The half-life for the first-order decomposition of nitramide, \(\mathrm{NH}_{2} \mathrm{NO}_{2}(\mathrm{aq}) \longrightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(1),\) is \(123 \min\) at \(15^{\circ} \mathrm{C} .\) If \(165 \mathrm{mL}\) of a \(0.105 \mathrm{M} \mathrm{NH}_{2} \mathrm{NO}_{2}\) solution is allowed to decompose, how long must the reaction proceed to yield \(50.0 \mathrm{mL}\) of \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) collected over water at \(15^{\circ} \mathrm{C}\) and a barometric pressure of \(756 \mathrm{mm} \mathrm{Hg} ?\) (The vapor pressure of water at \(15^{\circ} \mathrm{C}\) is \(12.8 \mathrm{mmHg} .)\)

Short Answer

Expert verified
The reaction needs to proceed for 6.6 minutes to yield 50.0 mL of N2O gas.

Step by step solution

01

Calculate moles of nitramide

The initial number of moles of nitramide can be calculated using the formula: Molarity (M) x Volume (L) = moles. Here, the molarity is 0.105 M and the volume is 165 mL or 0.165 L. So, the initial moles of nitramide = 0.105 M x 0.165 L = 0.0173 moles.
02

Calculate moles of N2O

Use the ideal gas law (PV=nRT) to calculate the moles of N2O. First, calculate the partial pressure of N2O by subtracting the vapor pressure of water from the total pressure (756 mmHg - 12.8 mmHg = 743.2 mmHg). Then convert pressure to atm by dividing by 760 (743.2/760 = 0.9774 atm). The volume of N2O gas collected is 50 mL or 0.05 L, and the temperature is 15°C or 288.15 K. R (ideal gas constant) = 0.0821 L.atm/mol.K. Solve the equation for n = PV/RT = (0.9774 atm x 0.05 L) / (0.0821 L.atm/mol.K x 288.15 K) = 0.00207 moles.
03

Calculate the time

The number of moles of nitramide that decayed to for the N2O is the difference between the initial and final moles of nitramide (0.0173 moles - 0.00207 moles = 0.01523 moles). Use the formula for the half-life of first order reactions ( t = (1/k)ln([A]0/[A])), where k is the rate constant (k = 0.693/half-life), [A]0 is the initial concentration and [A] is the final concentration. The half-life is given as 123 min or 2.05 hours. So, k = 0.693/2.05 = 0.338 hr^-1. Convert moles to molarity by dividing by the volume of the solution (0.165 L): [A]0 = 0.105 M and [A] = 0.01523/0.165 = 0.0923 M. Substituting the values in the formula, t = (1/0.338) ln(0.105/0.0923) = 0.11 hr or 6.6 min.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Kinetics
Chemical kinetics is the branch of physical chemistry that is concerned with understanding the rates of chemical reactions. It's essential to know how fast a reaction proceeds to control industrial processes, biological systems, and even the shelf life of many products. A key aspect of kinetics is the half-life of a reaction, which is the time it takes for half of the reactant to undergo the reaction.

In the problem given, the first-order decomposition of nitramide (H_2NO_2) has a half-life that greatly impacts how long it will take to get a certain amount of product. First-order reactions have a constant half-life regardless of the concentration of the reactants, which makes it relatively simple to calculate the time it takes to reach a particular product concentration using the given half-life. The step-by-step solution exemplifies how integral the knowledge of kinetics and the half-life concept is to deduce the duration needed for a specific amount of gas (N_2O) to form from the decomposition of nitramide.
Ideal Gas Law
The ideal gas law is a fundamental equation in physical chemistry that relates the pressure ( P), volume ( V), temperature ( T), and amount (number of moles, n) of an ideal gas. It is usually expressed as PV = nRT, with R being the universal gas constant. The ideal gas law assumes that the gas particles are points with no volume and that there's no interaction between them when in reality, these assumptions are approximations that work well under normal conditions.

