Question

As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant pressure and temperature, the volume of the product gases collected is twice the volume of \(\mathrm{NH}_{3}\) reacted. Explain. As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant volume and temperature, the total pressure increases by some factor. Why the increase in pressure and by what factor does the total pressure increase when reactants are completely converted into products? How do the partial pressures of the product gases compare to each other and to the initial pressure of \(\mathrm{NH}_{3}?\)

Short answer

In the decomposition of ammonia gas (\(NH_3\)) to nitrogen gas (\(N_2\)) and hydrogen gas (\(H_2\)), the volume of the product gases is twice the volume of the reacted ammonia gas under constant pressure and temperature. This is because the stoichiometry of the reaction dictates that one mole of \(NH_3\) produces 1/2 mole of \(N_2\) and 3/2 moles of \(H_2\), resulting in a volume ratio of 2. When the reaction occurs at constant volume and temperature, the total pressure increases by a factor of 2. This is due to the increase in the number of moles of gaseous products compared to the reactants. The partial pressure of \(N_2\) is half the initial pressure of \(NH_3\), and the partial pressure of \(H_2\) is three times the initial pressure of \(NH_3\).

Step-by-step solution

1. Relate the volume of product gases to the volume of NH3 under constant pressure and temperature

Since the pressure and temperature are constant, we can use the ideal gas law to relate the volume of product gases to the volume of NH3: \(n_1 = \dfrac{P_1V_1}{RT}\) \(n_2 = \dfrac{P_2V_2}{RT}\) However, the molar ratio of NH3 to N2 and H2 in the reaction is 1:1/2:3/2, so if one mole of NH3 decomposes, it produces 1/2 mole of N2 and 3/2 moles of H2. Hence: \(n_{NH_3} = n_{N_2} + n_{H_2}\)

2. Determine the volume ratio of product gases to NH3 under constant pressure and temperature

Divide the equations above to find the ratio of volumes in terms of n: \(\dfrac{V_2}{V_1} = \dfrac{n_{N_2} + n_{H_2}}{n_{NH_3}}\) To analyze the ratio further, replace the moles of N2 and H2 by the moles of NH3 according to the stoichiometry of the reaction: \(\dfrac{V_2}{V_1} = \dfrac{\frac{1}{2}n_{NH_3} + \frac{3}{2}n_{NH_3}}{n_{NH_3}}\) This simplifies to: \(\dfrac{V_2}{V_1} = 2\) This means that the volume of the product gases collected is twice the volume of NH3 reacted.

3. Analyze the total pressure change under constant volume and temperature

Now let's examine the scenario where the reaction occurs at constant volume and temperature. We will use the ideal gas law again to relate the initial and final pressures: \(P_1V = n_{NH_3}RT\) \(P_2V = (n_{N_2} + n_{H_2})RT\)

4. Determine the pressure increase factor

Again, divide the two equations above: \(\dfrac{P_2}{P_1} = \dfrac{n_{N_2} + n_{H_2}}{n_{NH_3}}\) Now, substitute the moles of N2 and H2 based on the stoichiometry of the reaction: \(\dfrac{P_2}{P_1} = \dfrac{\frac{1}{2}n_{NH_3} + \frac{3}{2}n_{NH_3}}{n_{NH_3}}\) This results in: \(\dfrac{P_2}{P_1} = 2\) So, the total pressure increases by a factor of 2 when the reactants are completely converted into products.

5. Compare the partial pressures of the product gases

Their relative partial pressures are equal to their respective mole fractions. \(P_{N_2} = x_{N_2}P_2\) \(P_{H_2} = x_{N_2}P_2\) The mole fraction can be calculated by dividing the stoichiometric coefficients: \(x_{N_2} = \dfrac{1/2}{1/2 + 3/2} = \dfrac{1}{4}\) \(x_{H_2} = \dfrac{3/2}{1/2 + 3/2} = \dfrac{3}{4}\) Now, multiply the mole fractions by the final pressure to obtain the partial pressures: \(P_{N_2} = \dfrac{1}{4}(2P_1) = \dfrac{1}{2}P_1\) \(P_{H_2} = \dfrac{3}{4}(2P_1) = \dfrac{3}{2}P_1\) Therefore, the partial pressure of N2 is half the initial pressure of NH3, and the partial pressure of H2 is three times the initial pressure of NH3.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental concept in chemistry that relates the properties of gases. It's represented by the equation \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles of gas, \( R \) is the gas constant, and \( T \) is temperature. This formula allows us to understand how gases behave under different conditions. For instance, when decomposition occurs and gases are produced, we can predict changes in volume or pressure as long as temperature and pressure are held constant. The equation demonstrates that the volume of gases produced is directly proportional to the number of moles generated, assuming constant temperature and pressure throughout the reaction.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It involves using balanced chemical equations to deduce relationships between quantities of different substances. In the decomposition of ammonia \( (\mathrm{NH}_3) \), stoichiometry allows us to calculate how much nitrogen \( (\mathrm{N}_2) \) and hydrogen \( (\mathrm{H}_2) \) are produced. The balanced chemical equation shows that 1 mole of \( \mathrm{NH}_3 \) decomposes to produce 1/2 mole of \( \mathrm{N}_2 \) and 3/2 moles of \( \mathrm{H}_2 \). By knowing these ratios, we can accurately predict the amounts of products formed in the reaction, which is crucial when considering changes in volume or pressure.

Partial Pressure

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. Each gas in a mixture contributes to the total pressure in proportion to its amount. When \( \mathrm{NH}_3 \) decomposes into \( \mathrm{N}_2 \) and \( \mathrm{H}_2 \), we can calculate the partial pressures of the resulting gases. The total pressure of the mixture is the sum of these individual pressures. The partial pressure of each gas is determined by its mole fraction multiplied by the total pressure. For example, with our stoichiometry, \( \mathrm{N}_2 \) and \( \mathrm{H}_2 \), having mole fractions of 1/4 and 3/4 respectively, influence their individual pressures. Thus, understanding partial pressures becomes key when predicting how each gas contributes to the overall pressure increase in a closed system.

Molar Ratio

Molar ratio plays a vital role in determining the proportions of reactants and products in a chemical reaction. Given the decomposition of ammonia \( (\mathrm{NH}_3) \), the molar ratio derived from the balanced equation tells us that for every mole of \( \mathrm{NH}_3 \) decomposed, 1/2 mole of \( \mathrm{N}_2 \) and 3/2 moles of \( \mathrm{H}_2 \) are formed. This ratio is essential since it helps predict the number of moles of each gas produced or required in a reaction. By understanding molar ratios, we can determine the relative amounts of product gases, which provides insight into the chemical dynamics and assists in practical applications like calculating changes in volume and pressure.