Zero-Order ReactionsA zero-order reaction is a reaction that proceeds at a rate that is independent of reactant concentration. Typically with increasing or decreasing reactants concentrations not afecting the observed reaction. This means that the rate of the reaction is equal to the rate constant, k, of that reaction. This property differs from both first-order reactions and second-order reactions. The Differential Form of the Rate Law\[\large Rate = - \dfrac{d[A]}{dt} = k[A]^0 = k = constant\] where \(Rate\) is the reaction rate and \(k\) is the reaction rate coefficient. In this example, the units of \(k\) are M/s. The units can vary with other types of reactions. For zero-order reactions, the units of the rate contants are always M/s. In higher order reactions we will see how \(k\) will have different units. The Integrated Form of the Rate LawReaction:A → P Rate Law \[Rate = k[A]^n\] First, write the differential form of the rate law with n=1 \[Rate = - \dfrac{d[A]}{dt} = k\] then rearrange \[{d}[A] = -kdt\] Second, integrate both sides of the equation. \[\displaystyle \int_{[A]_{0}}^{[A]} d[A] = - \int_{0}^{t} kdt\] Third, solve for \([A]\). This provides the integrated form of the rate law. \[[A] = [A]_0 -kt\] The integrated form of the rate law allows us to find the population of reactant at any time after the start of the reaction. For more information on differential and integrated rate laws, see rate laws and rate constants. This derived formula is the main important one to understand, the math to get to it does not necessarily need to be understood (though knowing it would always be better). Graphing Zero-order Reactions\[[A] = -kt + [A]_0\] is in the form y = mx+b where slope = m = -k and the y-intecercept = b = \([A]_0\) Zero-order reactions are only applicable for a very narrow region of time. Therefore, the linear graph shown below (Figure 1) is only realistic over a limited time range. If we were to extrapolate the line of this graph downward to represent all values of time for a given reaction, it would tell us that as time progresses, the concentration of our reactant becomes negative. We know from general chemistry that concentrations can never be negative, which is why zero-order reactions give us a brief window into looking at the reaction.
To understand where the above graph comes from, let us consider a catalyzed reaction. At the beginning of the reaction, and for small values of time, the rate of the reaction is constant. This is indicated by the blue line in both graph above and below (Figures 1 and 2, respectively). This situation typically happens when a catalyst is saturated with reactants. With respect to Michaelis-menton Kinetics, this point of catalyst saturation is related to the Vmax. As a reaction progresses through time, however, it is possible that less and less substrate will bind to the catalyst. As this occurs, the reaction slows and we see a tailing off of the graph (Figure 2). This portion of the reaction is represented by the dashed black line. In looking at this particular reaction, we can see that reactions are not zero-order under all conditions. They are only zero-order for a limited amount of time.
If we plot rate as a function of time, we obtain the graph below (Figure 3). Again, this only describes a narrow region of time. The slope of the graph is equal to k, the rate constant. Therefore, k is constant with time. In addition, we can see that the reaction rate is completely independent of how much reactant you put in.
Relationship Between Half-life and Zero-order ReactionsThe half-life is a timescale in which each half-life represents the reduction of the initial population to 50% of its original state. We can represent the relationship by the following equation. \[[A] = \dfrac{1}{2} [A]_o\] Using the integrated form of the rate law, we can develop a relationship between zero-order reactions and the half-life. \[[A] = [A]_o - kt\] Substitute \[\frac{1}{2}[A]_o = [A]_o - kt_{\frac{1}{2}}\] Solve for time, \(t_{\frac{1}{2}}\) \[t_{\frac{1}{2}} = \dfrac{[A]_o}{2k}\] Notice that, for zero-order reactions, the half-life depends on the initial concentration of reactant and the rate constant.
Practice Questions
Reaction A: k = 2.3 M-1s-1 Reaction B: k = 1.8 Ms-1 Reaction C: k = 0.75 s-1 Which reaction represents a zero-order reaction?
Answers1. The rate constant k is 0.00624 M/s 2. The half-life is 96 seconds. 3. Since this is a zero-order reaction, the half-life is dependent on the concentration. In this instance, the half-life is decreased when the original concentration is reduced to 1.0 M. The new half-life is 80 seconds. 4. Reaction B represents a zero-order reaction because the units are in M/s. Zero-order reactions always have rate constants that are represented by molars per unit of time. Higher order reactions, however, require the rate constant to be represented in different units. 5. True. When using the rate function \( rate = k[A]^n \) with n equal to zero in zero-order reactions. Therefore, rate is equal to the rate constant k. SummaryThe kinetics of any reaction depend on the reaction mechanism, or rate law, and the initial conditions. If we assume for the reaction A -> Products that there is an initial concentration of reactant of [A]0 at time t=0, and the rate law is an integral order in A, then we can summarize the kinetics of the zero-order reaction as follows:
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