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Chapter 14: Problem 10

At \(65^{\circ} \mathrm{C}\), the half-life for the first-order decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})\) is 2.38min. $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ If \(1.00 \mathrm{g}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is introduced into an evacuated \(15 \mathrm{L}\) flask at \(65^{\circ} \mathrm{C}\) (a) What is the initial partial pressure, in \(\mathrm{mmHg}\), of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) ?\) (b) What is the partial pressure, in \(\mathrm{mmHg}\), of \(\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g})\) after \(2.38 \mathrm{min} ?\) (c) What is the total gas pressure, in \(\mathrm{mm} \mathrm{Hg}\), after \(2.38 \mathrm{min} ?\)

Short Answer

Expert verified
The initial pressure of \(N_{2}O_{5}(g)\) is 11.92mmHg, the partial pressure after 2.38min is 5.96mmHg and the total gas pressure after 2.38min is 23.84mmHg.

Step by step solution

01

Calculation of initial partial pressure of N2O5

To calculate the initial partial pressure, the Ideal Gas Law is used. The formula for Ideal Gas Law is \(PV=nRT\). Here \(P\) is pressure, \(V\) is volume, \(n\) is number of moles, \(R\) is universal gas constant and \(T\) is the temperature. Given, mass of \(N_{2}O_{5}\) = 1.00 g, Volume = 15L, and temperature = \(65^{\circ}C = 65 + 273 = 338K\). Molecular weight of \(N_{2}O_{5} = 2*14(N) + 5*16(O) = 108 g/mol.\) Thus, number of moles \(n = \frac{Mass}{Molar\, mass} = \frac{1.00 g}{108 g/mol} = 0.00926 mol\). To get the pressure in \(\mathrm{mmHg}\), the value for \(R\) we are going to use is 62.4 \(L.mmHg/K.mol\). The equation then becomes \(P = \frac{nRT}{V} = \frac{0.00926 mol * 62.4 L.mmHg/K.mol * 338K}{15L} = 11.92mmHg\).
02

Calculation of partial pressure of N2O5 after 2.38 min

Given that \(N_{2}O_{5}(g)\) decomposes with a half-life of 2.38min, after one half-life, half of the original \(N_{2}O_{5}(g)\) remains. Therefore, the pressure of the remaining \(N_{2}O_{5}(g)\) would also be halved. Hence, partial pressure of \(N_{2}O_{5}\) after 2.38min = \(11.92mmHg / 2 = 5.96mmHg\).
03

Calculation of total gas pressure after 2.38 min

The reaction given implies that for each molecule of \(N_{2}O_{5}\) that decomposes, two molecules of \(NO_{2}(g)\) and half a molecule of \(O_{2}(g)\) are generated. Therefore, for half of the \(N_{2}O_{5}(g)\) that decomposes, an equal pressure of \(O_{2}(g)\) and twice that pressure of \(NO_{2}(g)\) is produced. Therefore, the total pressure after 2.38 min = remaining \(N_{2}O_{5}\) pressure + \(O_{2}(g)\) pressure + \(NO_{2}(g)\) pressure = 5.96mmHg (from Step 2) + 5.96mmHg + 2*5.96mmHg = 23.84mmHg.

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Most popular questions from this chapter

The following data are obtained for the initial rates of reaction in the reaction \(A+2B+C \longrightarrow 2 D+E.\) $$\begin{array}{lllll} \hline & \text { Initial } & \text { Initial } & & \\ \text { Expt } & \text { [A], M } & \text { [B],M } & \text { [C], M } & \text { Initial Rate } \\ \hline 1 & 1.40 & 1.40 & 1.00 & R_{1} \\ 2 & 0.70 & 1.40 & 1.00 & R_{2}=\frac{1}{2} \times R_{1} \\ 3 & 0.70 & 0.70 & 1.00 & R_{3}=\frac{1}{4} \times R_{2} \\ 4 & 1.40 & 1.40 & 0.50 & R_{4}=16 \times R_{3} \\ 5 & 0.70 & 0.70 & 0.50 & R_{5}=? \\ \hline \end{array}$$ (a) What are the reaction orders with respect to A, B, and C? (b) What is the value of \(R_{5}\) in terms of \(R_{1} ?\)

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For the reaction \(A \longrightarrow\) products the following data are obtained. $$\begin{array}{cll} \hline {\text { Experiment 1 }} & &{\text { Experiment 2 }} \\ \hline [\mathrm{A}]=1.204 \mathrm{M} & t=0 \mathrm{min} & {[\mathrm{A}]=2.408 \mathrm{M}} & t=0 \mathrm{min}\\\ {[\mathrm{A}]=1.180 \mathrm{M}} & t=1.0 \mathrm{min} & {[\mathrm{A}]=?} & t=1.0 \mathrm{min} \\ {[\mathrm{A}]=0.602 \mathrm{M}} & t=35 \mathrm{min} & {[\mathrm{A}]=?} & t=30 \mathrm{min} \\ \hline \end{array}$$ (a) Determine the initial rate of reaction in Experiment 1. (b) If the reaction is second order, what will be \([\mathrm{A}]\) at \(t=1.0\) min in Experiment 2? (c) If the reaction is first order, what will be \([\mathrm{A}]\) at 30 min in Experiment 2?

The reaction \(A+B \longrightarrow\) products is first order in \(A\) first order in \(\mathrm{B},\) and second order overall. Consider that the starting concentrations of the reactants are \([\mathrm{A}]_{0}\) and [ \(\mathrm{B}]_{0},\) and that \(x\) represents the decrease in these concentrations at the time \(t .\) That is, \([\mathrm{A}]_{t}=[\mathrm{A}]_{0}-x\) and \([\mathrm{B}]_{t}=[\mathrm{B}]_{0}-x .\) Show that the integrated rate law for this reaction can be expressed as shown below. $$\ln \frac{[\mathrm{A}]_{0} \times[\mathrm{B}]_{t}}{[\mathrm{B}]_{0} \times[\mathrm{A}]_{t}}=\left([\mathrm{B}]_{0}-[\mathrm{A}]_{0}\right) \times k t$$
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