Under certain conditions the reaction H2O2 + 3I + 2H*→ I3¯ + 2H2O occurs by thefollowing series of steps:kiStep 1. H2O2+H* =H;O2"+with K (equilibrium constant) = k1/k-1Step 2.H3O2* +I→ H2O + HOI(slow, rate constant k2)Step 3.HỘI + I→ OH¯ +I2(fast, rate constant k3)Step 4.OH + H*→ H2O(fast, rate constant k4)Step 5.I2 + I→ 13(fast, rate constant k5)Use approximations (RDS, eq, steady-state) if necessaryThe rate constant k for the reaction would be given by

Introduction: The rate of reaction is defined as the speed at which formation of product takes place…

Under certain conditions the reaction H2O2 + 3I + 2H*→ I3¯+ 2H2O occurs by the following series of steps: k1 Step 1. H202+H* = H3O2* + k.1 with K (equilibrium constant) = k1/k-1 %3D Step 2. H3O2* + I→ H2O + HOI (slow, rate constant k2) Step 3. HOI + I→ OH¯ + I2 (fast, rate constant k3) Step 4. OH + H*→ H2O (fast, rate constant k4) Step 5. I2 + I'→ I3 (fast, rate constant k5) Use approximations (RDS, eq, steady-state) if necessary The rate constant k for the reaction would be given by

Under certain conditions the reaction H2O2 + 3I + 2H→ I3¯+ 2H2O occurs by the following series of steps: k1 Step 1. H202+H = H3O2* + k.1 with K (equilibrium constant) = k1/k-1 %3D Step 2. H3O2* + I→ H2O + HOI (slow, rate constant k2) Step 3. HOI + I→ OH¯ + I2 (fast, rate constant k3) Step 4. OH + H*→ H2O (fast, rate constant k4) Step 5. I2 + I'→ I3 (fast, rate constant k5) Use approximations (RDS, eq, steady-state) if necessary The rate constant k for the reaction would be given by

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter13: Chemical Equilibrium

Section: Chapter Questions

Problem 16Q: Suppose a reaction has the equilibrium constant K = 1.7 108 at a particular temperature. Will there...

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Because carbonic acid undergoes a second ionization, the student in Exercise 12.39 is concerned that the hydrogen ion concentration she calculated is not correct. She looks up the equilibrium constant for the reaction HCO,-(aq) «=* H+(aq) + COf'(aq) Upon finding that the equilibrium constant for this reaction is 4.8 X 10“H, she decides that her answer in Exercise 12.39 is correct. Explain her reasoning. A student is simulating the carbonic acid—hydrogen carbonate equilibrium in a lake: H,CO,(aq) 5=6 H+(aq) + HCO,'(aq) K = 4.4 X 10'7She starts with 0.1000 A1 carbonic acid. W hat are the concentrations of all species at equilibrium?
Le Chteliers principle is stated (Section 12-7) as follows: If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. The systemN2(g)+3H2(g)2NH3(g) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in H2 concentration and an increase in NH3, concentration. In the experiment the volume is assumed to be constant. On the other hand, if N2 is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount o f NH3 actually could decrease, and the concentration of H2 would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas. holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.
An equilibrium involving the carbonate and bicarbonate ions exists in natural waters: HCO5_(aq) «=* H+(aq) + COf-(aq) Assuming that the reactions in both directions are elementary' processes: Write rate expressions for the forward and reverse reactions. Write an expression for the equilibrium constant based on the rates of the forward and reverse reactions.
Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal?
What is Le Chteliers principle? Consider the reaction 2NOCI(g)2NO(g)+Cl2(g) If this reaction is at equilibrium. what happens when the following changes occur? a. NOCI(g) is added. b. NO(g) is added. c. NOCI(g) is removed. d. Cl2(g) is removed. e. The container volume is decreased. For each of these changes, what happens to the value of K for the reaction as equilibrium is reached again? Give an example of a reaction for which the addition or removal of one of the reactants or products has no effect on the equilibrium position. In general, how will the equilibrium position of a gas-phase reaction be affected if the volume of the reaction vessel changes? Are there reactions that will not have their equilibria shifted by a change in volume? Explain. Why does changing the pressure in a rigid container by adding an inert gas not shift the equilibrium position for a gas-phase reaction?
In the reaction in Exercise 12.33, another trial was carried out. The reaction began with an initial concentration of N2 equal to the initial concentration of NO. Each had a concentration of 0.100 mol L-1. WTat were the equilibrium concentrations of all species? The following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) ^2 NO K = 4.1 X 10~4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L'1 ofO2, what were the equilibrium concentrations of all species?
A common type of reaction we will study is that having a very small K value (K 1). Solving for equilibrium concentrations in an equilibrium problem usually requires many mathematical operations to be perfomed. However, the math involved when solving equilibrium problems for reactions having small K values (K 1) is simplified. What assumption is made when solving the equilibrium concentrations for reactions with small K values? Whenever assumptions are made, they must be checked for validity. In general, the 5% rule is used to check the validity of assuming x (or 2 x, 3x, and so on) is very small compared to some number. When x (or 2 x. 3x. and so on) is less than 5% of the number the assumption was made against, then the assumption is said to be valid. If the 5% rule fails, what do you do to solve for the equilibrium concentrations?
. In general terms. what does the equilibrium constant for a reaction represent? What is the algebraic form of the equilibrium constant for a typical reaction? What do square brackets indicate when we write an equilibrium constant?
Again the experiment in Exercise 12.33 was redesigned. This time, 0.15 mol each of N, and O2 was injected into a 5.0-L container at 2500 K, at which the equilibrium constant is 3.6 X 10-?. What was the composition of the reaction mixture at equilibrium? l'he following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) *2 2 NO K = 4.1 X IO-4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L-’ ofO2, what were the equilibrium concentrations of all species?
The experiment in Exercise 12.33 was redesigned so that the reaction started with 0.15 mol each of N2 and O2 being injected into a 1.0-L container at 2500 K. The equilibrium constant at 2500 K is 3.6 X 10“’. What was the composition of the reaction mixture after equilibrium was attained? The following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) *2 2 NO K = 4.1 X IO-4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L-’ ofO2, what were the equilibrium concentrations of all species?
7-29 The following reaction was allowed to reach equilibrium at 25°C. Under each component is its equilibrium concentration. Calculate the equilibrium constant, K, for this reaction.
Equilibrium exists between butane and isobutane when [butane] = 0.020 M and [isobutane] = 0.050 M. An additional 0.0200 mol/L of isobutane is added to the mixture. What are the concentrations of butane and isobutane after equilibrium has again been attained?