Question

Hydrogen azide, \(\mathrm{HN}_{3},\) decomposes on heating by the following unbalanced equation: $$\mathrm{HN}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2}(g)$$ If 3.0 atm of pure \(\mathrm{HN}_{3}(g)\) 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 6.0 atm, with partial pressures of nitrogen gas and hydrogen gas being 1.5 atm and 4.5 atm, respectively.

Step-by-step solution

Balance the chemical equation

First, we need to balance the given chemical equation: $$\mathrm{HN}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2}(g)$$ Balancing the equation gives: $$2\mathrm{HN}_{3}(g) \longrightarrow \mathrm{N}_{2}(g) + 3\mathrm{H}_{2}(g)$$

Determine mole ratios

From the balanced equation, we can determine the mole ratios of the reactants and products: $$ \begin{cases} 2\, moles\, of\, HN_{3} \longrightarrow 1\, mole\, of\, N_{2} \\ 2\, moles\, of\, HN_{3} \longrightarrow 3\, moles\, of\, H_{2} \end{cases} $$

Determine change in moles

Since the initial pressure of HN3 is 3.0 atm and the volume and temperature are constant, we can use the mole ratios from step 2 to determine the change in moles for each substance: Change in moles of HN3: -3 atm (decreases as it decomposes) Change in moles of N2: +1.5 atm (half of the HN3 decomposes into N2) Change in moles of H2: +4.5 atm (1.5 times the HN3 decomposes into H2)

Calculate final total pressure

Now we add the changes in moles to the initial pressure to find the final total pressure: Final total pressure = Initial HN3 pressure + Change in N2 pressure + Change in H2 pressure Final total pressure = 3.0 atm - 3.0 atm + 1.5 atm + 4.5 atm = 6.0 atm

Calculate partial pressures

Finally, we can calculate the partial pressures of nitrogen and hydrogen gas using the mole fractions and total pressure: Partial pressure of N2 = (N2 moles / Total moles) * Total pressure Partial pressure of N2 = (1.5 / 6.0) * 6.0 atm = 1.5 atm Partial pressure of H2 = (H2 moles / Total moles) * Total pressure Partial pressure of H2 = (4.5 / 6.0) * 6.0 atm = 4.5 atm The final total pressure in the reaction container is 6.0 atm, the partial pressure of nitrogen gas is 1.5 atm, and the partial pressure of hydrogen gas is 4.5 atm.

Key concepts

Partial Pressure

In a mixture of gases, each gas contributes to the overall pressure. This is called partial pressure. Partial pressure is the force that gas molecules exert on the walls of a container. It is calculated as if the gas were the only one present. When gases are present in a mixture, the total pressure can be found by adding their partial pressures.

In the example of hydrogen azide decomposition, we start with 3 atm of \( \text{HN}_3 \), which decomposes to form nitrogen \( \text{N}_2 \) and hydrogen \( \text{H}_2 \). Each resulting gas exerts its own partial pressure. Using the mole fraction ratio from the balanced equation, we found:- Partial pressure of \( \text{N}_2 \) is 1.5 atm- Partial pressure of \( \text{H}_2 \) is 4.5 atm

These partial pressure values add up to give the final total pressure of 6 atm in the container. Understanding partial pressure is crucial for predicting how gases behave when mixed.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the balanced chemical equation, showing the proportionate amounts of each substance involved.

In the decomposition of \( \text{HN}_3 \), stoichiometry helps us understand how many moles of gases are produced. Initially, the reaction is not balanced. After balancing, we have 2 moles of \( \text{HN}_3 \) decomposing to form 1 mole of \( \text{N}_2 \) and 3 moles of \( \text{H}_2 \).

Using this stoichiometric ratio, we can convert between amounts of different substances. This directly influences the calculation of partial pressures. Knowing the relationship between reactants and products is essential for accurately predicting the quantities of materials consumed and produced in a reaction.

Gas Laws

Gas laws are sets of rules that describe the behavior of gases and their response to changes in temperature, volume, and pressure. They include concepts like Boyle's Law, Charles's Law, and the Ideal Gas Law.

In our scenario, the Ideal Gas Law is indirectly applied to understand how the pressure changes. The law is represented as: \[ PV = nRT \]where \( P \) is pressure, \( V \) is volume, \( n \) is number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature.

Though volume and temperature stayed constant in the decomposition reaction, this helps us recognize that pressure changes are predominantly due to changes in moles of gas. When \( \text{HN}_3 \) decomposes, the volume of gas in the container increases, and so does the pressure. The ease of use and predictability of gas laws make them indispensable tools for chemists when analyzing reactions that involve gaseous substances.