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Chapter 11: Problem 18

Hydrogen reacts explosively with oxygen. However, a mixture of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) can exist indefinitely at room temperature. Explain why \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) do not react under these conditions.

Short Answer

Expert verified
Hydrogen and oxygen do not react explosively at room temperature because the average kinetic energy of the molecules is not sufficient to overcome the activation energy barrier required for the reaction to occur. In the absence of a catalyst or a higher temperature, hydrogen and oxygen can coexist indefinitely without undergoing an explosive reaction.

Step by step solution

01

Recall characteristics of chemical reactions

In a chemical reaction, reactants need to collide with each other at an appropriate orientation and sufficient energy for the reaction to take place. The minimum energy required for a successful collision is called the activation energy.
02

Understand activation energy and reaction rates

Activation energy is the threshold energy that molecules need to attain in order for a chemical reaction to occur. If the molecules do not possess the required energy, they will not successfully react even if they collide with the proper orientation. The rate of a reaction depends on factors such as temperature, concentration, and presence of a catalyst.
03

Consider room temperature conditions

At room temperature, the average kinetic energy of the hydrogen and oxygen molecules is relatively low. Consequently, most collisions between the molecules do not possess sufficient energy to overcome the activation energy barrier for the explosive reaction to occur.
04

Role of a catalyst

A catalyst can lower the activation energy of a reaction, allowing it to occur more easily at lower temperatures. However, in this scenario, there is no catalyst present to initiate the explosive reaction between hydrogen and oxygen at room temperature.
05

Conclusion

Hydrogen and oxygen do not react explosively at room temperature because the average kinetic energy of the molecules at this temperature is not high enough to overcome the activation energy barrier. In the absence of a catalyst or a higher temperature, these two gases can coexist indefinitely without undergoing a reaction.

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Most popular questions from this chapter

The rate law for a reaction can be determined only from experiment and not from the balanced equation. Two experimental procedures were outlined in Chapter \(11 .\) What are these two procedures? Explain how each method is used to determine rate laws.

The activation energy of a certain uncatalyzed biochemical reaction is \(50.0 \space\mathrm{kJ} / \mathrm{mol} .\) In the presence of a catalyst at \(37^{\circ} \mathrm{C}\) the rate constant for the reaction increases by a factor of \(2.50 \times 10^{3}\) as compared with the uncatalyzed reaction. Assuming the frequency factor \(A\) is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

A first-order reaction has rate constants of \(4.6 \times 10^{-2} \mathrm{s}^{-1}\) and \(8.1 \times 10^{-2} \mathrm{s}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(20 .^{\circ} \mathrm{C},\) respectively. What is the value of the activation energy?

A proposed mechanism for a reaction is $$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9}^{+}+\mathrm{Br}^{-} \quad \text { Slow }$$ $$\mathrm{C}_{4} \mathrm{H}_{9}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}^{+} \quad \text { Fast }$$ $$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}+\mathrm{H}_{3} \mathrm{O}^{+}\quad \text { Fast }$$ Write the rate law expected for this mechanism. What is the overall balanced equation for the reaction? What are the intermediates in the proposed mechanism?

The activation energy for the decomposition of \(\mathrm{HI}(g)\) to \(\mathrm{H}_{2}(g)\) and \(\mathrm{I}_{2}(g)\) is \(186 \mathrm{kJ} / \mathrm{mol} .\) The rate constant at \(555 \space\mathrm{K}\) is \(3.52 \times\) \(10^{-7} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s} .\) What is the rate constant at \(645 \space\mathrm{K} ?\)
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