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Chapter 5: Problem 91

Hydrogen azide, \(\mathrm{HN}_{3}\), decomposes on heating by the following unbalanced reaction: $$ \mathrm{HN}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) $$ If \(3.0 \mathrm{~atm}\) of pure \(\mathrm{HN}_{3}(\mathrm{~g})\) is decomposed initially, what is the final total pressure in the reaction container? What are the partial pressures of nitrogen and hydrogen gas? Assume the volume and temperature of the reaction container are constant.

Short Answer

Expert verified
The final total pressure in the reaction container is 3.0 atm, with a partial pressure of 1.5 atm for nitrogen gas and a partial pressure of 4.5 atm for hydrogen gas.

Step by step solution

01

Balance the reaction equation

To balance the reaction equation, we need to ensure that the number of atoms of each element on each side of the reaction is equal: \[ 2\ \mathrm{HN}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+3\ \mathrm{H}_{2}(g) \] Now, the reaction equation is balanced, with 2 moles of hydrogen azide decomposing to 1 mole of nitrogen gas and 3 moles of hydrogen gas.
02

Identify the initial conditions

The initial conditions are given as follows: - The pressure of the HN3 (P₁) is 3.0 atm - The volume (V) and temperature (T) are constant and do not change during the reaction
03

Calculate the moles of HN3 using the Ideal Gas Law

We can use the Ideal Gas Law formula, PV=nRT, to calculate the moles of HN3 initially present in the container. Rearranging to solve for n (moles): \[ n = \frac{PV}{RT} \] As the volume and temperature remain constant throughout the reaction, we don't need to know their exact values. We can just compare the initial and final pressures to determine the partial pressures of the products: \[ \frac{n_{HN3}\times R \times T}{V} = 3.0\ \mathrm{atm} \]
04

Calculate the moles of N2 and H2 formed after the reaction is complete

From the balanced equation, we know that 2 moles of HN3 will produce 1 mole of N2 and 3 moles of H2. Since we do not know the exact number of moles of HN3 initially present, let's denote the number of moles of HN3 as x. Therefore, if x moles of HN3 decompose, the reaction products will be: - \( \frac{x}{2} \) moles of N2 - \( \frac{3x}{2} \) moles of H2
05

Calculate the final total pressure

Using the ideal gas law, we can determine the pressure of the reaction products formed: - \( P_{N2} = \frac{\frac{x}{2} \times R \times T}{V} \) - \( P_{H2} = \frac{\frac{3x}{2} \times R \times T}{V} \) The final total pressure can be determined by adding the partial pressures of the reaction products: \[ P_{final} = P_{N2} + P_{H2} \] Substituting the expressions for the partial pressures, we get: \[ P_{final} = \frac{xRT}{2V} + \frac{3xRT}{2V} = (\frac{1}{2} + \frac{3}{2})\frac{xRT}{V} = 3.0\ \mathrm{atm}\] It's important to notice that since the initial and final conditions didn't change in terms of the volume and temperature, the final total pressure is equal to the initial pressure, which is 3.0 atm.
06

Determine the partial pressures of N2 and H2

Since the final total pressure is 3.0 atm, we can find the partial pressures of N2 and H2 by using the mole ratios from the balanced equation: - \( P_{N2} = \frac{1}{2} \times P_{final} = \frac{1}{2}(3.0\ \mathrm{atm}) = 1.5\ \mathrm{atm} \) - \( P_{H2} = \frac{3}{2} \times P_{final} = \frac{3}{2}(3.0\ \mathrm{atm}) = 4.5\ \mathrm{atm} \) #Final answer: The final total pressure in the reaction container is 3.0 atm, with a partial pressure of 1.5 atm for nitrogen gas and a partial pressure of 4.5 atm for hydrogen gas.

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Most popular questions from this chapter

A chemist weighed out \(5.14 \mathrm{~g}\) of a mixture containing unknown amounts of \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\) and placed the sample in a \(1.50-\mathrm{L}\) flask containing \(\mathrm{CO}_{2}(g)\) at \(30.0^{\circ} \mathrm{C}\) and 750 . torr. After the reaction to form \(\mathrm{BaCO}_{3}(s)\) and \(\mathrm{CaCO}_{3}(s)\) was completed, the pressure of \(\mathrm{CO}_{2}(g)\) remaining was 230 . torr. Calculate the mass percentages of \(\mathrm{CaO}(s)\) and \(\mathrm{BaO}(s)\) in the mixture.

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One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3} .\) This resulted in a calculated atomic mass of \(13.5\) for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In 1894 , A. Combes (Comptes Rendus 1894, p. 1221\()\) reacted beryllium with the anion \(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows: If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3}\). Show how Combes's data help to confirm that beryllium is a divalent metal.

Consider the following reaction: $$ 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ It takes \(2.00 \mathrm{~L}\) pure oxygen gas at STP to react completely with a certain sample of aluminum. What is the mass of aluminum reacted?

Sulfur trioxide, \(\mathrm{SO}_{3}\), is produced in enormous quantities each year for use in the synthesis of sulfuric acid. $$ \begin{aligned} \mathrm{S}(s)+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{SO}_{2}(g) \\ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{SO}_{3}(g) \end{aligned} $$ What volume of \(\mathrm{O}_{2}(g)\) at \(350 .{ }^{\circ} \mathrm{C}\) and a pressure of \(5.25 \mathrm{~atm}\) is needed to completely convert \(5.00 \mathrm{~g}\) sulfur to sulfur trioxide?
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