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 \(3.50 \mathrm{~atm}\) 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 \(\mathrm{NO}_{2}(g) ?\)

To balance the equation, we have to ensure that the number of atoms of each element is equal on both sides. The balanced equation is: \[ N_{2}(g) + 2O_{2}(g) \longrightarrow 2NO_{2}(g) \]

We will use the ideal gas law to convert the volumes and pressures to moles using the given values of the gases: \(P = 3.50 \: \mathrm{atm}\) \(T =100+273.15 =373.15 \: \mathrm{K}\) (converting from Celsius to Kelvin) For nitrogen gas: \(V_{N_2} = 2.50 \times 10^2 \: \mathrm{mL} = 0.250 \: \mathrm{L}\) (converting from mL to L) For oxygen gas: \(V_{O_2} = 4.50 \times 10^2 \: \mathrm{mL} = 0.450 \: \mathrm{L}\) (converting from mL to L) Using the ideal gas law, \(PV=nRT\) Where \(R = 0.0821 \: \frac{\mathrm{atm \cdot L}}{\mathrm{mol \cdot K}}\) Solving for n (number of moles) separately for N₂ and O₂: \(n_{N_2} = \frac{P \cdot V_{N_2}}{R \cdot T} = \frac{3.50 \: \mathrm{atm} \cdot 0.250 \: \mathrm{L} }{0.0821 \: \frac{\mathrm{atm \cdot L}}{\mathrm{mol \cdot K}} \cdot 373.15 \: \mathrm{K}}\) \(n_{O_2} = \frac{P \cdot V_{O_2}}{R \cdot T} = \frac{3.50 \: \mathrm{atm} \cdot 0.450 \: \mathrm{L} }{0.0821 \: \frac{\mathrm{atm \cdot L}}{\mathrm{mol \cdot K}} \cdot 373.15 \: \mathrm{K}}\)

Now, let's find the mole ratio of reactants in the balanced equation \(N_2:O_2 = 1:2\) Comparing the mole ratio: \(\frac{n_{N_2}}{1} \) and \(\frac{n_{O_2}}{2}\) Now, determine the limiting reactant: If \(\frac{n_{N_2}}{1} < \frac{n_{O_2}}{2}\), then N₂ is the limiting reactant If \(\frac{n_{N_2}}{1} > \frac{n_{O_2}}{2}\), then O₂ is the limiting reactant

Using the limiting reactant, we can calculate the moles of NO₂ produced. For the balanced equation, the mole ratio is \(N_2:O_2:NO_2 = 1:2:2\) If N₂ is the limiting reactant, then: \(n_{NO_2} = 2 \times n_{N_2}\) If O₂ is the limiting reactant, then: \(n_{NO_2} = n_{O_2}\)

Now, we can calculate the heat input by using the enthalpy change and the calculated moles of NO₂: \(\Delta H = 67.7 \: \mathrm{kJ}\) (which is the enthalpy change per mole of N₂) The total heat input (Q) needed for the maximum yield of NO₂ is: \(Q = \Delta H \times n_{NO_2} = 67.7 \: \mathrm{kJ} \cdot n_{NO_2}\) Remember to consider the value of \(n_{NO_2}\) that corresponds to the limiting reactant that was determined in step 3.