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Chapter 14: Problem 24

The following first-order reaction occurs in \(\mathrm{CCl}_{4}(1)\) at \(45^{\circ} \mathrm{C}: \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) .\) The rate constant is \(k=6.2 \times 10^{-4} \mathrm{s}^{-1} .\) An \(80.0 \mathrm{g}\) sample of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in \(\mathrm{CCl}_{4}(\mathrm{l})\) is allowed to decompose at \(45^{\circ} \mathrm{C}.\) (a) How long does it take for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to be reduced to \(2.5 \mathrm{g} ?\) (b) How many liters of \(\mathrm{O}_{2},\) measured at \(745 \mathrm{mmHg}\) and \(45^{\circ} \mathrm{C},\) are produced up to this point?

Short Answer

Expert verified
It takes approximately 2,403 seconds for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to be reduced to \(2.5 g\). The volume of \(\mathrm{O}_{2}\) produced up to this point is approximately 0.384 L at 745 mmHg and \(45^{\circ}C\).

Step by step solution

01

Apply First-Order Reaction Formula

The integrated rate law for a first-order reaction is given by: \(\log[\mathrm{N}_{2} \mathrm{O}_{5}]_{t} = -kt + \log[\mathrm{N}_{2} \mathrm{O}_{5}]_{0}\). We need to calculate the time it takes for \(80.0 g\) of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to reduce to \(2.5 g\). The rate constant \(k\) is \(6.2 \times 10^{-4} s^{-1}\). Converting the masses to moles (molar mass of \(\mathrm{N}_{2} \mathrm{O}_{5} = 108 g/mol\)), we get \([\mathrm{N}_{2} \mathrm{O}_{5}]_{0} = 80.0/108\) mol and \([\mathrm{N}_{2} \mathrm{O}_{5}]_{t} = 2.5/108\) mol.
02

Solve for Time

Substitute the values into the integrated rate law and solve for \(t\). The obtained answer is the time it takes for the quantity of \(\mathrm{N}_{2} \mathrm{O}_{5}\) to reduce to 2.5 g.
03

Use the Ideal Gas Law

To find the volume of \(\mathrm{O}_{2}\) produced up to this point, use the ideal gas law: \(PV = nRT\). The pressure \(P\) can be converted to atm (745 mmHg = 0.98 atm). The temperature \(T\) should be in Kelvin, so convert \(45^{\circ}C\) to 318 K. The universal gas constant \(R\) is 0.0821 L·atm/K·mol. The number of moles \(n\) can be calculated as the difference between initial and final moles of \(\mathrm{N}_{2} \mathrm{O}_{5}\) divided by 2 (because the stoichiometry of the reaction shows that 1 mol \(\mathrm{N}_{2} \mathrm{O}_{5}\) produces \(\frac{1}{2}\) mol \(\mathrm{O}_{2}\)).

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Most popular questions from this chapter

The reaction \(A \longrightarrow\) products is first order in A. (a) If \(1.60 \mathrm{g} \mathrm{A}\) is allowed to decompose for 38 min, the mass of A remaining undecomposed is found to be 0.40 g. What is the half-life, \(t_{1 / 2}\), of this reaction? (b) Starting with \(1.60 \mathrm{g} \mathrm{A},\) what is the mass of \(\mathrm{A}\) remaining undecomposed after \(1.00 \mathrm{h} ?\)

The decomposition of nitric oxide occurs through two parallel reactions: $$\mathrm{NO}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad k_{1}=25.7 \mathrm{s}^{-1}$$ $$\mathrm{NO}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\frac{1}{4} \mathrm{O}_{2}(\mathrm{g}) \quad k_{2}=18.2 \mathrm{s}^{-1}$$ (a) What is the reaction order for these reactions? (b) Which reaction is the slow reaction? (c) If the initial concentration of \(\mathrm{NO}(\mathrm{g})\) is \(2.0 \mathrm{M},\) what is the concentration of \(\mathrm{N}_{2}(\mathrm{g})\) after 0.1 seconds? (d) If the initial concentration of \(\mathrm{NO}(\mathrm{g})\) is \(4.0 \mathrm{M},\) what is the concentration of \(\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) after 0.025 seconds?

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 \(12.8 \mathrm{mmHg} .)\)

According to collision theory, chemical reactions occur through molecular collisions. A unimolecular elementary process in a reaction mechanism involves dissociation of a single molecule. How can these two ideas be compatible? Explain.

For the reaction \(A+2 B \longrightarrow 2 C\), the rate of reaction is \(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\) at the time when \([\mathrm{A}]=0.3580 \mathrm{M}.\) (a) What is the rate of formation of \(\mathrm{C}\) ? (b) What will \([\mathrm{A}]\) be 1.00 min later? (c) Assume the rate remains at \(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\) How long would it take for \([\mathrm{A}]\) to change from 0.3580 to \(0.3500 \mathrm{M} ?\)
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