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Learning Objectives
You learned that the integrated rate law for each common type of reaction (zeroth, first, or second order in a single reactant) can be plotted as a straight line. Using these plots offers an alternative to the methods described for showing how reactant concentration changes with time and determining reaction order. We will illustrate the use of these graphs by considering the thermal decomposition of NO2 gas at elevated temperatures, which occurs according to the following reaction: \[\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \label{14.26}\] Experimental data for this reaction at 330°C are listed in Table \(\PageIndex{1}\); they are provided as [NO2], ln[NO2], and 1/[NO2] versus time to correspond to the integrated rate laws for zeroth-, first-, and second-order reactions, respectively. Table \(\PageIndex{1}\): Concentration of NO2 as a Function of Time at 330°CTime (s)[NO2] (M)ln[NO2]1/[NO2] (M−1)01.00 × 10−2−4.605100606.83 × 10−3−4.9861461205.18 × 10−3−5.2631931804.18 × 10−3−5.4772392403.50 × 10−3−5.6552863003.01 × 10−3−5.8063323602.64 × 10−3−5.937379The actual concentrations of NO2 are plotted versus time in part (a) in Figure \(\PageIndex{1}\). Because the plot of [NO2] versus t is not a straight line, we know the reaction is not zeroth order in NO2. A plot of ln[NO2] versus t (part (b) in Figure \(\PageIndex{1}\)) shows us that the reaction is not first order in NO2 because a first-order reaction would give a straight line. Having eliminated zeroth-order and first-order behavior, we construct a plot of 1/[NO2] versus t (part (c) in Figure \(\PageIndex{1}\)). This plot is a straight line, indicating that the reaction is second order in NO2.  We have just determined the reaction order using data from a single experiment by plotting the concentration of the reactant as a function of time. Because of the characteristic shapes of the lines shown in Figure \(\PageIndex{2}\), the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method of initial rates required multiple experiments at different NO2 concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions. Example \(\PageIndex{1}\) Dinitrogen pentoxide (N2O5) decomposes to NO2 and O2 at relatively low temperatures in the following reaction: \[2N_2O_5(soln) → 4NO_2(soln) + O_2(g)\] This reaction is carried out in a CCl4 solution at 45°C. The concentrations of N2O5 as a function of time are listed in the following table, together with the natural logarithms and reciprocal N2O5 concentrations. Plot a graph of the concentration versus t, ln concentration versus t, and 1/concentration versus t and then determine the rate law and calculate the rate constant. Time (s)[N2O5] (M)ln[N2O5]1/[N2O5] (M−1)00.0365−3.31027.46000.0274−3.59736.512000.0206−3.88248.518000.0157−4.15463.724000.0117−4.44885.530000.00860−4.75611636000.00640−5.051156Given: balanced chemical equation, reaction times, and concentrations Asked for: graph of data, rate law, and rate constant Strategy: A Use the data in the table to separately plot concentration, the natural logarithm of the concentration, and the reciprocal of the concentration (the vertical axis) versus time (the horizontal axis). Compare the graphs with those in Figure \(\PageIndex{1}\) to determine the reaction order. B Write the rate law for the reaction. Using the appropriate data from the table and the linear graph corresponding to the rate law for the reaction, calculate the slope of the plotted line to obtain the rate constant for the reaction. Solution A Here are plots of [N2O5] versus t, ln[N2O5] versus t, and 1/[N2O5] versus t: The plot of ln[N2O5] versus t gives a straight line, whereas the plots of [N2O5] versus t and 1/[N2O5] versus t do not. This means that the decomposition of N2O5 is first order in [N2O5]. B The rate law for the reaction is therefore rate = k[N2O5] Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a first-order reaction is −k. We can calculate the slope using any two points that lie on the line in the plot of ln[N2O5] versus t. Using the points for t = 0 and 3000 s, \[\begin{align*} \textrm{slope} &=\dfrac{\ln[\mathrm{N_2O_5}]_{3000}-\ln[\mathrm{N_2O_5}]_0}{3000\textrm{ s}-0\textrm{ s}} \\[4pt] &=\dfrac{(-4.756)-(-3.310)}{3000\textrm{ s}} \\[4pt] &=-4.820\times10^{-4}\textrm{ s}^{-1} \end{align*}\] Thus \(k = 4.820 \times 10^{−4}\, s^{−1}\). Exercise \(\PageIndex{1}\) 1,3-Butadiene (CH2=CH—CH=CH2; C4H6) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C4H6 as a function of time at 326°C are listed in the following table along with ln[C4H6] and the reciprocal concentrations. Graph the data as concentration versus t, ln concentration versus t, and 1/concentration versus t. Then determine the reaction order in C4H6, the rate law, and the rate constant for the reaction. Time (s)[C4H6] (M)ln[C4H6]1/[C4H6] (M−1)01.72 × 10−2−4.06358.19001.43 × 10−2−4.24769.918001.23 × 10−2−4.39881.336009.52 × 10−3−4.65410560007.30 × 10−3−4.920137Answersecond order in C4H6; rate = k[C4H6]2; k = 1.3 × 10−2 M−1·s−1 SummaryFor a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of −k. For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of −k. For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of k. This page titled 5.7: Using Graphs to Determine Integrated Rate Laws is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous. When a graph is plotted for log k vs 1 t?If we plot a graph between log K and `(1)/(T)` by Arrhenius equation , the slope is. Solution : ln k = ln `- (E_(a))/(RT)` is Arrhenius equation . Thus plots of ln k vs 1/T will give slope = `-E_(a)//RT` or `-E_(a)// 2.303R`.
Which of the following curves between log k and 1 t?The graph between `log k` versus `1//T` is a straight line.
What is the slope of graph of log K vs 1 T?Therefore Slope of logk vs 1/T graph is given by 2.
What is the slope of LNK vs 1 T?So now we can use it to calculate the Activation Energy by graphing lnk versus 1/T. When the lnk (rate constant) is plotted versus the inverse of the temperature (kelvin), the slope is a straight line. The value of the slope (m) is equal to -Ea/R where R is a constant equal to 8.314 J/mol-K.
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What is the shape of the graph between log k vs 1 T?
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