/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 $$\mathrm{H}_{2}(g)+\mathrm{I}_{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 9

$$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{HI}(g)$$ When the reaction given above takes place in a sealed isothermal container, the rate law is $$\text { Rate }=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]$$ If a mole of \(\mathrm{H}_{2}\) gas is added to the reaction chamber and the temperature remains constant, which of the following will be true? (A) The rate of reaction and the rate constant will increase. (B) The rate of reaction and the rate constant will not change. (C) The rate of reaction will increase and the rate constant will decrease. (D) The rate of reaction will increase and the rate constant will not change.

Short Answer

Expert verified
The correct answer is (D). The rate of reaction will increase due to the addition of a mole of H2, and the rate constant will not change.

Step by step solution

01

Understand the rate law equation

The rate law equation for the reaction: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{HI}(g)\) is given as Rate = k [H2] [I2]. This represents the rate of the chemical reaction as being proportional to the product of the concentration of Hydrogen gas [H2] and Iodine gas [I2]. The rate constant (k) is a proportionality constant that depends on factors like the temperature and catalysts, but not the concentration of reactants. So, when the concentration of a reactant is increased, the rate of the reaction increases, but the rate constant remains unchanged.
02

Interpret the effect of adding a mole of H2

In our case, a mole of H2 gas is added to the reaction chamber so the concentration of H2 is increased. Because H2 is a reactant in this equation and appears in the rate law as such, increasing the concentration of H2 will increase the rate of the reaction. But the rate constant (k) will remain the same because it does not depend on the concentration of reactants.
03

Identify the validity of the options

Now based on the analysis given, it can deducted that when a mole of H2 gas is added to the reactant chamber, the rate of reaction will increase but the rate constant will not change. So option (D) is the correct answer. The rate constant k is a property of the reaction at a given temperature and is not affected by changes in concentrations of reactants or products.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Rate Law
The rate law is essential for understanding how the speed of a chemical reaction depends on the concentration of the reactants. It is expressed as an equation: Rate = k [Reactant1] [Reactant2] ... This equation signifies that the reaction rate is multiplied by the product of the concentrations of reactants raised to some power. The powers, usually represented by exponents in the rate law, denote the reaction order with respect to each reactant. For our specific reaction, the rate law is written as Rate = k [H2][I2]. Here, the reaction is first order in hydrogen gas, H2, and first order in iodine gas, I2.
The overall order of the reaction is the sum of these individual orders. In this case, it's second order because 1 (for H2) + 1 (for I2) = 2. Understanding rate laws helps in predicting how a change in concentrations will affect the reaction rate.
Rate Constant
The rate constant, denoted as k, is a crucial part of the rate law equation. It's a proportionality constant that ties the concentration of reactants to the rates of reaction. Even though it sounds complex, the rate constant simplifies the relationship between [Reactants] and the Rate.
  • The value of k provides insights on the reaction speed at a given temperature.
  • The rate constant changes with changes in temperature and the presence of a catalyst.
  • It's specific to a particular reaction at a specific set of conditions.

In our equation, the rate constant remains unchanged when concentrations change, emphasizing its nature as a factor reflecting conditions rather than concentration itself.
Concentration Effect
Concentration effect refers to how the amount of each reactant can influence the speed of a reaction. In chemical kinetics, concentration significantly impacts the reaction rate, as reflected in the rate law equation. When concentrations of reactants increase, they often lead to an increase in the rate of reaction.
For example, in our reaction, if the concentration of H2 or I2 increases, the rate of the reaction will increase proportionally. This is because more reactants imply more collision occurrences, which enhances the likelihood of products forming.
However, it is important to remember that while concentration changes affect the rate of reaction, they do not alter the rate constant itself.
Reaction Rate
Understanding the reaction rate is vital, as it tells how fast a chemical reaction occurs. The reaction rate can vary and depends on several factors, including concentration, temperature, and presence of catalysts.
  • In the context of our problem, the reaction rate increases with the addition of more H2, due to its concentration increase.
  • This occurs even though other factors, like the rate constant, stay constant.
  • Theres a direct correlation between increased concentrations of reactants and a faster reaction rate, provided that other conditions remain stable.

