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The propane used in a gas grill does not usually react with air in a combustion reaction unless first initiated with a spark. Explain using collision theory.

Short Answer

Expert verified
Propane does not usually react with air in a combustion reaction, because without a spark, the propane and oxygen molecules typically lack the necessary activation energy for successful collisions that lead to a reaction, as per collision theory. A spark, however, provides the required activation energy by heating a small amount of propane and oxygen, which initiates the reaction and releases heat, leading to further reactions.

Step by step solution

01

Understanding the Collision Theory

Collision theory states that in order for a reaction to occur, it's not enough for reactant molecules to simply collide. These molecules must also have proper orientation and enough energy (activation energy) to break the initial bonds and form new ones, leading to the end products.
02

Applying Collision Theory to Propane and Air

In the case of propane (C3H8) and air (which contains oxygen - O2), without an initial spark or source of ignition, the molecules of propane and oxygen do not typically have sufficient energy to break their existing bonds and form new ones. While propane and oxygen can have numerous collisions, ambient conditions do not provide the activation energy necessary for a successful reaction.
03

Role of the Spark

Supplying a spark can provide this needed activation energy. The spark heats a small amount of propane and oxygen, creating high energy molecules that can successfully collide to initiate the reaction. Once the reaction is started, it releases heat which triggers more reactions. That is why once it's ignited with a spark, the propane will continue to burn in oxygen.

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Key Concepts

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

Activation Energy
In the world of chemistry, activation energy is like a hurdle that reactant molecules need to leap over to start a chemical reaction. It represents the minimum amount of energy needed for a reaction to occur. When molecules collide, they must possess this energy to break existing bonds and create new bonds leading to product formation.

Think of it this way: imagine reactant molecules are cars speeding toward each other. For them to crash and change into something else (the products), they must be going fast enough. This speed is representative of activation energy. Without it, molecules simply bounce off each other like bumper cars with no resulting reaction. Thus, collision alone isn’t enough; they must also "hit the gas" to supply the necessary energy threshold.

This concept is crucial in understanding why some reactions require an initial input of energy, like a spark, to overcome this energy barrier and allow the reaction to proceed.
Propane Combustion
Propane combustion is a chemical reaction where propane gas reacts with oxygen to form carbon dioxide and water, accompanied by the release of heat and light. This process can be expressed in a simplified chemical equation:
  • \[C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O + \,\text{energy (heat and light)}\]
Initially, at room temperatures, the molecules of propane (C\(_3\)H\(_8\)) and the oxygen in the air do not have enough energy on their own to start the reaction. This is due to the high activation energy required.

Propane combustion only begins efficiently with the help of a spark. The initial spark heats the gas molecules, providing them with the necessary activation energy. Once ignition occurs, the compound reaction releases heat, sustaining the burning of propane until the fuel is exhausted. This is why once propane is ignited, it keeps burning as long as oxygen is available. It’s a great example of how energy release during combustion can maintain a chain reaction.
Molecular Orientation
Molecular orientation is just as important as energy in the collision theory context. When molecules collide, their alignment can determine whether a reaction will occur. For a successful reaction, molecules must be oriented in a manner that favors bond breaking and formation.

Think of it like two puzzle pieces that must fit together perfectly to connect. Even if molecules have enough energy, if they are aligned improperly, they will not react. Not only do propane and oxygen need the correct amount of energy, but they also need to collide with the right side "up" – the molecules must be oriented in a way that allows the rejeweling of atoms.
  • This is why not every collision results in a reaction.
  • Successful reactions depend on both energy and correct orientation.
In propane combustion, the oxygen should ideally be aligned in a proper angle with respect to carbon and hydrogen atoms for them to form new bonds efficiently.

Thus, both activation energy and molecular orientation are essential factors ensured by the initial spark that sequentially leads to a chain reaction in controlled combustion.

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Most popular questions from this chapter

The decomposition of HI into its elements is an exothermic reaction: $$ 2 \mathrm{HI}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ Normally this reaction has an activation energy of \(180 \mathrm{~kJ} / \mathrm{mol}\). When the reaction is carried out in the presence of platinum metal, the activation energy is lowered to \(80 \mathrm{~kJ} / \mathrm{mol}\). Draw an energy diagram for the uncatalyzed reaction. Label reactants, products, and a possible activated complex. Use a dotted line to show how the energy changes are different when platinum metal is added to the reaction.

What must you know about a reaction before you can predict the effect of a temperature change on the position of the equilibrium?

A sample of hydrogen iodide gas is placed in a reaction container, heated to \(450^{\circ} \mathrm{C}\), and allowed to reach a state of equilibrium: $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ The equilibrium concentrations were determined to be $$ \begin{aligned} [\mathrm{HI}] &=0.195 M \\ \left[\mathrm{H}_{2}\right] &=0.0275 M \\ \left[\mathrm{I}_{2}\right] &=0.0275 M \end{aligned} $$ (a) Write the equilibrium constant expression for this reaction. (b) Calculate the value of the equilibrium constant at \(450^{\circ} \mathrm{C}\). (c) Describe the position of the equilibrium.

Consider the following exothermic reaction: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ Which of the following changes will increase the number of moles of NO at equilibrium? Explain why or why not for each change. (a) remove \(\mathrm{H}_{2} \mathrm{O}\) (b) decrease volume (c) decrease temperature (d) add \(\mathrm{O}_{2}\) (e) add a catalyst

Consider the following reaction: $$ \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2}(g) $$ At a specific temperature, the equilibrium concentrations were determined to be \(\left[\mathrm{CH}_{4}\right]=0.049 M,\left[\mathrm{H}_{2} \mathrm{O}\right]=0.048 M\), \(\left[\mathrm{CO}_{2}\right]=0.00090 \mathrm{M}\), and \(\left[\mathrm{H}_{2}\right]=0.0036 \mathrm{M}\). (a) What is the value of the equilibrium constant? (b) Describe the position of the equilibrium.

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