero order in substrate.

In a Lineweaver-Burk plot, the equation is 1/v = (K_M/V_{max})(1/[S]) + 1/V_{max}. The y-intercept equals 1/V_{max}, and the slope equals K_M/V_{max}, so K_M = \text{slope} / \text{intercept}.

A competitive inhibitor resembles the substrate and binds reversibly to the substrate-binding (active) site of the enzyme, so it competes with the substrate for that site. It does not bind to the substrate itself. Because the competition can be overcome by adding more substrate, a competitive inhibitor becomes less effective as the substrate concentration rises. Its effect is to increase the apparent K_M, K_{M,app} = K_M(1 + [I]/K_i), while V_{max} remains unchanged.