Questions: Hydrogen azide, HN3, decomposes on heating by the following unbalanced equation: HN3(g) → N2(g)+H2(g) If 2.0 atm of pure HN3(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. Total pressure = atm Partial pressure of N2 = atm Partial pressure of H2 = atm

Solution Steps

First, we need to balance the given chemical equation: \[ \mathrm{HN}_{3}(g) \rightarrow \mathrm{N}_{2}(g) + \mathrm{H}_{2}(g) \]

Balancing the nitrogen atoms: \[ \mathrm{HN}_{3}(g) \rightarrow \mathrm{1.5N}_{2}(g) + \mathrm{H}_{2}(g) \]

To avoid fractional coefficients, we multiply the entire equation by 2: \[ 2\mathrm{HN}_{3}(g) \rightarrow 3\mathrm{N}_{2}(g) + \mathrm{H}_{2}(g) \]

We start with 2.0 atm of pure $\mathrm{HN}_{3}(g)$.

Since the volume and temperature are constant, the pressure is directly proportional to the number of moles of gas. According to the balanced equation, 2 moles of $\mathrm{HN}_{3}$ decompose to produce 3 moles of $\mathrm{N}_{2}$ and 1 mole of $\mathrm{H}_{2}$, totaling 4 moles of gas.

The initial pressure of $\mathrm{HN}_{3}$ is 2.0 atm. When it decomposes, the total number of moles of gas increases from 2 moles to 4 moles. Therefore, the final total pressure is: \[ \text{Final Total Pressure} = 2.0 \, \text{atm} \times \frac{4 \, \text{moles}}{2 \, \text{moles}} = 4.0 \, \text{atm} \]

The partial pressures are proportional to the number of moles of each gas produced.

• Partial pressure of $\mathrm{N}_{2}$: \[ \text{Partial Pressure of } \mathrm{N}_{2} = 4.0 \, \text{atm} \times \frac{3 \, \text{moles of } \mathrm{N}_{2}}{4 \, \text{moles total}} = 3.0 \, \text{atm} \]

• Partial pressure of $\mathrm{H}_{2}$: \[ \text{Partial Pressure of } \mathrm{H}_{2} = 4.0 \, \text{atm} \times \frac{1 \, \text{mole of } \mathrm{H}_{2}}{4 \, \text{moles total}} = 1.0 \, \text{atm} \]

Final Answer