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Chapter 6: Problem 39

One of the components of polluted air is NO. It is formed in the high- temperature environment of internal combustion engines by the following reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) \quad \Delta H=180 \mathrm{~kJ} $$ Why are high temperatures needed to convert \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) to NO?

Short Answer

Expert verified
High temperatures are needed to convert N₂ and O₂ to NO because it is an endothermic reaction, requiring the absorption of heat. High temperatures provide the necessary activation energy for the stable N₂ and O₂ molecules to react by increasing their kinetic energy, resulting in more frequent and higher energy collisions. Additionally, high temperatures facilitate the breaking of stable N₂ and O₂ bonds and the formation of new NO bonds.

Step by step solution

01

Identifying the endothermic reaction

An endothermic reaction is one where energy is absorbed from the surroundings in the form of heat. In this case, the reaction absorbs 180 kJ of heat to proceed. This means that the reaction requires high temperatures for N₂ and O₂ to form NO. This can be observed in the enthalpy change provided by the equation: \[ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g), \quad \Delta H = 180 \mathrm{~kJ} \]
02

Collision theory and reaction rate

The collision theory states that chemical reactions occur when molecules collide with enough energy and in the correct orientation. The rate of a chemical reaction increases with an increase in temperature because it increases the kinetic energy of the reacting molecules, resulting in more frequent collisions with higher energy. This increases the probability of successful collisions (ones that lead to a reaction).
03

Activation energy requirement

N₂ and O₂ molecules are very stable, and their bonds are difficult to break. The energy barrier that must be overcome for them to react is called the activation energy (Ea). High temperatures provide this activation energy for a larger fraction of colliding molecules. When the energy of the collisions is greater than or equal to the activation energy, the reaction can proceed and produce NO.
04

Bond formation and breaking

The high temperatures in the combustion environment cause the bonds in the N₂ and O₂ molecules to break, providing them with enough energy to form new bonds and create NO molecules. Since the formation of new bonds in NO molecules releases less energy than what was required to break the initial N₂ and O₂ bonds, the net energy change for the reaction is positive, indicating an endothermic reaction. In summary, high temperatures are needed to convert N₂ and O₂ to NO because it is an endothermic reaction that requires the input of energy (heat), provides the necessary activation energy for the stable N₂ and O₂ molecules to react, and facilitates the breaking and formation of new chemical bonds.

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Most popular questions from this chapter

For the process \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) at \(298 \mathrm{~K}\) and \(1.0 \mathrm{~atm}, \Delta H\) is more positive than \(\Delta E\) by \(2.5 \mathrm{~kJ} / \mathrm{mol}\). What does the \(2.5 \mathrm{~kJ} / \mathrm{mol}\) ouantity renresent?

Combustion of table sugar produces \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) When \(1.46 \mathrm{~g}\) table sugar is combusted in a constant-volume (bomb) calorimeter, \(24.00 \mathrm{~kJ}\) of heat is liberated. a. Assuming that table sugar is pure sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\), write the balanced equation for the combustion reaction. b. Calculate \(\Delta E\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) for the combustion reaction of sucrose. c. Calculate \(\Delta I I\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) for the combustion reaction of sucrose at \(25^{\circ} \mathrm{C}\).

Using the following data, calculate the standard heat of formation of \(\operatorname{ICl}(g)\) in \(\mathrm{kJ} / \mathrm{mol}\) : $$ \begin{aligned} \mathrm{Cl}_{2}(g) & \longrightarrow 2 \mathrm{Cl}(g) & \Delta H^{\circ} &=242.3 \mathrm{~kJ} \\ \mathrm{I}_{2}(g) & \longrightarrow 2 \mathrm{I}(g) & \Delta H^{\circ} &=151.0 \mathrm{~kJ} \\ \mathrm{ICl}(g) & \longrightarrow \mathrm{I}(g)+\mathrm{Cl}(g) & \Delta H^{\circ} &=211.3 \mathrm{~kJ} \\ \mathrm{I}_{2}(s) & \Delta H^{\circ}=62.8 \mathrm{~kJ} \end{aligned} $$

Consider \(5.5 \mathrm{~L}\) of a gas at a pressure of \(3.0 \mathrm{~atm}\) in a cylinder with a movable piston. The external pressure is changed so that the volume changes to \(10.5 \mathrm{~L}\). a. Calculate the work done, and indicate the correct sign. b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is \(7.0 \mathrm{~L}\). The second step results in a final volume of \(10.5\) L. Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is \(8.0 \mathrm{~L}\) and the second step leads to a volume of \(10.5 \mathrm{~L}\). Does the work differ from that in part b? Explain.

The bomb calorimeter in Exercise 108 is filled with \(987 \mathrm{~g}\) water. The initial temperature of the calorimeter contents is \(23.32^{\circ} \mathrm{C}\). A \(1.056-\mathrm{g}\) sample of benzoic acid \(\left(\Delta E_{\text {comb }}=-26.42 \mathrm{~kJ} / \mathrm{g}\right)\) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?
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