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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 \(\mathrm{NO}_{2}(g) ?\)

Short Answer

Expert verified
The heat input necessary to synthesize the maximum yield of NOâ‚‚(g) is calculated using the balanced equation, the ideal gas law, stoichiometry, and the enthalpy change. To do this: 1. Balance the equation: \(N_{2}(g) + 2O_{2}(g) \longrightarrow 2NO_{2}(g)\) 2. Convert the volumes of Nâ‚‚(g) and Oâ‚‚(g) to moles using the ideal gas law. 3. Determine the limiting reactant using the mole ratios from the balanced equation. 4. Calculate the moles of NOâ‚‚ produced using the limiting reactant and the stoichiometry of the balanced equation. 5. Calculate the heat input needed (Q) using the enthalpy change and the calculated moles of NOâ‚‚: \(Q = \Delta H \times n_{NO_2} = 67.7 \: \mathrm{kJ} \cdot n_{NO_2}\). Make sure to use the appropriate value of \(n_{NO_2}\) based on the limiting reactant found in step 3.

Step by step solution

01

Balance the chemical reaction

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: \[ N_{2}(g) + 2O_{2}(g) \longrightarrow 2NO_{2}(g) \]
02

Convert the volumes of gases to moles using ideal gas law

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 Nâ‚‚ and Oâ‚‚: \(n_{N_2} = \frac{P \cdot V_{N_2}}{R \cdot T} = \frac{3.50 \: \mathrm{atm} \cdot 0.250 \: \mathrm{L} }{0.0821 \: \frac{\mathrm{atm \cdot L}}{\mathrm{mol \cdot K}} \cdot 373.15 \: \mathrm{K}}\) \(n_{O_2} = \frac{P \cdot V_{O_2}}{R \cdot T} = \frac{3.50 \: \mathrm{atm} \cdot 0.450 \: \mathrm{L} }{0.0821 \: \frac{\mathrm{atm \cdot L}}{\mathrm{mol \cdot K}} \cdot 373.15 \: \mathrm{K}}\)
03

Determine the limiting reactant

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 Nâ‚‚ is the limiting reactant If \(\frac{n_{N_2}}{1} > \frac{n_{O_2}}{2}\), then Oâ‚‚ is the limiting reactant
04

Calculate the moles of NOâ‚‚ produced

Using the limiting reactant, we can calculate the moles of NOâ‚‚ produced. For the balanced equation, the mole ratio is \(N_2:O_2:NO_2 = 1:2:2\) If Nâ‚‚ is the limiting reactant, then: \(n_{NO_2} = 2 \times n_{N_2}\) If Oâ‚‚ is the limiting reactant, then: \(n_{NO_2} = n_{O_2}\)
05

