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In a classroom demonstration, hydrogen gas and oxygen gas are mixed in a balloon. The mixture is stable under normal conditions, but, if a spark is applied to it or some powdered metal is added, the mixture explodes. (a) Is the spark acting as a catalyst? Explain. (b) Is the metal acting as a catalyst? Explain.

Short Answer

Expert verified
No, the spark is not a catalyst. Yes, the powdered metal can be a catalyst.

Step by step solution

01

- Understand what a catalyst is

A catalyst is a substance that speeds up a chemical reaction without being consumed in the process. It lowers the activation energy needed for the reaction to proceed.
02

- Determine if the spark acts as a catalyst

Examine whether the spark is consumed in the reaction or simply initiates it. A spark provides the activation energy required for the hydrogen and oxygen reaction but is not a substance nor reusable in multiple reactions. It initiates but doesn't lower the activation energy itself.
03

- Conclusion on the spark

The spark is not a catalyst because it is not a substance that changes the reaction rate by lowering activation energy. It's simply an external energy source to start the reaction.
04

- Determine if the metal acts as a catalyst

Check if the powdered metal is consumed in the reaction or remains unchanged. A catalyst should be reusable and lower the activation energy. The metal, if acting as a catalytic surface, can speed up the reaction without being used up in the process.
05

- Conclusion on the metal

If the powdered metal remains unchanged after initiating the explosion and can be reused, it acts as a catalyst. It lowers the activation energy required for the reaction between hydrogen and oxygen.

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Key Concepts

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

Activation Energy
To understand catalysts in chemical reactions, we first need to discuss the concept of activation energy. Activation energy is the minimum amount of energy required to initiate a chemical reaction. Think of it as the initial push needed to start a car's engine. Without this push, the reactants will not have enough energy to break their bonds and form new ones.
In the case of the hydrogen and oxygen reaction in the balloon, the activation energy is the energy needed to break the bonds in the hydrogen and oxygen molecules. Once the activation energy is provided, the reaction can proceed and produce water and energy in the form of an explosion.
For many reactions, activation energy is a crucial factor that determines the reaction rate. Higher activation energy means that fewer molecules will have enough energy to react at a given temperature. Catalysts play a crucial role here by lowering the activation energy, making it easier for the reaction to occur.
Hydrogen and Oxygen Reaction
The specific reaction we are looking at is the combination of hydrogen (H₂) and oxygen (O₂) gases, forming water (H₂O). The chemical equation for this reaction is:
2H₂(g) + O₂(g) → 2H₂O(l) + Energy
Under normal conditions, hydrogen and oxygen gases are quite stable. They do not react with each other quickly because the activation energy for this reaction is relatively high.
When a spark is introduced to the mixture, it provides the necessary activation energy. This energy excites the hydrogen and oxygen molecules, allowing them to overcome the activation energy barrier. Once the initial molecules react, they release a lot of energy, which triggers other hydrogen and oxygen molecules to react in a chain reaction, leading to an explosion. The spark serves to start this exothermic reaction but is not involved in lowering the activation energy.
Chemical Reaction Initiation
Chemical reactions often need an external force or agent to initiate, especially if the reactants are stable under normal conditions. In our exercise, both a spark and powdered metal were proposed as initiators.
A spark acts as an energy provider. It supplies the necessary activation energy to start the reaction. However, it is not a catalyst because it does not lower the activation energy nor is it reusable in multiple reactions.
On the other hand, if powdered metal is used, it can act differently. Metals like platinum or palladium can serve as catalysts in such reactions. Here’s how they work:
  • The metal provides a surface for the hydrogen and oxygen molecules to come closer together.
  • It weakens the bonds in the reactant molecules, effectively lowering the activation energy.
  • The metal remains unchanged at the end of the reaction, meaning it can be reused.
Thus, metal can act as a true catalyst by lowering the activation energy needed for the hydrogen and oxygen reaction, unlike the spark, which merely ignites the reaction.

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Most popular questions from this chapter

16.103 Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]\). The long-accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) In the 1960 s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism: (1) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) [fast] (2) \(\mathrm{H}_{2}(g)+\mathrm{I}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{I}(g)\) [fast] (3) \(\mathrm{H}_{2} \mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) \quad[\) slow \(]\) Show that this mechanism is consistent with the rate law.

Nitrification is a biological process in which \(\mathrm{NH}_{3}\) in wastewater is converted to \(\mathrm{NH}_{4}^{+}\) and then removed according to the following reaction: \(\mathrm{NH}_{4}^{+}+2 \mathrm{O}_{2} \longrightarrow \mathrm{NO}_{3}^{-}+2 \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{O}\) The first-order rate constant is given as \(k_{1}=0.47 e^{0.095\left(T-15^{\circ} \mathrm{C}\right)}\) where \(k_{1}\) is in day \(^{-1}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). (a) If the initial concentration of \(\mathrm{NH}_{3}\) is \(3.0 \mathrm{~mol} / \mathrm{m}^{3},\) how long will it take to reduce the concentration to \(0.35 \mathrm{~mol} / \mathrm{m}^{3}\) in the spring \(\left(T=20^{\circ} \mathrm{C}\right) ?\) (b) In the winter \(\left(T=10^{\circ} \mathrm{C}\right) ?\) (c) Using your answer to part (a), what is the rate of \(\mathrm{O}_{2}\) consumption?

Define the half-life of a reaction. Explain on the molecular level why the half-life of a first-order reaction is constant.

Arrhenius proposed that each reaction has an energy threshold that must be reached for the particles to react. The kinetic theory of gases proposes that the average kinetic energy of the particles is proportional to the absolute temperature. How do these concepts relate to the effect of temperature on rate?

The mathematics of the first-order rate law can be applied to any situation in which a quantity decreases by a constant fraction per unit of time (or unit of any other variable). (a) As light moves through a solution, its intensity decreases per unit distance traveled in the solution. Show that \(\ln \left(\frac{\text { intensity of light leaving the solution }}{\text { intensity of light entering the solution }}\right)\) \(=-\) fraction of light removed per unit of length \(\times\) distance traveled in solution (b) The value of your savings declines under conditions of constant inflation. Show that \(\ln \left(\frac{\text { value remaining }}{\text { initial value }}\right)\) \(=-\) fraction lost per unit of time \(\times\) savings time interval

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