In solving our textbook problem, the ideal gas law enables us to calculate the number of moles of nitrogen monoxide (N_2O) by knowing the pressure, volume, and temperature in which the gas is collected. The exercise sets a context that requires us to adjust for the water vapor pressure to find the effective pressure exerted by the gas we're interested in. Once we have the correct pressure, the ideal gas law facilitates the conversion to moles, which is an essential step to figure out how long the nitramide must decompose to yield the 50.0 mL of N_2O.
Rate Constant
The rate constant, often symbolized by k, is a proportionality constant in the rate equation that relates the reaction rate to the concentrations of reactants. For a first-order reaction, the rate constant connects the half-life of the reaction to the concentration of reactants, allowing us to calculate either one if the other is known.

In our exercise, the rate constant is derived from the half-life given for the decomposition of nitramide. By knowing the half-life, we can determine the rate constant using the formula k = 0.693 / t_{1/2}. With this constant, we utilize the integrated rate law for first-order reactions to find out how much time is necessary for a certain amount of reactant to decompose into the product. The rate constant is pivotal as it incorporates both the nature of the chemical reaction and the influence of temperature, making it an invaluable tool when studying the kinetics of a reaction.

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Most popular questions from this chapter

We have seen that the unit of \(k\) depends on the overall order of a reaction. Derive a general expression for the units of \(k\) for a reaction of any overall order, based on the order of the reaction (o) and the units of concentration (M) and time (s).

One example of a zero-order reaction is the decomposition of ammonia on a hot platinum wire, \(2 \mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If the concentration of ammonia is doubled, the rate of the reaction will (a) be zero; (b) double; (c) remain the same; (d) exponentially increase.

The following data are for the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow\) products. Establish the order of this reaction with respect to A and to B. $$\begin{array}{cccc} \hline \text { Expt 1, }[\mathrm{B}]=1.00 \mathrm{M} & & {\text { Expt 2, }[\mathrm{B}]=0.50 \mathrm{M}} \\ \hline \begin{array}{cccc} \text { Time, } \\ \text { min } \end{array} & \begin{array}{c} \text { [A], M } \\ \end{array} & \text { Time, } \text { min } &\text { [A], M } \\ \hline 0 & 1.000 \times 10^{-3} & 0 & 1.000 \times 10^{-3} \\ 1 & 0.951 \times 10^{-3} & 1 & 0.975 \times 10^{-3} \\ 5 & 0.779 \times 10^{-3} & 5 & 0.883 \times 10^{-3} \\ 10 & 0.607 \times 10^{-3} & 10 & 0.779 \times 10^{-3} \\ 20 & 0.368 \times 10^{-3} & 20 & 0.607 \times 10^{-3} \\ \hline \end{array}$$

In the reaction \(A(g) \longrightarrow 2 B(g)+C(g),\) the total pressure increases while the partial pressure of \(\mathrm{A}(\mathrm{g})\) decreases. If the initial pressure of \(\mathrm{A}(\mathrm{g})\) in a vessel of constant volume is \(1.000 \times 10^{3} \mathrm{mmHg}\) (a) What will be the total pressure when the reaction has gone to completion? (b) What will be the total gas pressure when the partial pressure of \(\mathrm{A}(\mathrm{g})\) has fallen to \(8.00 \times 10^{2} \mathrm{mmHg} ?\)

