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Ch.14 - Chemical Kinetics

Chapter 14, Problem 127c

Consider the reversible, first-order interconversion of two molecules A and B: where kf = 3.0⨉10-3 s-1 is the rate constant for the forward reaction and kr = 1.0⨉10-3 s-1 is the rate constant for the reverse reaction. We'll see in Chapter 15 that a reaction does not go to completion but instead reaches a state of equilibrium with comparable concentrations of reactants and products if the rate constants kf and kr have comparable values.

(c) What are the relative concentrations of B and A when the rates of the forward and reverse reactions become equal?

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Hello. In this problem we are told the following reaction is a reversible reaction of first order. So why produces E. In the forward direction and Z produces Y in the reverse direction is forward and reverse reactions have corresponding rate constants of 1. minus four per second and one times 10 to minus four per second. When K. four and K reverse have comparable values. The reaction does not proceed to completion but instead reaches equilibrium. Where the concentration of reactant and products are equal. When the rates of the forward reaction and reverse reaction are equal. We wanna what are the relative concentrations of Y. N. C. So we begin then by re writing the rate of the forward reaction and put subscript S. F. To stanford forward. It's unequal to our reaction rate constant for the forward reaction and this is dependent on the concentration Y. And it is first order. So it's to the one power. The rate of the reverse reaction to subscript R is equal to K. Subscript. Our reaction rate constant for the reverse reaction. And this is dependent on the concentration of Z. It's the first order. So it also has a power of one. So when we reach equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. So that means that K4 times the concentration of Y. Is equal to K. Reverse times the concentration of Z. We get our reaction rate constants on one side and our concentrations to the other side. So we divide both sides by Kr and we divide both sides by why? So now we have K forward over K. Reverse is equal to Z over Y. We are told. The reaction rate constant for the forward reaction is 1.5 times 10 to the minus four per second, And that of the reverse is one times 10 to the - per second. So this works out to 1.5 over one. So our concentration of Z to Y Is 1.5 - one. So that means then the relative concentration of Z and why Our 1. and one respectively. Hope this helps. Thanks for watching.
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Textbook Question

Consider the following concentration–time data for the reaction of iodide ion and hypochlorite ion (OCl-). The products are chloride ion and hypoiodite ion (OI-).

(d) Propose a mechanism that is consistent with the rate law, and express the rate constant in terms of the rate constants for the elementary steps in your mechanism. (Hint: Transfer of an H+ ion between H2O and OCl- is a rapid reversible reaction.)

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Textbook Question

Consider the reversible, first-order interconversion of two molecules A and B: where kf = 3.0⨉10-3 s-1 is the rate constant for the forward reaction and kr = 1.0⨉10-3 s-1 is the rate constant for the reverse reaction. We'll see in Chapter 15 that a reaction does not go to completion but instead reaches a state of equilibrium with comparable concentrations of reactants and products if the rate constants kf and kr have comparable values.

(a) What are the rate laws for the forward and reverse reactions?

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Textbook Question

Consider the reversible, first-order interconversion of two molecules A and B: where kf = 3.0⨉10-3 s-1 is the rate constant for the forward reaction and kr = 1.0⨉10-3 s-1 is the rate constant for the reverse reaction. We'll see in Chapter 15 that a reaction does not go to completion but instead reaches a state of equilibrium with comparable concentrations of reactants and products if the rate constants kf and kr have comparable values.

(b) Draw a qualitative graph that shows how the rates of the forward and reverse reactions vary with time.

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Textbook Question
Assume that you are studying the first-order conversion of a reactant X to products in a reaction vessel with a constant volume of 1.000 L. At 1 p.m., you start the reaction at 25 °C with 1.000 mol of X. At 2 p.m., you find that 0.600 mol of X remains, and you immediately increase the temperature of the reaction mixture to 35 °C. At 3 p.m., you discover that 0.200 mol of X is still present. You want to finish the reaction by 4 p.m. but need to continue it until only 0.010 mol of X remains, so you decide to increase the temperature once again. What is the minimum temperature required to convert all but 0.010 mol of X to products by 4 p.m.?
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Textbook Question
The half-life for the first-order decomposition of N2O4 is 1.3 * 10-5 s. N2O41g2S 2 NO21g2 If N2O4 is introduced into an evacuated flask at a pressure of 17.0 mm Hg, how many seconds are required for the pressure of NO2 to reach 1.3 mm Hg?
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Textbook Question

Some reactions are so rapid that they are said to be diffusion-controlled; that is, the reactants react as quickly as they can collide. An example is the neutralization of H3O+ by OH-, which has a second-order rate constant of 1.3⨉1011 M-1 s-1 at 25 °C. (a) If equal volumes of 2.0 M HCl and 2.0 M NaOH are mixed instantaneously, how much time is required for 99.999% of the acid to be neutralized?

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