Question

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

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

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.

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

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: \[ \frac{n_{HN3}\times R \times T}{V} = 3.0\ \mathrm{atm} \]

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

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.

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.

Key concepts

Partial Pressures

To understand the concept of partial pressures in the context of gas reactions, it is important to know that in a mixture of gases, each gas contributes to the total pressure of the system. This contribution is called the "partial pressure" of that gas. Each gas in a mix exerts its pressure as if it alone occupied the entire volume. The sum of these partial pressures gives us the total pressure.

In the exercise, we're asked to determine the partial pressures of nitrogen ( _2") and hydrogen ( _2") produced from the decomposition of hydrogen azide ( _2H_{3}"). The reaction proceeds until all reactants are converted to products, and since the volume and temperature don't change, partial pressures can be directly calculated using stoichiometry from the balanced equation.

Partial pressures are useful for gauging the amount of each gas present. They can be found using the formula for each gas: P_{i} = rac{n_iRT}{V} or, in our context, relating it to the mole fraction of each gas in the total reaction.

Reaction Stoichiometry

Reaction stoichiometry involves using the balanced chemical equation to find the relationships between the quantities of reactants and products. In our given exercise, the initial unbalanced equation is: \(_3 \rightarrow_2 +_2\).

When balancing, we obtain: 2 ext{HN}_{3} ightarrow ext{N}_{2} + 3 ext{H}_{2}. This tells us that two moles of hydrogen azide yield one mole of nitrogen and three moles of hydrogen. Such mole ratios are essential in calculating the amount of reactants needed or products formed in a chemical reaction.

By knowing the initial condition of 3.0 atm of 2, stoichiometry helps us understand how much 2 and 2 are produced when the reaction reaches completion. These ratios also let us determine the individual partial pressures by multiplying them with the mole ratio fraction to the total pressure.

Gas Reactions

Gas reactions, particularly those involving ideal gases, are interesting because their behaviors are well-described by the Ideal Gas Law: PV = nRT. This law interrelates pressure ( P"), volume ( V"), and temperature ( T") with the amount of gas in moles ( )", and the gas constant ( R").

In the gas reaction described in the exercise, HN_{3} decomposes into N_{2} and H_{2}. Since the container's volume and temperature are constant, these reaction conditions make it possible to determine the pressure of the gases using stoichiometry without needing the exact amounts of gas in moles.

Gas reactions are fundamental in chemistry because they illustrate how changes in conditions, like pressure, can affect reaction outcomes. This concept helps in predicting how a reaction will behave in different environmental conditions, which is crucial for industrial chemical processes.