Thus, when studying reaction rates, focusing on how the concentration of reactants plays a crucial role is often the simplest approach to understanding kinetics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ What is the required voltage to make this cell function? (A) 0.34 V (B) 0.42 V (C) 0.76 V (D) 1.10 V

$$\begin{array}{|c|c|}\hline \text { Time (Hours) } & {[\mathrm{A}] M} \\\ \hline 0 & {0.40} \\ \hline 1 & {0.20} \\ \hline 2 & {0.10} \\ \hline 3 & {0.05} \\ \hline\end{array}$$ Reactant A underwent a decomposition reaction. The concentration of A was measured periodically and recorded in the chart above. Based on the data in the chart, which of the following is the rate law for the reaction? (A) Rate \(=k[\mathrm{A}]\) (B) Rate \(=k[\mathrm{A}]^{2}\) (C) Rate \(=2 k[\mathrm{A}]\) (D) Rate \(=\frac{1}{2} k[\mathrm{A}]\)

Directions: Questions 4-7 are short free-response questions that require about 9 minutes each to answer and are worth 4 points each. Write your response in the space provided following each question. Examples and equations may be included in your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures. Hyprobromous acid, HBrO, is a weak monoprotic acid with a \(K_{\mathrm{a}}\) value of \(2.0 \times 10^{-9} \mathrm{at} 25^{\circ} \mathrm{C} .\) (a) Write out the equilibrium reaction of hyprobromous acid with water, identifying any conjugate acid/based pairs present. (b) (i) What would be the percent dissociation of a 0.50 M solution of hyprobromous acid? (ii) If the 0.50 M solution were diluted, what would happen to the percent dissociation of the HBrO? Why?

Directions: Questions 1-3 are long free-response questions that require about 23 minutes each to answer and are worth 10 points each. Write your response in the space provided following each question. Examples and equations may be included in your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures. A student is tasked with determining the identity of an unknown carbonate compound with a mass of 1.89 g. The compound is first placed in water, where it dissolves completely. The \(K_{s p}\) value for several carbonate-containing compounds are given below. $$\begin{array}{|c|c|}\hline \text { Compound } & {K_{s p}} \\ \hline \text { Lithium carbonate } & {8.15 \times 10^{-4}} \\ \hline \text { Nickel (II) carbonate } & {1.42 \times 10^{-7}} \\ \hline \text { Strontium carbonate } & {5.60 \times 10^{-10}} \\ \hline\end{array}$$ (a) In order to precipitate the maximum amount of the carbonate ions from solution, which of the following should be added to the carbonate solution: \(\operatorname{LiNO}_{3}, \mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2},\) or \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2} ?\) Justify your answer. (b) For the carbonate compound that contains the cation chosen in part (a), determine the concentration of each ion of that compound in solution at equilibrium. (c) When mixing the solution, should the student ensure the carbonate solution or the nitrate solution is in excess? Justify your answer. (d) After titrating sufficient solution to precipitate out all of the carbonate ions, the student filters the solution before placing it in a crucible and heating it to drive off the water. After several heatings, the final mass of the precipitate remains constant and is determined to be 2.02 g. (i) Determine the number of moles of precipitate. (ii) Determine the mass of carbonate present in the precipitate. (e) Determine the percent, by mass, of carbonate in the original sample. (f) Is the original compound most likely lithium carbonate, sodium carbonate, or potassium carbonate? Justify your answer.

Which of the following compounds would have the highest lattice energy? (A) LiF (B) MgCl_ (C) CaBr_ (D) \(\mathrm{C}_{2} \mathrm{H}_{6}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.