Question

Nitric acid is a key industrial chemical, largely used to make fertilizers and explosives, the first step in its synthesis is the oxidiation of ammonia. In this reaction, gaseous ammonia reacts with dioxygen gas to produce nitrogen monoxide gas and water.

Suppose a chemical engineer studying a new catalyst for the oxidation of ammonia reactions finds that 613. liters per second of dioxygen are consumed when the reaction is run at 175. C and 0.16 atn=m. Calculate the rate at which nitrogen monoxide is being produced. Give your answer in Kg/second. Round your answer to 2 significant digits.

Answer 1

To find the rate at which nitrogen monoxide is produced, we use the balanced chemical equation for the oxidation of ammonia and the ideal gas law. We convert the given rate of dioxygen consumption to moles using the reaction conditions, apply the stoichiometry of the reaction, and calculate the mass of NO produced, finally converting it to kilograms per second.

Step-by-step explanation:

To determine the rate at which nitrogen monoxide (NO) is being produced from the oxidation of ammonia (NH3), we first need to look at the balanced chemical equation for the reaction:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

We see that for every 5 moles of dioxygen (O2), 4 moles of nitrogen monoxide are produced. Therefore, we can calculate the rate of NO production by using the stoichiometry of the reaction and the given rate of O2 consumption.

The rate of dioxygen consumption is 613 liters per second. Since the reaction conditions include a temperature of 175°C and a pressure of 0.16 atn/m, we can use the ideal gas law PV=nRT to find the moles of O2 consumed per second and then calculate the mass of NO produced per second.

Let's do the calculation:

1. First, convert the pressure from atn/m to atmospheres (atm) to use the ideal gas law: 0.16 atn/m is equivalent to 0.16 atm.
2. Then, we can use the ideal gas law with R = 0.0821 L·atm/(K·mol) and T = 175°C, which is 448.15K (since we have to convert Celsius to Kelvin by adding 273.15).
3. The moles of O2 that are consumed per second (n) equals PV divided by RT. So, n = (0.16 atm * 613 L) / (0.0821 L·atm/(K·mol) * 448.15 K).
4. Now, calculate the moles of NO produced per second using the mole ratio from the balanced equation: For every 5 moles of O2, 4 moles of NO are produced.
5. Multiply the moles of NO by its molar mass to get the mass of NO produced per second. The molar mass of NO is 30.01 g/mol.
6. Finally, convert the grams of NO to kilograms by dividing by 1000.

After completing these steps, you'll have the production rate of NO in kg/s, which is the answer to this problem.