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Chapter 13: Problem 59

How does a catalyst increase the rate of a reaction?

Short Answer

Expert verified
A catalyst increases the rate of a chemical reaction by reducing its activation energy, thereby enabling a larger proportion of reactant molecules to react. The catalyst does this without being consumed or altered, allowing it to continue facilitating the reaction.

Step by step solution

01

Understanding Catalysts

A catalyst is a substance that can speed up a chemical reaction by lowering the activation energy, but itself remains unchanged at the end of the reaction. It can be a solid, liquid, or gas and does not effect the position of the equilibrium of the reaction.
02

Describing Catalysts' Effect on Activation Energy

In order for a reaction to occur, reactant molecules must collide with enough energy to break existing bonds and form new ones. This required energy is called the activation energy. A catalyst works by providing an alternative reaction pathway with a lower activation energy.
03

Explaining Catalysts' Role in Reaction

In the presence of a catalyst, more reactant molecules have the necessary energy to react, because the activation energy is lower. As a result, the rate of the reaction increases. However, the catalyst itself is not consumed in the reaction, so it can continue to facilitate the reaction repeatedly.
04

Summarizing How Catalyst Increases Reaction Rate

A catalyst increases the rate of a reaction by lowering the activation energy and providing an alternative reaction pathway. In doing so, the catalyst allows more reactant atoms or molecules to have enough energy to undergo the reaction, thereby speeding up the reaction. It's important to remember, though, that the catalyst itself is not used up in the reaction, so it can continue acting as a catalyst for as long as the reaction continues.

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Most popular questions from this chapter

Consider the following mechanism for the enzymecatalyzed reaction: $$ \begin{array}{r} \mathrm{E}+\mathrm{S} \underset{k-1}{\stackrel{k_{1}}{\sum_{-1}}} \mathrm{ES} \\\ \mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} \end{array} $$ Derive an expression for the rate law of the reaction in terms of the concentrations of \(\mathrm{E}\) and \(\mathrm{S}\). (Hint: To solve for [ES], make use of the fact that, at equilibrium, the rate of forward reaction is equal to the rate of the reverse reaction.

Define activation energy. What role does activation energy play in chemical kinetics?

Some reactions are described as parallel in that the reactant simultaneously forms different products with different rate constants. An example is and $$ \begin{array}{l} \mathrm{A} \stackrel{k_{1}}{k_{2}} \mathrm{~B} \\ \mathrm{~A} \stackrel{\longrightarrow}{\longrightarrow} \mathrm{C} \end{array} $$ The activation energies are \(45.3 \mathrm{~kJ} / \mathrm{mol}\) for \(k_{1}\) and \(69.8 \mathrm{~kJ} / \mathrm{mol}\) for \(k_{2}\). If the rate constants are equal at \(320 \mathrm{~K},\) at what temperature will \(k_{1} / k_{2}=2.00 ?\)

Consider the following elementary steps for a consecutive reaction: $$ \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{C} $$ (a) Write an expression for the rate of change of \(\mathrm{B}\). (b) Derive an expression for the concentration of \(\mathrm{B}\) under steady- state conditions; that is, when \(\mathrm{B}\) is decomposing to \(\mathrm{C}\) at the same rate as it is formed from \(A\).

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 \(\left[\mathrm{NO}_{2}^{-}\right]=0.080 M\)
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