Calculate the heat input needed

Now, we can calculate the heat input by using the enthalpy change and the calculated moles of NOâ‚‚: \(\Delta H = 67.7 \: \mathrm{kJ}\) (which is the enthalpy change per mole of Nâ‚‚) The total heat input (Q) needed for the maximum yield of NOâ‚‚ 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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
Enthalpy change is a key concept in chemistry, particularly in the study of reactions. It refers to the heat absorbed or released during a chemical reaction at constant pressure. This parameter is crucial when predicting the energy flow in a reaction. In the exercise, the reaction between nitrogen (\( \mathrm{N}_{2}(g) \)) and oxygen (\( \mathrm{O}_{2}(g) \)) to form nitrogen dioxide (\( \mathrm{NO}_{2}(g) \)) requires an input of energy because it is endothermic. The enthalpy change (\( \Delta H \)) is given as 67.7 kJ per mole of \( \mathrm{N}_{2} \).When calculating the heat input for the reaction, understanding that \( \Delta H \) is the energy required to convert reactants into products is essential. In endothermic reactions like this one, energy is absorbed, meaning the system gains energy from its surroundings. Calculating the total energy involves determining the moles of product formed from the limiting reactant in the reaction and multiplying by the \( \Delta H \).This gives a complete picture of the energy dynamics in the reaction, helping in predicting how much heat is required to achieve the desired product yield. Remember, knowing whether a reaction is endothermic or exothermic is crucial for understanding how it will interact with its environment.
Limiting Reactant
The concept of a limiting reactant is vital for determining the maximum possible yield of a chemical reaction. Essentially, the limiting reactant is the substance that is completely used up first, stopping the reaction from continuing and thus determining how much product can be formed.In our scenario, nitrogen gas (\( \mathrm{N}_{2}\)) and oxygen gas (\( \mathrm{O}_{2}\)) are mixed together. To find the limiting reactant, you measure the initial amounts of each reactant, in moles. Then, based on the balanced reaction equation, you compare the mole ratios required. For this equation, the ratio is 1 mole of \( \mathrm{N}_{2}\)) to 2 moles of \( \mathrm{O}_{2}\)).
  • If the quantity of \( \mathrm{N}_{2}\)) is less than half of \( \mathrm{O}_{2}\)), then \( \mathrm{N}_{2}\)) is the limiting reactant.
  • Conversely, if the available amount of \( \mathrm{O}_{2}\)) is less than twice that of \( \mathrm{N}_{2}\)), \( \mathrm{O}_{2}\)) becomes limiting.
Determining the limiting reactant allows us to identify which reactant will dictate the total amount of product formed, in this case, \( \mathrm{NO}_{2}\)). Recognizing this helps in calculating the maximum potential yield and the necessary energy involved in the reaction.
Chemical Reaction Balancing
Balancing chemical reactions is an essential skill in chemistry. It ensures that the law of conservation of mass is observed, where no atoms are lost or gained in a chemical reaction. In the given exercise, we begin with an unbalanced equation: \[ \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \]To balance it, each type of atom must have the same number on both sides of the equation. For example, nitrogen atoms should be balanced by adjusting the coefficients in the equation. Oxygen, which is present in two molecules on each side, should also be accounted for.The balanced equation becomes:\[ \mathrm{N}_{2}(g) + 2\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{NO}_{2}(g) \]Here, one molecule of \( \mathrm{N}_{2} \) reacts with two molecules of \( \mathrm{O}_{2} \) to form two molecules of \( \mathrm{NO}_{2} \).Balancing involves a simple trial and error method or the use of algebraic techniques, and it’s crucial for understanding reaction stoichiometry. This process ensures that calculations about the reaction, including the amount of reactants used and the product formed, are accurate and reliable.

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Most popular questions from this chapter

A \(110 .-\mathrm{g}\) sample of copper (specific heat capacity \(=0.20 \mathrm{~J} /{ }^{\circ} \mathrm{C}\). \(\mathrm{g}\) ) is heated to \(82.4^{\circ} \mathrm{C}\) and then placed in a container of water at \(22.3^{\circ} \mathrm{C}\). The final temperature of the water and copper is \(24.9^{\circ} \mathrm{C}\). What is the mass of the water in the container, assum- ing that all the heat lost by the copper is gained by the water?

Calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ \begin{array}{cr} \text { Equation } & \Delta H(\mathrm{k}\rfloor) \\ 2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(I) & -1010 \\ \mathrm{~N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(I)+\mathrm{H}_{2} \mathrm{O}(I) & -317 \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(I)+\mathrm{H}_{2} \mathrm{O}(I) & -143 \\ \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(I) & -286 \end{array} $$

Calculate \(w\) and \(\Delta E\) when 1 mole of a liquid is vaporized at its boiling point \(\left(80 .{ }^{\circ} \mathrm{C}\right)\) and \(1.00\) atm pressure. \(\Delta H_{\text {vap }}\) for the liquid is \(30.7 \mathrm{~kJ} / \mathrm{mol}\) at \(80 .{ }^{\circ} \mathrm{C}\).

In a bomb calorimeter, the reaction vessel is surrounded by water that must be added for each experiment. Since the amount of water is not constant from experiment to experiment, the mass of water must be measured in each case. The heat capacity of the calorimeter is broken down into two parts: the water and the calorimeter components. If a calorimeter contains \(1.00 \mathrm{~kg}\) water and has a total heat capacity of \(10.84 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\), what is the heat capacity of the calorimeter components?

A system releases \(125 \mathrm{~kJ}\) of heat while \(104 \mathrm{~kJ}\) of work is done on it. Calculate \(\Delta E\).

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