The object is to study the kinetics of the reaction between peroxodisulfate and iodide ions. $$\begin{aligned} &\text { (a) } \mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq}) \end{aligned}$$ The \(I_{3}^{-}\) formed in reaction (a) is actually a complex of iodine, \(\mathrm{I}_{2},\) and iodide ion, \(\mathrm{I}^{-}\). Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) also present in the reaction mixture, reacts with \(\mathrm{I}_{3}^{-}\) just as fast as it is formed. $$\text { (b) } 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}+3 \mathrm{I}^{-}(\mathrm{aq})$$ When all of the thiosulfate ion present initially has been consumed by reaction (b), a third reaction occurs between \(\mathrm{I}_{3}^{-}(\mathrm{aq})\) and starch, which is also present in the reaction mixture. $$\text { (c) } \mathrm{I}_{3}^{-}(\mathrm{aq})+\operatorname{starch} \longrightarrow \text { blue complex }$$ The rate of reaction (a) is inversely related to the time required for the blue color of the starch-iodine complex to appear. That is, the faster reaction (a) proceeds, the more quickly the thiosulfate ion is consumed in reaction (b), and the sooner the blue color appears in reaction (c). One of the photographs shows the initial colorless solution and an electronic timer set at \(t=0 ;\) the other photograph shows the very first appearance of the blue complex (after 49.89 s). Tables I and II list some actual student data obtained in this study. $$\begin{array}{l} \hline\text { TABLE I } \\ \text { Reaction conditions at } 24^{\circ} \mathrm{C}: 25.0 \mathrm{mL} \text { of the } \\ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}_{2} \mathrm{O}_{8}(\text { aq) listed, } 25.0 \mathrm{mL} \text { of the } \mathrm{KI}(\mathrm{aq}) \\ \text { listed, } 10.0 \mathrm{mL} \text { of } 0.010 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(\mathrm{aq}), \text { and } 5.0 \mathrm{mL} \\ \text { starch solution are mixed. The time is that of the } \\ \text { first appearance of the starch-iodine complex. } \\ \hline & \text { Initial Concentrations, } \mathrm{M} \\ \hline \text { Experiment } & \left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}_{2} \mathrm{O}_{8} & \mathrm{KI} & \text { Time, s } \\ \hline 1 & 0.20 & 0.20 & 21 \\ 2 & 0.10 & 0.20 & 42 \\ 3 & 0.050 & 0.20 & 81 \\ 4 & 0.20 & 0.10 & 42 \\ 5 & 0.20 & 0.050 & 79 \\ \hline \end{array}$$ $$\begin{array}{l} \hline \text { TABLE II } \\ \text { Reaction conditions: those listed in Table I for } \\ \text { Experiment } 4, \text { but at the temperatures listed. } \\ \hline \text { Experiment } & \text { Temperature, }^{\circ} \mathrm{C} & \text { Time, } \mathrm{s} \\ \hline 6 & 3 & 189 \\ 7 & 13 & 88 \\ 8 & 24 & 42 \\ 9 & 33 & 21 \\ \hline \end{array}$$ (a) Use the data in Table I to establish the order of reaction (a) with respect to \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) and to I \(^{-}\). What is the overall reaction order? [Hint: How are the times required for the blue complex to appear related to the actual rates of reaction? (b) Calculate the initial rate of reaction in Experiment 1 expressed in \(\mathrm{M} \mathrm{s}^{-1} .\) [Hint: You must take into account the dilution that occurs when the various solutions are mixed, as well as the reaction stoichiometry indicated by equations \((a),(b), \text { and }(c) .]\) (c) Calculate the value of the rate constant, \(k,\) based on experiments 1 and 2 (d) Calculate the rate constant, \(k\), for the four different temperatures in Table II. (e) Determine the activation energy, \(E_{\mathrm{a}}\), of the peroxodisulfate- iodide ion reaction. (f) The following mechanism has been proposed for reaction (a). The first step is slow, and the others are fast. $$\begin{array}{c} \mathrm{I}^{-}+\mathrm{S}_{2} \mathrm{O}_{8}^{2-} \longrightarrow \mathrm{IS}_{2} \mathrm{O}_{8}^{3-} \\ \mathrm{IS}_{2} \mathrm{O}_{8}^{3-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}+\mathrm{I}^{+} \\ \mathrm{I}^{+}+\mathrm{I}^{-} \longrightarrow \mathrm{I}_{2} \\ \mathrm{I}_{2}+\mathrm{I}^{-} \longrightarrow \mathrm{I}_{3}^{-} \end{array}$$ Show that this mechanism is consistent with both the stoichiometry and the rate law of reaction (a). Explain why it is reasonable to expect the first step in the mechanism to be slower than the others.

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