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Indicate whether each statement is true or false. (a) If you measure the rate constant for a reaction at different temperatures, you can calculate the overall enthalpy change for the reaction. (b) Exothermic reactions are faster than endothermic reactions. (c) If you double the temperature for a reaction, you cut the activation energy in half.

Short Answer

Expert verified
(a) False, (b) False, (c) False.

Step by step solution

01

Analyze Statement (a)

Statement (a) claims that measuring the rate constant at different temperatures enables the calculation of the overall enthalpy change. This is false. The rate constant variation with temperature provides information about the activation energy, not directly about the overall enthalpy change of the reaction. The overall enthalpy change is related to the enthalpy of reactants and products, not the rate constant.
02

Analyze Statement (b)

Statement (b) states that exothermic reactions are faster than endothermic reactions. This is false. The speed of a reaction depends on the activation energy, not whether the reaction is exothermic or endothermic. Exothermic reactions release energy, but this release does not necessarily correlate with a faster reaction rate.
03

Analyze Statement (c)

Statement (c) suggests that doubling the temperature halves the activation energy. This is false. Doubling the temperature generally increases the rate of reaction but does not directly alter the activation energy value. The activation energy is a barrier to reaction that remains constant unless the reaction pathway is altered by a catalyst.

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Key Concepts

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

Rate Constant
In the study of chemical kinetics, the rate constant is an essential factor. It is a number that helps determine how fast a reaction occurs. The rate constant is influenced by various factors, but particularly by the temperature at which a reaction takes place. According to the Arrhenius equation, the rate constant \( k \) can be expressed as: \[ k = A e^{-\frac{E_a}{RT}} \]
Here, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.
  • An increase in temperature generally leads to an increase in the rate constant, making reactions proceed faster.
  • The rate constant provides information about the speed of a reaction at a given temperature but does not directly relate to the total energy change of the system.

Thus, while knowing how the rate constant changes with temperature can give insights into the reaction’s activation energy, it doesn’t directly inform us about the enthalpy change, or the heat absorbed or released in the reaction.
Exothermic and Endothermic Reactions
Chemical reactions are classified into two major types based on their energy changes: exothermic and endothermic reactions. This classification is driven primarily by the enthalpy change associated with a reaction.
  • Exothermic Reactions: These reactions release energy to their surroundings, often in the form of heat. This can make the surroundings feel warmer as the reaction takes place.
  • Endothermic Reactions: These reactions absorb energy from their surroundings. They require an input of energy to proceed, sometimes resulting in the cooling of their surroundings.

It is crucial to understand that the speed of a reaction is not inherently determined by whether it is exothermic or endothermic. Rather, the rate of a reaction depends more on its activation energy and other kinetic factors.
Therefore, as opposed to common belief, exothermic reactions aren't necessarily faster than endothermic ones. The speed depends on the specific conditions and energy barriers involved in the reaction process.
Activation Energy
Activation energy is a critical concept in chemical kinetics. It refers to the minimum amount of energy needed for a chemical reaction to proceed. Think of activation energy as a hill that reactants need to climb over for the reaction to occur.
  • Reactions with lower activation energy usually proceed faster because they require less energy to get started.
  • On the other hand, reactions with high activation energy need more energy input and typically occur slowly without additional factors like catalysts.

Temperature plays a crucial role in overcoming activation energy. As temperature increases, particles move more energetically, which can lead to more frequent and successful collisions that surpass the activation energy barrier. However, it's important to note that altering the temperature affects how often these collisions occur and their energy but does not change the activation energy itself.
In some cases, catalysts can be used to lower the activation energy required, allowing the reaction to proceed faster without the need for additional heat. Ultimately, understanding activation energy is key to controlling and predicting the speed at which chemical reactions happen.

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Most popular questions from this chapter

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law, \((\mathbf{b})\) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\) what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathbf{c})\) What is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of NO is increased to \(0.10 \mathrm{M}\) while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M} ?\) (d) What is the reaction rate at \(1000 \mathrm{~K}\) if \([\mathrm{NO}]\) is decreased to \(0.010 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]\) is increased to \(0.030 \mathrm{M} ?\)

(a) For the generic reaction \(\mathrm{A} \rightarrow \mathrm{B}\) what quantity, when graphed versus time, will yield a straight line for a first- order reaction? (b) How can you calculate the rate constant for a first-order reaction from the graph you made in part (a)?

The gas-phase decomposition of ozone is thought to occur by the following two- step mechanism. Step \(1: \quad \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{O}_{2}(g)+\mathrm{O}(g)\) (fast) Step 2: \(\mathrm{O}(g)+\mathrm{O}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{O}_{2}(g)\) (slow) (a) Write the balanced equation for the overall reaction. (b) Derive the rate law that is consistent with this mechanism. (Hint: The product appears in the rate law.) (c) Is \(\mathrm{O}\) a catalyst or an intermediate? (d) If instead the reaction occurred in a single step, would the rate law change? If so,

The following kinetic data are collected for the initial rates of a reaction \(2 \mathrm{X}+\mathrm{Z} \longrightarrow\) products: \begin{tabular}{llll} \hline Experiment & {\([\mathrm{X}]_{0}(M)\)} & {\([\mathrm{Z}]_{0}(\mathrm{M})\)} & Rate \((\mathrm{M} / \mathrm{s})\) \\ \hline 1 & 0.25 & 0.25 & \(4.0 \times 10^{1}\) \\ 2 & 0.50 & 0.50 & \(3.2 \times 10^{2}\) \\ 3 & 0.50 & 0.75 & \(7.2 \times 10^{2}\) \\ \hline \end{tabular} (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(\mathrm{X}\) is \(0.75 \mathrm{M}\) and that of \(\mathrm{Z}\) is \(1.25 \mathrm{M} ?\)

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{SiO}_{2}\right) .\) (a) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) How does the surface area affect the rate of reaction?

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