The burning of methane in oxygen is a highly exothermic reaction. Yet a mixture of methane and oxygen gas can be kept indefinitely without any apparent change. Explain.

Short Answer

Expert verified
A mixture of methane and oxygen gas can be kept indefinitely without any apparent change because the reaction between them requires an initial investment of energy, known as activation energy, to begin. In absence of a spark or flame to provide this energy, the reaction doesn't occur, and the gases remain unreacted.

Step by step solution

01

Understanding The Concepts

First, acknowledge that chemical reactions, despite the potential to release or absorb energy, don't occur spontaneously. They require an initial investment of energy known as the activation energy. For the methane and oxygen reaction, this comes from an external source, like a spark or flame.
02

Acknowledge the Role of Activation Energy

Understand that the activation energy acts like a barrier that prevents the reaction from happening spontaneously. Despite a reaction being exothermic, without this initial energy, the reactants - in this case, methane and oxygen - will not transform into products.
03

Apply to Given Scenario

Finally, apply these understanding to the given scenario. The methane and oxygen gases mixture can be kept indefinitely without any change because there's no source of activation energy to start the reaction.

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Key Concepts

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

Exothermic Reactions
When we talk about exothermic reactions, we're referring to processes that release energy, typically in the form of heat, into their surroundings. This is the kind of reaction that makes you feel warm near a burning candle or why a fireplace can heat up a room. What's crucial here is that this released energy is originally part of the chemical bonds in the reactants, and when they transform into products, they let go of that energy.

Take, for example, the burning of methane, a common natural gas. This process combines methane molecules with oxygen to produce carbon dioxide and water, releasing energy in the process. This release is so significant that it can be used to heat homes or even cook food. The practical aspect is fascinating because by harnessing this energy, we can do work—boiling water, perhaps, or generating electricity.

What can be confusing, however, is why these reactions don't just occur spontaneously even though they're energetically favorable. This leads us to a crucial concept in understanding chemical processes: the need for activation energy. Without a little push to get the reaction started, even the most energetically favorable reactions won't proceed on their own.
Chemical Kinetics
Chemical kinetics dives into the speed or rate at which a chemical reaction proceeds. For students, imagine it as setting a timer for a race; some reactions are a sprint, quick and over before you know it, while others are more like a marathon, taking their sweet time to reach the finish line. The rate of a chemical reaction can tell us a lot about how the reaction occurs and under what conditions it can be sped up or slowed down.

Several factors influence the speed of a reaction—temperature, for instance, can give molecules more energy to move around and collide with each other, potentially increasing the rate. Concentration is another factor; more molecules in a given space can lead to more frequent collisions, hence a faster reaction. And let's not forget catalysts—these are the special players that can speed up a reaction without being consumed by it. They work by lowering the activation energy needed for a reaction. It's like they're lowering the height of a hurdle in a hurdle race, making it easier for the runners—our reactants—to get over and reach the product side.
Energy Barriers
Energy barriers, often referred to as activation energy, are akin to the hill you have to climb at the start of a hike. It requires effort (energy) to start the journey, but once you're over the peak, it's a smooth walk—or even a downhill stroll. In chemical reactions, this 'hill' is the minimum amount of energy that's necessary for the reactants to transform into products.

Even in an exothermic reaction like the combustion of methane and oxygen, we need a spark or a little heat to get things going. Why? Because the molecules in our reactants are stable enough on their own; they need that extra energy to break apart before they can rearrange into something new and more stable—our products. This is why you can store a mixture of methane and oxygen without any issue; without that initial energy to overcome the barrier, there's no reaction.

The idea of energy barriers is incredibly useful in everything from designing safer storage for reactive substances to creating efficient reactions for industrial processes. By understanding and manipulating these barriers, chemists are able to control when and how a reaction will take place.

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Most popular questions from this chapter

The activation energy \(\left(E_{\mathrm{a}}\right)\) for the reaction $$ 2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta H^{\circ}=-164 \mathrm{~kJ} / \mathrm{mol} $$ is \(240 \mathrm{~kJ} / \mathrm{mol} .\) What is \(E_{\mathrm{a}}\) for the reverse reaction?

The rate law for the reaction $$ \mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ is given by rate \(=k\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{NO}_{2}^{-}\right] .\) At \(25^{\circ} \mathrm{C},\) the rate constant is \(3.0 \times 10^{-4} / M \cdot\) s. Calculate the rate of the reaction at this temperature if \(\left[\mathrm{NH}_{4}^{+}\right]=0.26 M\) and \(\left[\mathrm{NO}_{2}^{-}\right]=0.080 M\)

Variation of the rate constant with temperature for the first-order reaction $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{O}_{4}(g)+\mathrm{O}_{2}(g) $$ is given in the following table. Determine graphically the activation energy for the reaction. $$ \begin{array}{cc} \hline T(K) & k\left(\mathrm{~s}^{-1}\right) \\ \hline 298 & 1.74 \times 10^{-5} \\ 308 & 6.61 \times 10^{-5} \\ 318 & 2.51 \times 10^{-4} \\ 328 & 7.59 \times 10^{-4} \\ 338 & 2.40 \times 10^{-3} \\ \hline \end{array} $$

The first-order rate constant for the decomposition of dimethyl ether $$ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{H}_{2}(g)+\mathrm{CO}(g) $$ is \(3.2 \times 10^{-4} \mathrm{~s}^{-1}\) at \(450^{\circ} \mathrm{C} .\) The reaction is carried out in a constant-volume flask. Initially only dimethyl ether is present and the pressure is 0.350 atm. What is the pressure of the system after 8.0 min? Assume ideal behavior.

The reaction \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{I}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}+\mathrm{I}_{2}\) proceeds slowly in aqueous solution, but it can be catalyzed by the \(\mathrm{Fe}^{3+}\) ion. Given that \(\mathrm{Fe}^{3+}\) can oxidize \(\mathrm{I}^{-}\) and \(\mathrm{Fe}^{2+}\) can reduce \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-},\) write a plausible two-step mechanism for this reaction. Explain why the uncatalyzed reaction is slow.

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