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Chapter 14: Problem 66

The decomposition of \(\mathrm{N}_{2} \mathrm{O}\) to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is a first-order reaction. At \(730^{\circ} \mathrm{C}\) the half-life of the reaction is \(3.58 \times\) \(10^{3} \mathrm{~min}\). If the initial pressure of \(\mathrm{N}_{2} \mathrm{O}\) is \(2.10 \mathrm{~atm}\) at \(730^{\circ} \mathrm{C},\) calculate the total gas pressure after one halflife. Assume that the volume remains constant.

Short Answer

Expert verified
The total gas pressure after one half-life is \(3.15\) atm.

Step by step solution

01

Understand the reaction and half-life

First, it’s vital to understand the reaction. The reaction is: \(N_2O \rightarrow N_2 + O_2\). On decomposition, one molecule of \(N_2O\) produces two molecules of gas, one each of \(N_2\) and \(O_2\). So, in terms of pressure, one unit of pressure of \(N_2O\) should result in two units of pressure of the resultant gases given volume remains constant. Also, after one half-life, half of the \(N_2O\) will have decomposed.
02

Calculate the remaining \(N_2O\)

Let's calculate the pressure of the remaining \(N_2O\). After one half-life, half of the \(N_2O\) will have decomposed. Thus, there remains \(2.10 / 2 = 1.05\) atm of \(N_2O\).
03

Calculate the amount of \(N_2\) and \(O_2\) formed

Half of the original \(N_{2}O\) that is, \(2.10/2 = 1.05\) atm of \(N_{2}O\), has decomposed. By the stoichiometry of the reaction, this amount would produce double the amount of gas pressure, that is, \(2 \times 1.05 = 2.10\) atm.
04

Calculate the total pressure

Now that we have the pressure of the undecomposed \(N_2O\) and the pressure of the gas molecules formed, we need to add these two to get the total pressure. So, \(1.05 + 2.10 = 3.15\) atm.

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