Question

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)(\text { unbalanced }) $$ 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

The heat required to synthesize the maximum yield of NO₂ when given volumes and pressures of N₂ and O₂ is 1.938 kJ.

Step-by-step solution

1. Balance the Chemical Equation

To balance the equation, start by counting the number of atoms of each element on the reactant and product sides. Based on the given reaction, we have: N₂(g) + O₂(g) → NO₂(g) (unbalanced) Adjust the coefficients to balance the number of atoms on each side of the equation: 2 N₂(g) + 2 O₂(g) → 4 NO₂(g) The balanced coefficient is 2:2:4 for N₂, O₂ and NO₂, respectively.

2. Convert mL Gas to Moles Using Ideal Gas Law

Next, we will convert the given volumes and pressures of N₂ and O₂ into moles using the Ideal Gas Law, PV=nRT. For N₂: - Volume (V) = 2.50 x 10² mL = 0.25 L - Pressure (P) = 3.50 atm - Temperature (T) = 100 °C = 373.15 K - Gas constant (R) = 0.0821 L atm / (K mol) - n (moles N₂) = ? PV = nRT --> n = PV / (RT) n(N₂) = (3.50 atm * 0.25 L) / (0.0821 L atm / (K mol) * 373.15 K) n(N₂) = 0.0286 mol For O₂: - Volume (V) = 4.50 x 10² mL = 0.45 L - Pressure (P) = 3.50 atm - Temperature (T) = 100 °C = 373.15 K - Gas constant (R) = 0.0821 L atm / (K mol) - n (moles O₂) = ? PV = nRT --> n = PV / (RT) n(O₂) = (3.50 atm * 0.45 L) / (0.0821 L atm / (K mol) * 373.15 K) n(O₂) = 0.0514 mol

3. Determine the Limiting Reactant

Determine the limiting reactant by comparing the stoichiometric coefficients of N₂ and O₂ according to the balanced chemical equation: 2 N₂(g) + 2 O₂(g) → 4 NO₂(g) n(N₂) = 0.0286 mol n(O₂) = 0.0514 mol Ratio of moles of N₂ to O₂: (0.0286 mol N₂/0.0514 mol O₂) = 0.556 Since the ratio of moles N₂:O₂ is less than the balanced equation's stoichiometric coefficient (2:2), N₂ is the limiting reactant.

4. Calculate the Maximum Yield of NO₂

The maximum yield of NO₂ can be found using the stoichiometric ratio from the balanced chemical equation. 2 N₂(g) + 2 O₂(g) → 4 NO₂(g) Using the limiting reactant N2 (0.0286 mol), calculate the moles of NO₂: n(NO₂) = 0.0286 mol N₂ * (4 mol NO₂ / 2 mol N₂) n(NO₂) = 0.0572 mol NO₂

5. Calculate the Heat Required

Finally, calculate the heat required for the reaction, using the enthalpy change of reaction, ΔH, provided: ΔH = 67.7 kJ (1 mol N₂ reacts with 1 mol O₂) Since there are 2 moles of N₂ as coefficient, we have: Heat required for 2 mole of N₂ = 67.7 kJ\(\times2\)=135.4 kJ Calculate the heat required for 0.0286 moles of N₂ (limiting reactant) by proportion: Heat Required = (0.0286 mol N₂ * 135.4 kJ) / 2 mol N₂ Heat Required = 1.938 kJ Thus, 1.938 kJ of heat is required to synthesize the maximum yield of NO₂.

Key concepts

ideal gas law

The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of gases under different conditions of pressure, volume, and temperature. It is written as \( PV = nRT \), where:

• \( P \) is the pressure of the gas
• \( V \) is the volume of the gas
• \( n \) is the number of moles of the gas
• \( R \) is the ideal gas constant (approximately 0.0821 L atm / K mol for our purposes)
• \( T \) is the temperature in Kelvin

To convert Celsius to Kelvin, add 273.15 to the Celsius temperature.
In this problem, the Ideal Gas Law assists in converting the given volumes and pressures of nitrogen (\( \mathrm{N}_2 \)) and oxygen (\( \mathrm{O}_2 \)) into moles. By knowing the moles, we can proceed to other steps like determining the stoichiometric balances and calculating enthalpy changes.

stoichiometry

Stoichiometry is the area of chemistry that pertains to the quantitative relationships between reactants and products in a balanced chemical equation. It requires balancing chemical equations so that the number of atoms of each element is equal on the reactant and product sides.
For the chemical reaction in the original exercise, the equation is balanced as:2 \( \mathrm{N}_2(g) + 2 \mathrm{O}_2(g) \rightarrow 4 \mathrm{NO}_2(g) \)This means every two moles of nitrogen gas (\( \mathrm{N}_2 \)) reacts with two moles of oxygen gas (\( \mathrm{O}_2 \)) to produce four moles of nitrogen dioxide (\( \mathrm{NO}_2 \)).
Stoichiometry helps determine how much of each reactant is required, and how much product can be made based on the balanced equation.

limiting reactant

A limiting reactant in a chemical reaction is the reactant that is completely consumed first, which limits the amount of product that can be formed. To find the limiting reactant, you need to compare the mole ratio of the reactants used to the mole ratio in the balanced equation.
In this problem, with 0.0286 mol of \( \mathrm{N}_2 \) and 0.0514 mol of \( \mathrm{O}_2 \) calculated using the Ideal Gas Law, \( \mathrm{N}_2 \) is the limiting reactant. Although the balanced equation suggests a 1:1 ratio for \( \mathrm{N}_2 \) and \( \mathrm{O}_2 \), in reality, there isn't enough \( \mathrm{N}_2 \) to fully react with the available \( \mathrm{O}_2 \).
Thus, the amount of \( \mathrm{NO}_2 \) produced is determined by the quantity of \( \mathrm{N}_2 \) present.

endothermic reactions

Endothermic reactions are chemical reactions that absorb energy from the surroundings. This energy, often in the form of heat, is required for the reaction to proceed. The enthalpy change (\( \Delta H \)) for an endothermic reaction is positive, indicating that heat is absorbed.
In the context of the exercise, creating \( \mathrm{NO}_2 \) from \( \mathrm{N}_2 \) and \( \mathrm{O}_2 \) is endothermic, requiring 67.7 kJ of energy per mole of \( \mathrm{N}_2 \) and \( \mathrm{O}_2 \).
For our specific calculation, we scaled the energy requirement to the actual amount of \( \mathrm{N}_2 \) involved, determining that approximately 1.938 kJ of heat is necessary to achieve the maximum yield of \( \mathrm{NO}_2 \) from the available reactants.