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Chapter 1: Problem 4

A multi-step reaction takes place with the following elementary steps: $\begin{array}{ll}{\text { Step I. }} & {A+B=C} \\ {\text { Step II. }} & {C+A \rightarrow D} \\ {\text { Step III. }} & {C+D \rightarrow B+E}\end{array}$ Why would increasing the temperature make the reaction rate go up? (A) It is an endothermic reaction that needs an outside energy source to function. (B) The various molecules in the reactions will move faster and collide more often. (C) The overall activation energy of the reaction will be lowered. (D) A higher fraction of molecules will have the same activation energy.

Short Answer

Expert verified
The correct answer is (B) The various molecules in the reactions will move faster and collide more often.

Step by step solution

01

Understanding The Role of Temperature in Chemical Reactions

Temperature is directly related to the kinetic energy of molecules. As temperature increases, the kinetic energy of the molecules also increases. This means that the molecules move faster.
02

Influence of Faster Molecular Movement

When molecules move faster, they are likely to collide more frequently. More collisions increase the chances of reactant particles hitting each other with the right orientation and with enough energy to overcome the activation energy of the reaction.
03

Understanding Activation Energy

The activation energy of a reaction is the minimum amount of energy required for a reaction to occur. However, not all molecules in a system will have the same energy - it follows a distribution. At higher temperatures, more molecules will have energy equal to or greater than the activation energy, hence more successful collisions leading to a reaction are possible.
04

Evaluating the Provided Options

\(A\) is incorrect because the endothermic or exothermic nature of a reaction has no direct relationship with the frequency of molecular collisions. \(C\) is incorrect because the activation energy of a reaction is intrinsic and doesn't decrease with higher temperature. \(D\) is incorrect because a larger fraction of molecules will have energy greater than the activation energy, but they won't necessarily have the same activation energy. Therefore, \(B\) is correct: with increased temperature, molecules will move faster and collide more often.

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Most popular questions from this chapter

Which substance would have the highest vapor pressure? (A) Ethanol, because of the hybridization of its carbon atoms (B) Actone, because it exhibits the weakest intermolecular forces (C) Ethylene glycol, because it has the most lone pairs assigned to individual atoms (D) All three substances would have similar vapor pressure because they have a similar number of electrons.

A 0.1-molar solution of which of the following acids will be the best conductor of electricity? (A) \(\mathrm{H}_{2} \mathrm{CO}_{3}\) (B) \(\mathrm{H}_{2} \mathrm{S}\) (C) \(\mathrm{HF}\) (D) \(\mathrm{HNO}_{3}\)

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Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ If the \(\mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu}(s)\) half-reaction has a standard reduction potential of \(+0.52 \mathrm{V},\) what is the standard reduction potential for the \(\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}(s)\) half-reaction? (A) \(+0.76 \mathrm{V}\) (B) \(-0.76 \mathrm{V}\) (C) \(+0.24 \mathrm{V}\) (D) \(-0.24 \mathrm{V}\)

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. How much formic acid is necessary to reach the equivalence point? (A) 10.0 mL (B) 20.0 mL (C) 30.0 mL (D) 40.0 mL
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