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Which of the following reactions would you expect to proceed at a faster rate at room temperature? Why? (Hint: Think about which reaction would have the lower activation energy.) $$2 \mathrm{Ce}^{4+}(a q)+\mathrm{Hg}_{2}^{2+}(a q) \longrightarrow 2 \mathrm{Ce}^{3+}(a q)+2 \mathrm{Hg}^{2+}(a q)$$ $$\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)$$

Short Answer

Expert verified
The second reaction, H₃O⁺(aq) + OH⁻(aq) → 2 H₂O(l), is expected to proceed at a faster rate at room temperature, as it is an acid-base neutralization reaction that generally has a lower activation energy compared to the heavy metal redox reaction involving cerium and mercury ions in the first reaction.

Step by step solution

01

Reaction 1: Cerium and Mercury Reaction

In this reaction, we have cerium ions with a +4 charge reacting with a mercury ion with a +2 charge. The result of the reaction will be cerium ions with a +3 charge and mercury ions with a +2 charge. This reaction involves the change of oxidation states of both cerium and mercury ions.
02

Reaction 2: Hydronium and Hydroxide Reaction

In this reaction, hydronium ions (H_3O^+), which are essentially water molecules with an extra H^+ attached, react with hydroxide ions (OH^-), forming two water molecules (H_2O). This reaction is actually a classic example of an acid-base neutralization reaction, as hydronium ions are acidic and hydroxide ions are basic.
03

Comparing Activation Energies

In general, neutralization reactions, such as Reaction 2, have relatively low activation energies because acid-base reactions tend to be fast. On the other hand, Reaction 1 involves the change of oxidation states of heavy metal ions, such as cerium and mercury, which may require a higher activation energy for the reaction to proceed.
04

Conclusion

Considering the two reactions and the general trend of activation energies for acid-base reactions and heavy metal redox reactions, we can conclude that Reaction 2 (H_3O^+ + OH^- -> 2 H_2O) is expected to have a lower activation energy and therefore proceed at a faster rate at room temperature.

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Key Concepts

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

Activation Energy
Activation energy is a critical concept in chemical kinetics. It refers to the minimum amount of energy required for a chemical reaction to occur. Think of it as the initial "push" needed to get substances to react.
Imagine a rollercoaster. The activation energy is like the initial climb up the track. Once you reach the top, the ride can continue down the other side with the energy gained from the climb.
For a reaction to happen, molecules must collide with sufficient energy to overcome the activation barrier. Low activation energy means reactions can occur rapidly even at room temperature.
  • Reactions with low activation energy often occur more quickly.
  • Temperature variations can affect the speed of reactions by influencing this energy threshold.
In contrast, reactions with high activation energy might need more energy input, like heat, to occur.
Acid-Base Reactions
Acid-base reactions are fundamental in chemistry, often resulting in the formation of water and a salt. The essential players here are acids, which donate protons (\( H^+ \) ions), and bases, which accept them.
In the example above, hydronium (\( H_3O^+ \)) acts as the acid and hydroxide (\( OH^- \)) as the base. When they react, they form water (\( H_2O \)), which is a neutralization process.
Why do these reactions occur so quickly?
  • Acid-base reactions typically have low activation energy.
  • These reactions are often exothermic, releasing energy, which further propels the reaction.
This is why these reactions usually proceed faster and are often more straightforward to observe in a lab setting.
Redox Reactions
Redox reactions, or reduction-oxidation reactions, are processes where electrons are transferred between substances. This electron transfer leads to changes in the oxidation states of elements involved in the reaction.
In the example provided, cerium and mercury ions are undergoing a redox process. Cerium reduces from \( ext{Ce}^{4+} \) to \( ext{Ce}^{3+} \), and mercury splits into separate ions each with a \( ext{Hg}^{2+} \) charge.
Redox reactions can be more complex due to the involvement of electron transfer:
  • They often have higher activation energies compared to acid-base reactions.
  • Reactions may involve changes in chemical structure and physical conditions.
These complexities may slow down these reactions at room temperature unless specific catalysts or conditions are used to lower the activation energy.

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Most popular questions from this chapter

Consider the reaction $$4 \mathrm{PH}_{3}(g) \longrightarrow \mathrm{P}_{4}(g)+6 \mathrm{H}_{2}(g)$$ If, in a certain experiment, over a specific time period, 0.0048 mole of \(\mathrm{PH}_{3}\) is consumed in a 2.0 - \(\mathrm{L}\) container each second of reaction, what are the rates of production of \(\mathbf{P}_{4}\) and \(\mathbf{H}_{2}\) in this experiment?

Draw a rough sketch of the energy profile for each of the following cases: a. \(\Delta E=+10 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=25 \mathrm{kJ} / \mathrm{mol}\) b. \(\Delta E=-10 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{kJ} / \mathrm{mol}\) c. \(\Delta E=-50 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{kJ} / \mathrm{mol}\)

A reaction of the form $$aA \longrightarrow Products$$gives a plot of \(\ln [\mathrm{A}]\) versus time (in seconds), which is a straight line with a slope of \(-7.35 \times 10^{-3} .\) Assuming \([\mathrm{A}]_{0}=\) \(0.0100 M,\) calculate the time (in seconds) required for the reaction to reach \(22.9 \%\) completion.

The central idea of the collision model is that molecules must collide in order to react. Give two reasons why not all collisions of reactant molecules result in product formation.

Two isomers \((A \text { and } B)\) of a given compound dimerize as follows: $$\begin{aligned} &2 \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{A}_{2}\\\ &2 \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{B}_{2} \end{aligned}$$ Both processes are known to be second order in reactant, and \(k_{1}\) is known to be 0.250 \(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C} .\) In a particular experiment \(\mathrm{A}\) and \(\mathrm{B}\) were placed in separate containers at \(25^{\circ} \mathrm{C}\) where \([\mathrm{A}]_{0}=1.00 \times 10^{-2} \mathrm{M}\) and \([\mathrm{B}]_{0}=2.50 \times 10^{-2} \mathrm{M} .\) It was found that after each reaction had progressed for 3.00 min, \([\mathrm{A}]=3.00[\mathrm{B}] .\) In this case the rate laws are defined as $$\begin{array}{l} \text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k_{1}[\mathrm{A}]^{2} \\ \text { Rate }=-\frac{\Delta[\mathrm{B}]}{\Delta t}=k_{2}[\mathrm{B}]^{2} \end{array}$$ a. Calculate the concentration of \(\mathrm{A}_{2}\) after 3.00 min. b. Calculate the value of \(k_{2}\) c. Calculate the half-life for the experiment involving A.

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