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Chapter 14: Problem 49

How does a catalyst increase the rate of a reaction?

Short Answer

Expert verified
A catalyst increases the rate of a reaction by reducing the activation energy required for the reaction to proceed, providing an alternative pathway. This means more molecules have the necessary energy to react, hence the reaction happens faster.

Step by step solution

01

Define Catalyst

A catalyst is a substance that can significantly change the rate (speed up or slow down) of a chemical reaction without being consumed or changed itself. It is important in numerous chemical processes.
02

Understand the Action of a Catalyst

A catalyst functions by providing an alternative reaction pathway with a lower activation energy compared to the non-catalyzed reaction. The activation energy is the minimum energy required for a chemical reaction to process. It achieves this by forming transient, intermediate compounds with the reactants, effectively 'holding' them in a particular orientation that encourages the desired interaction.
03

Describe how this Increases the Reaction Rate

Because the catalyst provides an alternative pathway with a lower activation energy, more molecules have the necessary energy to engage in the reaction. Consequently, the reaction happens faster, which means the rate of the reaction increases.

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Most popular questions from this chapter

The rate law for this reaction $$ \mathrm{CO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{NO}(g) $$ is rate \(=k\left[\mathrm{NO}_{2}\right]^{2}\). Suggest a plausible mechanism for the reaction, given that the unstable species \(\mathrm{NO}_{3}\) is an intermediate.

The rate of the reaction $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(a q) &+\mathrm{H}_{2} \mathrm{O}(l) \\ \longrightarrow & \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \end{aligned} $$ shows first-order characteristics-that is, rate \(=\) \(k\left[\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\right]\) - even though this is a second- order reaction (first order in \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\) and first order in \(\mathrm{H}_{2} \mathrm{O}\) ). Explain.

To carry out metabolism, oxygen is taken up by hemoglobin (Hb) to form oxyhemoglobin \(\left(\mathrm{HbO}_{2}\right)\) according to the simplified equation $$ \mathrm{Hb}(a q)+\mathrm{O}_{2}(a q) \stackrel{k}{\longrightarrow} \mathrm{HbO}_{2}(a q) $$ where the second-order rate constant is \(2.1 \times\) \(10^{6} / M \cdot \mathrm{s}\) at \(37^{\circ} \mathrm{C}\). (The reaction is first order in \(\mathrm{Hb}\) and \(\mathrm{O}_{2} .\) ) For an average adult, the concentrations of \(\mathrm{Hb}\) and \(\mathrm{O}_{2}\) in the blood at the lungs are \(8.0 \times 10^{-6} M\) and \(1.5 \times 10^{-6} M,\) respectively. (a) Calculate the rate of formation of \(\mathrm{HbO}_{2}\). (b) Calculate the rate of consumption of \(\mathrm{O}_{2}\). (c) The rate of formation of \(\mathrm{HbO}_{2}\) increases to \(1.4 \times 10^{-4} \mathrm{M} / \mathrm{s}\) during exercise to meet the demand of increased metabolism rate. Assuming the Hb concentration to remain the same, what must be the oxygen concentration to sustain this rate of \(\mathrm{HbO}_{2}\) formation?

Consider this mechanism for the enzyme-catalyzed reaction $$ \mathrm{E}+\mathrm{S} \stackrel{k_{1}}{\rightleftharpoons_{-1}} \mathrm{ES} \quad \text { (fast equilbrium) } $$ $$ \mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} \quad(\text { slow }) $$ 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 the forward reaction is equal to the rate of the reverse reaction.)

Are enzyme-catalyzed reactions examples of homogeneous or heterogeneous catalysis?
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