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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

For the reaction \(A \longrightarrow 2 B+C\), the following data are obtained for \([\mathrm{A}]\) as a function of time: \(t=0 \mathrm{min}\) \([\mathrm{A}]=0.80 \mathrm{M} ; 8 \mathrm{min}, 0.60 \mathrm{M} ; 24 \mathrm{min}, 0.35 \mathrm{M} ; 40 \mathrm{min}\) \(0.20 \mathrm{M}\) (a) By suitable means, establish the order of the reaction. (b) What is the value of the rate constant, \(k ?\) (c) Calculate the rate of formation of \(\mathrm{B}\) at \(t=30 \mathrm{min}\).

The following statements about catalysis are not stated as carefully as they might be. What slight modifications would you make in them? (a) A catalyst is a substance that speeds up a chemical reaction but does not take part in the reaction. (b) The function of a catalyst is to lower the activation energy for a chemical reaction.

The rate constant for the reaction \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \longrightarrow\) \(2 \mathrm{HI}(\mathrm{g})\) has been determined at the following temperatures: \(599 \mathrm{K}, k=5.4 \times 10^{-4} \mathrm{M}^{-1} \mathrm{s}^{-1} ; 683 \mathrm{K}, k=2.8 \times 10^{-2} \mathrm{M}^{-1} \mathrm{s}^{-1} .\) Calculate the activation energy for the reaction.

In the first-order reaction \(A \longrightarrow\) products, \([\mathrm{A}]=0.816 \mathrm{M}\) initially and \(0.632 \mathrm{M}\) after \(16.0 \mathrm{min}.\) (a) What is the value of the rate constant, \(k ?\) (b) What is the half-life of this reaction? (c) At what time will \([\mathrm{A}]=0.235 \mathrm{M} ?\) (d) What will [A] be after 2.5 h?

If even a tiny spark is introduced into a mixture of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g}),\) a highly exothermic explosive reaction occurs. Without the spark, the mixture remains unreacted indefinitely. (a) Explain this difference in behavior. (b) Why is the nature of the reaction independent of the size of the spark?

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