16. (EX-14-8B) The following data are obtained for the reaction A \( \rightarrow \) P. What are the order of this reaction and its rate constant, \( k \) ? Xác ??nh b?c và h?ng s? k c?a ph?n ?ng \( A \rightarrow P \), v?i d? li?u ??ng h?c sau: \begin{tabular}{ccccccc} \( \mathrm{t} \) (phút) & 0 & \( 4.22 \) & \( 6.6 \) & \( 10.61 \) & \( 14.48 \) & \( 18.00 \) \\ \hline\( [\mathrm{A}] \) (M) & \( 0.250 \) & \( 0.210 \) & \( 0.188 \) & \( 0.150 \) & \( 0.114 \) & \( 0.083 \) \end{tabular} 17. (EX-14-9A) What is the half-life of the first-order decomposition of \( \mathrm{N}_{2} \mathrm{O}_{5} \) at \( 75.0^{\circ} \mathrm{C} \) ? With \( E_{a}=1,06 \times 10^{5} \mathrm{~J} / \mathrm{mol} \) and \( k=3,46 \times 10^{-5} \mathrm{~s}^{-1} \) at \( 298 \mathrm{~K} \) Ph?n ?ng phân h?y \( N_{2} O_{5} \) trong \( C C l_{4} \) tuân theo ??ng h?c b?c 1. Có \( E_{a}=1,06 \times 10^{5} \mathrm{~J} / \mathrm{mol}, \mathrm{k}= \) \( 3,46 \times 10^{-5} \mathrm{~s}^{-1} \) ? \( 298 \mathrm{~K} \). Tìm th?i gian bán h?y c?a ph?n ?ng phân h?y \( \mathrm{N}_{2} \mathrm{O}_{5} \) ? \( 75,0^{\circ} \mathrm{C} \). \( \mathrm{N}_{2} \mathrm{O}_{5}\left(\right. \) in \( \left.\mathrm{CCl}_{4}\right) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}\left(\right. \) in \( \left.\mathrm{CCl}_{4}\right)+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \) 18. (EX-14-9B) At what temperature will it take \( 1.50 \mathrm{~h} \) for two-thirds of a sample of \( \mathrm{N}_{2} \mathrm{O}_{5} \) in to decompose in Example \( 17 . \) Ph?n ?ng phân h?y \( N_{2} O_{5} \) trong \( C C l_{4} \) tuân theo ??ng h?c b?c \( 1 . \) Có \( E_{a}=1,06 \times 10^{5} \mathrm{~J} / \mathrm{mol}, \mathrm{k}= \) \( 3,46 \times 10^{-5} \mathrm{~s}^{-1} \) ? \( 298 \mathrm{~K} \). Xác ??nh nhi?t ?? ph?n ?ng ?? \( 2 / 3 \) l??ng m?u \( \mathrm{N}_{2} \mathrm{O}_{5} \) trong \( \mathrm{CCl}_{4} \) phân h?y trong \( 1,5 \mathrm{~h} \) ? 19. (EX-14-10A) In a proposed two-step mechanism for the reaction \( \mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{~g}) \rightarrow \) \( \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{NO}(\mathrm{g}) \), the second, fast step is \( \mathrm{NO}_{2}(\mathrm{k})+\mathrm{CO}(\mathrm{k}) \rightarrow \mathrm{CO}_{2}(\mathrm{k})+\mathrm{NO}_{2}(\mathrm{k}) . \) What must be the slow step? What would you expect the rate law of the reaction to be? Explain. C? ch? c?a ph?n ?ng \( \mathrm{CO}(k)+\mathrm{NO}_{2}(k) \rightarrow \mathrm{CO}_{2}(k)+\mathrm{NO}(k) \) ???c ?? xu?t g?m 2 giai ?o?n. Giai ?o?n nhanh là \( \mathrm{NO}_{3}(k)+\mathrm{CO}(k) \rightarrow \mathrm{CO}_{2}(k)+\mathrm{NO}_{2}(k) \). Theo b?n, ph?n ?ng giai ?o?n ch?m x?y ra nh? th? nào? Ph??ng trình ??ng h?c c?a ph?n ?ng này là gì? 20. (EX-14-10B) Show that the proposed mechanism for the reaction \( 2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow 2 \) \( \mathrm{NO}_{2} \mathrm{~F}(\mathrm{~g}) \) \( \mathrm{NO}_{2}(\mathrm{~g}) \) Sã chép
