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Chapter 6: Problem 41

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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Most popular questions from this chapter

A biology experiment requires the preparation of a water bath at \(37.0^{\circ} \mathrm{C}\) (body temperature). The temperature of the cold tap water is \(22.0^{\circ} \mathrm{C}\), and the temperature of the hot tap water is \(55.0^{\circ} \mathrm{C}\). If a student starts with \(90.0 \mathrm{~g}\) cold water, what mass of hot water must be added to reach \(37.0^{\circ} \mathrm{C}\) ?

Assuming gasoline is pure \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\), predict the signs of \(q\) and \(w\) for the process of combusting gasoline into \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\).

A sample of an ideal gas at \(15.0 \mathrm{~atm}\) and \(10.0 \mathrm{~L}\) is allowed to expand against a constant external pressure of \(2.00 \mathrm{~atm}\) at a constant temperature. Calculate the work in units of \(\mathrm{kJ}\) for the gas expansion. (Hint: Boyle's law applies.)

Calculate \(\Delta E\) for each of the following. a. \(q=-47 \mathrm{~kJ}, w=+88 \mathrm{~kJ}\) b. \(q=+82 \mathrm{~kJ}, w=-47 \mathrm{~kJ}\) c. \(q=+47 \mathrm{~kJ}, w=0\) d. In which of these cases do the surroundings do work on the system?

Consider a balloon filled with helium at the following conditions. $$ \begin{array}{l} 313 \text { g He } \\ 1.00 \text { atm } \\ \text { 1910. L } \\ \text { Molar Heat Capacity }=20.8 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{mol} \end{array} $$ The temperature of this balloon is decreased by \(41.6^{\circ} \mathrm{C}\) as the volume decreases to \(1643 \mathrm{~L}\), with the pressure remaining constant. Determine \(q, w\), and \(\Delta E\) (in \(\mathrm{kJ}\) ) for the compression of the balloon.
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