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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 \(\mathrm{NO}\) ?

Short Answer

Expert verified
High temperatures are required to convert N₂ and O₂ to NO because the reaction is endothermic, with a high activation energy. At higher temperatures, the kinetic energy of reactant molecules increases, leading to more effective collisions that can overcome the activation energy barrier. This results in an increased reaction rate and a higher yield of NO.

Step by step solution

01

Examining the Reaction

The reaction is given as: \[ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) \quad \Delta H=180 \mathrm{~kJ} \] This reaction has a positive enthalpy change, which means it is endothermic - it requires energy to proceed forward.
02

Activation Energy

Activation energy is the minimum amount of energy required for a reaction to occur. In an endothermic reaction, the activation energy is higher than the energy of the starting reactants. For the given reaction to proceed, the molecules of N₂ and O₂ must collide with enough energy to overcome the activation energy barrier.
03

Collision Theory

According to the collision theory, the rate of a chemical reaction depends on the frequency of effective collisions between reactant molecules. For a collision to be effective, the molecules must collide with the correct orientation and with sufficient energy to overcome the activation energy barrier.
04

High Temperatures

Increasing the temperature of the reactants increases their average kinetic energy. This means that, at high temperatures, a greater proportion of the molecules possess enough energy for effective collisions to occur and overcome the activation energy barrier. As a result, more molecules of N₂ and O₂ can react to form NO, thus increasing the rate of the reaction.
05

Conclusion

High temperatures are needed to convert N₂ and O₂ to NO because the reaction is endothermic and requires high activation energy. At elevated temperatures, more collisions between the reactant molecules have sufficient energy to overcome the activation energy barrier, leading to a faster reaction rate and higher yield of NO.

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Key Concepts

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

Endothermic Reaction
In every chemical reaction, energy plays a crucial role. In an endothermic reaction, energy is absorbed from the surroundings. Such reactions have a positive enthalpy change, noted as \( \Delta H > 0 \).This simply means that the products have more energy than the reactants. The reaction between nitrogen \(\mathrm{N}_{2} \)and oxygen \(\mathrm{O}_{2} \) to form nitrogen monoxide \(\mathrm{NO} \)is a perfect example. It has an enthalpy change of \(180 \mathrm{~kJ}\),which means it consumes energy.For students, think of it as the reaction needing an 'energy boost' to proceed. Without this energy input, the reaction would not occur.One of the common ways to supply this energy is through heat, which raises the temperature and allows the reaction to absorb the necessary energy from its surroundings.
Collision Theory
Collision theory is essential to understand why reactions happen at all. It describes that molecules must collide with each other to react. However, not all collisions lead to a reaction. For a collision to result in the formation of products:
  • The molecules must collide with enough energy.
  • The orientation of the molecules during the collision must be correct.
In simpler terms, imagine two students trying to high-five; if they miss or do not use enough force, then the high-five doesn't happen. This concept applies to molecules too. In the case of forming NO, nitrogen and oxygen molecules must collide effectively, overcoming an activation energy barrier. The energy they need comes from the surrounding heat, which increases their motion and likelihood of interacting correctly.
High Temperatures in Reactions
Temperature significantly influences chemical reactions. At higher temperatures, the kinetic energy of molecules increases. This means:
  • Molecules move faster and collide more often.
  • The proportion of molecules with enough energy to surpass the activation energy barrier increases.
The reaction between \(\mathrm{N}_{2} \)and \(\mathrm{O}_{2} \)to produce \(\mathrm{NO} \) illustrates this well. At standard temperatures, the energy barrier is too high for the molecules to react often. However, as temperature rises, more molecules gain the energy to overcome this barrier through more frequent and energetic collisions, leading to a faster reaction rate.For students, imagine it as going from a leisurely walk to a brisk jog; with more speed, you cover more ground. Similarly, high temperatures allow more reactions to occur in a given time frame.

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Most popular questions from this chapter

Consider the following equations: $$ \begin{array}{rlrl} 3 \mathrm{~A}+6 \mathrm{~B} \longrightarrow & 3 \mathrm{D} & & \Delta H=-403 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{E}+2 \mathrm{~F} & \Delta H & =-105.2 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C} & \mathrm{A} & & \Delta H= & 64.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ Suppose the first equation is reversed and multiplied by \(\frac{1}{6}\), the second and third equations are divided by 2 , and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?

The enthalpy of combustion of solid carbon to form carbon dioxide is \(-393.7 \mathrm{~kJ} / \mathrm{mol}\) carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is \(-283.3 \mathrm{~kJ} /\) mol CO. Use these data to calculate \(\Delta H\) for the reaction $$ 2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}(g) $$

Give the definition of the standard enthalpy of formation for a substance. Write separate reactions for the formation of \(\mathrm{NaCl}\), \(\mathrm{H}_{2} \mathrm{O}, \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), and \(\mathrm{PbSO}_{4}\) that have \(\Delta H^{\circ}\) values equal to \(\Delta H_{\mathrm{f}}^{\circ}\) for each compound.

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 \(10.5 \mathrm{~L}\). 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 overall reaction in a commercial heat pack can be represented as $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \quad \Delta H=-1652 \mathrm{~kJ} $$ a. How much heat is released when \(4.00\) moles of iron are reacted with excess \(\mathrm{O}_{2}\) ? b. How much heat is released when \(1.00\) mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is produced? c. How much heat is released when \(1.00 \mathrm{~g}\) iron is reacted with excess \(\mathrm{O}_{2}\) ? d. How much heat is released when \(10.0 \mathrm{~g} \mathrm{Fe}\) and \(2.00 \mathrm{~g} \mathrm{O}_{2}\) are reacted?

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