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- A chemical reaction with stoichiometry $\mathrm{A} \rightarrow$ products is said to follow an $n^{\text {th }}$ -order rate law if $\mathrm{A}$ is consumed at a rate proportional to the $n$ th power of its concentration in the reaction mixture. If $r_{\mathrm{A}}$ is the rate of consumption of A per unit reactor volume, then $$r_{\mathrm{A}}[\operatorname{mol} /(\mathbf{L} \cdot \mathbf{s})]=k C_{\mathrm{A}}^{n}$$ where $C_{A}(m o l / L)$ is the reactant concentration, and the constant of proportionality $k$ is the reaction rate constant. A reaction that follows this law is referred to as an $n^{\text {th }}$ order reaction. The rate constant is a strong function of temperature but is independent of the reactant concentration. (a) Suppose a first-order reaction ( $n=1$ ) is carried out in an isothermal batch reactor of constant volume $V$. Write a material balance on $A$ and integrate it to derive the expression $$C_{\mathrm{A}}=C_{\mathrm{A} 0} \exp (-k t)$$ where $C_{\mathrm{A} 0}$ is the concentration of $\mathrm{A}$ in the reactor at $t=0$ (b) The gas-phase decomposition of sulfuryl chloride $$\mathrm{SO}_{2} \mathrm{Cl}_{2} \rightarrow \mathrm{SO}_{2}+\mathrm{Cl}_{2}$$ is thought to follow a first-order rate law. The reaction is carried out in a constant-volume isothermal batch reactor and the concentration of $\mathrm{SO}_{2} \mathrm{Cl}_{2}$ is measured at several reaction times, with the following results: Verify the proposed rate law graphically [i.e., demonstrate that the expression given in Part (a) fits the data for $\left.C_{\mathrm{A}}(t)\right]$ and determine the rate constant $k,$ giving both its value and its units.
A gas-phase decomposition reaction with stoichiometry $2 \mathrm{A} \rightarrow 2 \mathrm{B}+\mathrm{C}$ follows a second-order rate law (see Problem 10.19): $$r_{\mathrm{d}}\left[\operatorname{mol} /\left(\mathrm{m}^{3} \cdot \mathrm{s}\right)\right]=k C_{\mathrm{A}}^{2}$$ where $C_{\mathrm{A}}$ is the reactant concentration in $\mathrm{mol} / \mathrm{m}^{3}$. The rate constant $k$ varies with the reaction temperature according to the Arrhenius law $$k\left[\mathrm{m}^{3} /(\mathrm{mol} \cdot \mathrm{s})\right]=k_{0} \exp (-E / R T)$$ where $k_{0}\left[\mathrm{m}^{3} /(\mathrm{mol} \cdot \mathrm{s}]\right)=$ the preexponential factor $E(\mathrm{J} / \mathrm{mol})=$ the reaction activation energy $R=$ the gas constant $T(\mathrm{K})=$ the reaction temperature (a) Suppose the reaction is carried out in a batch reactor of constant volume $V\left(\mathrm{m}^{3}\right)$ at a constant temperature $T(\mathrm{K}),$ beginning with pure $\mathrm{A}$ at a concentration $C_{\mathrm{A} 0} .$ Write a differential balance on A and integrate it to obtain an expression for $C_{\mathrm{A}}(t)$ in terms of $C_{\mathrm{A} 0}$ and $k$ (b) Let $P_{0}(\text { atm })$ be the initial reactor pressure. Prove that $t_{1 / 2}$, the time required to achieve a $50 \%$ conversion of $\mathrm{A}$ in the reactor, equals $R T / k P_{0},$ and derive an expression for $P_{1 / 2},$ the reactor pressure at this point, in terms of $P_{0} .