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How does an increase in temperature affect the rate of a reaction? Explain the two factors involved.

Short Answer

Expert verified
An increase in temperature increases the rate of a reaction by increasing both the frequency of collisions and the number of molecules with sufficient energy to surpass the activation energy.

Step by step solution

01

- Understand the Collision Theory

The Collision Theory states that chemical reactions can only occur when reactant molecules collide with sufficient energy and proper orientation. An increase in temperature causes molecules to move faster and collide more frequently, which can increase the rate of the reaction.
02

- Activation Energy

Activation energy is the minimum amount of energy required for a reaction to occur. With higher temperatures, more molecules have the energy needed to surpass the activation energy barrier. This results in more effective collisions leading to an increased reaction rate.
03

- Kinetic Energy Distribution

At higher temperatures, the distribution of kinetic energy among molecules increases. This means a greater proportion of molecules have higher kinetic energy, which further increases the number of collisions that have enough energy to overcome the activation energy barrier.

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Key Concepts

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

Collision Theory
The Collision Theory is key to understanding how temperature impacts the rate of chemical reactions. According to this theory, for a reaction to occur, reactant molecules must collide with each other. However, not just any collision will do the trick. The collisions need to have sufficient energy and the correct orientation. When the temperature increases, molecules gain kinetic energy, causing them to move faster. This increased movement leads to more frequent collisions. As the number of collisions increases, the chances of these collisions having the necessary energy and proper orientation also go up. This boosts the rate at which reactions occur.

Think of it like a crowded dance floor: the more people there are (molecules) and the faster they move (higher temperature), the more often they bump into each other (collide) in the right way to dance (react).
Activation Energy
Activation energy is the minimum energy required for a chemical reaction to happen. Every reaction needs a certain amount of this energy to get started. When the temperature increases, molecules gain more energy. This means more molecules have the required activation energy to break existing bonds and form new ones—essential steps in a chemical reaction. The higher the temperature, the more molecules can surpass this energy barrier, leading to more successful collisions and thus, a faster reaction rate.

Imagine pushing a boulder up a hill: the activation energy is the effort needed to get the boulder to the top. At higher temperatures, it's as if you have more people pushing, making it easier to get the boulder over the hill.
Kinetic Energy Distribution
Kinetic energy distribution describes how energy spreads among molecules in a system. As the temperature of a system increases, the spread of kinetic energy among its molecules also increases. This means a larger number of molecules have higher kinetic energy. At higher temperatures, the proportion of molecules with enough energy to overcome the activation energy barrier increases. Consequently, this leads to more effective collisions, further raising the rate of the reaction.

Think of it as spreading butter on toast: at higher temperatures, the butter (kinetic energy) spreads more evenly and easily due to its melting, covering more area (molecules) and making the toast (reaction) better.

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Most popular questions from this chapter

Express the rate of this reaction in terms of the change in concentration of each of the reactants and products: $$ 2 \mathrm{~A}(g) \rightarrow \mathrm{B}(g)+\mathrm{C}(g) $$ When \([\mathrm{C}]\) is increasing at \(2 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s},\) how fast is \([\mathrm{A}]\) decreasing?

For the reaction \(\mathrm{A}(\mathrm{g}) \longrightarrow \mathrm{B}(\mathrm{g})\), sketch two curves on the same set of axes that show (a) The formation of product as a function of time (b) The consumption of reactant as a function of time

Reactions between certain haloalkanes (alkyl halides) and water produce alcohols. Consider the overall reaction for \(t\) -butyl bromide ( 2 -bromo- 2 -methylpropane): \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}(a q)+\mathrm{H}^{+}(a q)+\mathrm{Br}^{-}(a q)\) The experimental rate law is rate \(=k\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\right] .\) The accepted mechanism for the reaction is \((1)\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}(a q) \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(a q)+\mathrm{Br}^{-}(a q) \quad[\mathrm{slow}]\) (2) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}_{2}^{+}(a q) \quad[\) fast\(]\) (3) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}_{2}^{+}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}(a q) \quad[\) fast\(]\) (a) Why doesn't \(\mathrm{H}_{2} \mathrm{O}\) appear in the rate law? (b) Write rate laws for the elementary steps. (c) What reaction intermediates appear in the mechanism? (d) Show that the mechanism is consistent with the experimental rate law.

In a kinctics experiment, a chemist places crystals of iodine in a closed reaction vessel, introduces a given quantity of \(\mathrm{H}_{2}\) gas, and obtains data to calculate the rate of HI formation. In a second experiment, she uses the same amounts of iodine and hydrogen but first warms the flask to \(130^{\circ} \mathrm{C}\), a temperature above the sublimation point of iodine. In which of these experiments does the reaction proceed at a higher rate? Explain.

The effect of substrate concentration on the first-order growth rate of a microbial population follows the Monod equation: \(\mu=\frac{\mu_{\max } S}{K_{\mathrm{s}}+S}\) where \(\mu\) is the first-order growth rate \(\left(\mathrm{s}^{-1}\right), \mu_{\max }\) is the maximum growth rate \(\left(\mathrm{s}^{-1}\right), S\) is the substrate concentration \(\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) and \(K_{\mathrm{s}}\) is the value of \(S\) that gives one-half of the maximum growth rate (in \(\mathrm{kg} / \mathrm{m}^{3}\) ). For \(\mu_{\max }=1.5 \times 10^{-4} \mathrm{~s}^{-1}\) and \(K_{\mathrm{s}}=0.03 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Plot \(\mu\) vs. \(S\) for \(S\) between 0.0 and \(1.0 \mathrm{~kg} / \mathrm{m}^{3}\). (b) The initial population density is \(5.0 \times 10^{3}\) cells \(/ \mathrm{m}^{3}\). What is the density after \(1.0 \mathrm{~h}\), if the initial \(S\) is \(0.30 \mathrm{~kg} / \mathrm{m}^{3} ?\) (c) What is it if the initial \(S\) is \(0.70 \mathrm{~kg} / \mathrm{m}^{3}\) ?

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