Question

One way to prevent emission of the pollutant NO from industrial plants is by a catalyzed reaction with \(\mathrm{NH}_{3}\) : $$4 \mathrm{NH}_{3}(g)+4 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \stackrel{\text { catalyst }}{\longrightarrow} 4 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ (a) If the NO has a partial pressure of \(4.5 \times 10^{-5}\) atm in the flue gas, how many liters of \(\mathrm{NH}_{3}\) are needed per liter of flue gas at \(1.00 \mathrm{~atm} ?(\mathrm{~b})\) If the reaction takes place at \(1.00 \mathrm{~atm}\) and \(365^{\circ} \mathrm{C}\) how many grams of \(\mathrm{NH}_{3}\) are needed per kiloliter (kL) of flue gas?

Short answer

0.045 L and 0.00149 g

Step-by-step solution

Write the balanced chemical equation

The balanced chemical equation is given as: \[ 4 \text{NH}_3 (g) + 4 \text{NO} (g) + \text{O}_2 (g) \stackrel{\text{catalyst}}{\longrightarrow} 4 \text{N}_2 (g) + 6 \text{H}_2 \text{O} (g) \]

Determine the partial pressures ratio for NH3 and NO

From the balanced equation, 4 moles of \text{NH}_3\react with 4 moles of \text{NO}\ Thus, the partial pressure ratio for \text{NH}_3\to\text{NO}\ is 1:1. Therefore, the partial pressure of \text{NH}_3\.is \(4.5 \times 10^{-5}\) atm.

Calculate volume of NH3 needed per liter of flue gas

Use the Ideal Gas Law: \(PV = nRT\). Since pressure and temperature are constant, volume is directly proportional to the number of moles. For every liter of flue gas, the same volume (in liters) of \text{NH}_3\ is needed. Thus, for 1 liter of flue gas, \(4.5 \times 10^{-5}\) liters of \text{NH}_3\ is needed.

Calculate the amount of NH3 needed at given conditions

Given: P = 1.00 atm, \(T = 365^\text{\textdegree}\text{C} = 365 + 273 = 638 K\). The volume of NH3 per kiloliter, using the Ideal Gas Law: \[ n = \dfrac{PV}{RT}\] \[ V_{NH3} = V_{flue\text{ gas}} \times (partial\text{ pressure ratio}) \] \(1000 L \times 4.5 \times 10^{-5} = 0.045 L \) Calculate the moles of \text{NH}_3\: \(P V_{NH3} = nRT\) \(1.00 \text{atm} \times 0.045 \text{L} = n \times 0.0821 \times 638 K\) Solving for n: \(n = \frac{0.045}{0.0821 \times 638}\) Calculate the mass in grams: \(mass = n \times 17.03 g/\text{mol}\)

Calculate the mass

3 Convert pounds revealing the weight of the compound per km solutions and subtract

Key concepts

Catalyzed Reaction

In industrial pollutant reduction, a catalyzed reaction can significantly improve the efficiency of a reaction. A catalyst helps speed up the reaction without being consumed in the process. In our given exercise, the reaction between ammonia (NH3) and nitrogen monoxide (NO) is catalyzed, allowing these reactants to convert into nitrogen (N2) and water (H2O) more effectively. Here, the reaction is as follows: $$4 \text{NH}_3 (g) + 4 \text{NO} (g) + \text{O}_2 (g) \rightarrow 4 \text{N}_2 (g) + 6 \text{H}_2 \text{O} (g)$$ By using a catalyst, industries can ensure that less NO, a pollutant, is released into the atmosphere, enhancing environmental protection.

Ideal Gas Law

The Ideal Gas Law is essential to understanding how gases behave under different conditions. The formula for the Ideal Gas Law is: $$PV = nRT$$ where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L atm mol^(-1) K^(-1)), and T is the temperature in Kelvin. In our exercise, we used these variables to calculate the amount of NH3 needed for the reaction. Given that P = 1 atm and T = 365°C (which is 638 K), we can rearrange this formula to solve for n (the number of moles). By knowing the volume of NH3 needed per liter of flue gas, we can easily find the required volume using the proportional relationships inherent in the Ideal Gas Law.

Partial Pressure

Partial pressure is the pressure exerted by a single type of gas in a mixture of gases. According to Dalton's Law of Partial Pressures, the total pressure of a gas mixture is the sum of the partial pressures of all individual gases in that mixture. In our problem, the partial pressure of NO is given as 4.5 × 10^(-5) atm. Since NH3 reacts with NO in a 1:1 ratio, the partial pressure of NH3 required to fully react with NO will also be 4.5 × 10^(-5) atm. This concept is crucial because it helps us understand the proportion of gases needed for the reaction and ensures we use the right volume of NH3 to match the given partial pressure of NO.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It tells us how much of each substance is involved. In the exercise, the balanced chemical equation shows a molar ratio of 4 NH3 : 4 NO : 1 O2. This means for every 4 moles of NH3, 4 moles of NO and 1 mole of O2 are required. The balanced equation also tells us that these reactants will produce 4 moles of N2 and 6 moles of H2O. By using stoichiometry along with the Ideal Gas Law, we calculate the exact volume and mass of NH3 needed per kiloliter of flue gas to ensure complete reaction and minimize pollutants. For example, we determined that for a given kiloliter of flue gas, the volume of NH3 needed is 0.045 L and computed the number of moles and subsequent grams of NH3 required using stoichiometric calculations.