$ Assume ideal-gas behavior. (c) The decomposition of nitrous oxide $\left(\mathrm{N}_{2} \mathrm{O}\right)$ to nitrogen and oxygen is carried out in a 5.00 -liter batch reactor at a constant temperature of $1015 \mathrm{K},$ beginning with pure $\mathrm{N}_{2} \mathrm{O}$ at several initial pressures. The reactor pressure $P(t)$ is monitored, and the times $\left(t_{1 / 2}\right)$ required to achieve $50 \%$ conversion of $\mathrm{N}_{2} \mathrm{O}$ are noted. $$\begin{array}{|c|c|c|c|c|} \hline P_{0}(\mathrm{atm}) & 0.135 & 0.286 & 0.416 & 0.683 \\ \hline t_{1 / 2}(\mathrm{s}) & 1060 & 500 & 344 & 209 \\ \hline \end{array}$$ Use these results to verify that the $\mathrm{N}_{2} \mathrm{O}$ decomposition reaction is second-order and determine the value of $k$ at $T=1015 \mathrm{K}$ (d) The same experiment is performed at several other temperatures at a single initial pressure of 1.00 atm, with the following results: $$\begin{array}{|c|c|c|c|c|} \hline T(\mathrm{K}) & 900 & 950 & 1000 & 1050 \\ \hline t_{1 / 2}(\mathrm{s}) & 5464 & 1004 & 219 & 55 \\ \hline \end{array}$$ Use a graphical method to determine the Arrhenius law parameters ( $k_{0}$ and $E$ ) for the reaction. (e) Suppose the reaction is carried out in a batch reactor at $T=980 \mathrm{K},$ beginning with a mixture at 1.20 atm containing 70 mole $\%$ N $_{2}$ O and the balance a chemically inert gas. How long (minutes) will it take to achieve a $90 \%$ conversion of $\mathrm{N}_{2} \mathrm{O} ?$
The graphing calculator can run a program that can tell you the order of a chemical reaction, provided you indicate the reactant concentrations and reaction rates for two experiments involving the same reaction. Go to Appendix C. If you are using a TI-83 Plus, you can download the program RXNORDER and run the application as directed. If you are using another calculator, your teacher will provide you with key-strokes and data sets to use. At the prompts, enter the reactant concentrations and reaction rates. Run the program as needed to find the order of the following reactions. (All rates are given in M/s.) a. $2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$ $\mathrm{N}_{2} \mathrm{O}_{5} :$ conc. $1=0.025 \mathrm{M} ;$ conc. $2=0.040 \mathrm{M}$ rate $1=8.1 \times 10^{-5} ;$ rate $2=1.3 \times 10^{-4}$ b. $2 \mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$ $\mathrm{NO}_{2} : \mathrm{conc.} 1=0.040 \mathrm{M} ; \mathrm{conc} .2=0.080 \mathrm{M}$ rate $1=0.0030 ;$ rate $2=0.012$ c. $2 \mathrm{H}_{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g)$ $\mathrm{H}_{2} \mathrm{O}_{2} :$ conc. $1=0.522 \mathrm{M} ;$ conc. $2=0.887 \mathrm{M}$ rate $1=1.90 \times 10^{-4} ;$ rate $2=3.23 \times 10^{-4}$ d. $2 \mathrm{NOBr}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$ NOBr: conc. $1=1.27 \times 10^{-4} \mathrm{M} ;$ conc. $2=$ $4.04 \times 10^{-4} \mathrm{M}$ rate $1=6.26 \times 10^{-5} ;$ rate $2=6.33 \times 10^{-4}$ e. $2 \mathrm{HI}(g) \rightarrow \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)$ HI: conc. $1=4.18 \times 10^{-4} \mathrm{M} ;$ conc. $2=$ $8.36 \times 10^{-4} \mathrm{M}$ rate $1=3.86 \times 10^{-5} ;$ rate $2=1.54 \times 10^{-4}$
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