Question

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \quad \text { (unbalanced) }$$ The enthalpy change of reaction for the balanced equation (with lowest whole-number coefficients) is \(\Delta H=67.7 \mathrm{kJ} .\) If \(2.50 \times\) \(10^{2} \mathrm{mL} \mathrm{N}_{2}(g)\) at \(100 .^{\circ} \mathrm{C}\) and \(3.50\) atm and \(4.50 \times 10^{2} \mathrm{mL} \mathrm{O}_{2}(g)\) at \(100 .^{\circ} \mathrm{C}\) and \(3.50\) atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)

Short answer

To synthesize the maximum yield of $\mathrm{NO}_{2}(g)$ from the given amounts of $\mathrm{N}_{2}(g)$ and $\mathrm{O}_{2}(g)$ under the given conditions, 778.4 kJ of heat is required.

Step-by-step solution

Balance the chemical equation

We are given the unbalanced equation: $$N_2(g) + O_2(g) \longrightarrow NO_2(g)$$ To balance this equation, we can see that for every N₂ molecule one O₂ molecule is combined to form two NO₂ molecules. $$N_2(g) + 2O_2(g) \longrightarrow 2NO_2(g)$$

Determine limiting reactant

We are given the initial conditions for N₂ and O₂ in volume and pressure: For N₂(g): 2.50 x 10² mL, 3.50 atm, and 100°C For O₂(g): 4.50 x 10² mL, 3.50 atm, and 100°C To determine the limiting reactant, we need to find the moles of each reactant using the ideal gas law and the given conditions. \(PV = nRT\) Where P is pressure, V is volume, n represents moles, R is the ideal gas constant (\(0.0821 \frac{L \cdot atm}{mol \cdot K}\)), and T is temperature in Kelvin. Firstly, convert temperatures and volumes into their respective units (K and L): \(Temperature = 100 + 273.15 = 373.15K\) \(2.50 \times 10^2 mL = 0.250 L\) and \(4.50 \times 10^2 mL = 0.450 L\) Now, let's find the moles for N₂ and O₂: For N₂(g): \(3.50 atm \cdot 0.250 L = n \cdot 0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 373.15 K\) For O₂(g): \(3.50 atm \cdot 0.450 L = n \cdot 0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 373.15 K\) Calculate the moles of each reactant, and determine the limiting reactant using the balanced equation.

Calculate theoretical yield of NO₂

Once the limiting reactant is determined, we can use stoichiometry to find the moles of NO₂ that can be produced by the reaction. According to the balanced equation, 1 mole of N₂ reacts with 2 moles of O₂ to produce 2 moles of NO₂. Calculate the theoretical yield of moles of NO₂ based on the limiting reactant.

Calculate the heat required

Now that we have the theoretical yield of NO₂, we can find the amount of heat required for the reaction. We are given the enthalpy change of the reaction: ΔH = 67.7 kJ. Since ΔH refers to the heat required for 1 mole of N₂ and 2 moles of O₂ to produce 2 moles of NO₂, Heat required for the reaction = (Moles of limiting reactant / Stoichiometric coefficient of limiting reactant in balanced equation) × ΔH Calculate the heat required for synthesizing the maximum yield of NO₂.

Key concepts

Chemical Equation Balancing

Understanding how to balance chemical equations is essential in the study of chemistry. It's a way of ensuring that the law of conservation of mass is respected, meaning that atoms are neither created nor destroyed during a chemical reaction. Balancing an equation involves comparing the number of atoms of each element on the reactant side with the number on the product side and using coefficients to ensure they are equal.

For example, our equation, when balanced, is \(N_{2}(g) + 2O_{2}(g) \longrightarrow 2NO_{2}(g)\). This means for every nitrogen molecule, we need two oxygen molecules to react and produce two nitrogen dioxide molecules. This balanced equation is critical when it comes to further calculations such as finding the limiting reactant, calculating theoretical yield, or determining the enthalpy change of a reaction.

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. Once this reactant is used up, no further reaction can occur. Identifying the limiting reactant is vital because it dictates the maximum amount of product that can be formed from the reactants. This concept is akin to a recipe where the ingredient in the shortest supply limits the amount of finished dish you can create.

In the above example, we first used the ideal gas law \( PV = nRT \) to determine the moles of each reactant. Once the moles are known, we compare these to the balanced equation to find which reactant would run out first, thereby limiting the overall reaction.

Theoretical Yield

The theoretical yield is the amount of product that will be generated if a reaction proceeds perfectly and goes to completion without any losses. It's calculated based on stoichiometry, from the amounts of reactants and the balanced equation. In our example, the theoretical yield of \( NO_{2}(g) \) depends on the limiting reactant we identified previously. It's a theoretical maximum because, in practice, some reactants may not react fully, side reactions may occur, or products might be lost during collection.

To calculate the theoretical yield, we apply the mole ratio from the balanced equation. Since one mole of \( N_{2} \) reacts with two moles of \( O_{2} \) to produce two moles of \( NO_{2} \) according to the balanced equation, we can calculate the expected amount of \( NO_{2} \) based on the amount of the limiting reactant.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the volume, pressure, temperature, and amount of moles of a gas. Given by the formula \( PV = nRT \) where \( P \) stands for pressure, \( V \) for volume, \( n \) for moles, \( R \) is the ideal gas constant, and \( T \) for temperature in Kelvin, it is an invaluable tool for chemists to predict the behavior of gases under different conditions.

In the exercise, we use the ideal gas law to convert volume and pressure measurements into moles of \( N_{2}(g) \) and \( O_{2}(g) \) at a given temperature. It is important to note that the ideal gas law assumes gases behave perfectly, meaning they have no intermolecular forces and occupy no volume, which is a good approximation under many conditions but can deviate at high pressures or low temperatures. Knowing the mole amount of each gas allows us to proceed with stoichiometric calculations such as determining the limiting reactant or the theoretical yield.