For a simple one-step reversible reaction at equilibrium, the relationship between the forward and reverse rate constants and the equilibrium constant is: A k, k_1 Blog [AJog K. %D Show that this is also true for the following multi-step equilibrium. Suppose that the equilibrium A B, has a mechanism involving two reversible steps, shown below. Show that the equilibrium constant for the reaction will be the quotient of the product of the forward rate constants of the individual steps divided by the product of the reverse rate constants of the steps. A + M C+ M k, k. k2, k-2 hint : If the reaction is at equilibrium, all reversible steps are also at equilihrium

For a simple one-step reversible reaction at equilibrium, the relationship between the forward and reverse rate constants and the equilibrium constant is: A k, k_1 Blog [AJog K. %D Show that this is also true for the following multi-step equilibrium. Suppose that the equilibrium A B, has a mechanism involving two reversible steps, shown below. Show that the equilibrium constant for the reaction will be the quotient of the product of the forward rate constants of the individual steps divided by the product of the reverse rate constants of the steps. A + M C+ M k, k. k2, k-2 hint : If the reaction is at equilibrium, all reversible steps are also at equilihrium

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

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Magnitude of the Equilibrium Constant 7 of 22 Review | Constants | Periodic Ta Part D For a reaction For the reaction aA + bB = cC + dD CO(g) + H2O(g) = H2(g) + CO2(g), K = 4.24 at 800 K the equilibrium constant K is defined as the ratio of products to reactants at equilibrium: what can be said about this reaction at this temperature? K [C]°D]* • View Available Hint(s) [A]*[B]* Because K represents the ratio of products to reactants, the magnitude of K is an indicator of the levels of products and reactants present when the reaction is at equilibrium. Some chemical reactions proceed almost fully to product, whereas other chemicals hardly react with each other at all, or react so as to have significant amounts of both reactants and products at equilibrium. O The equilibrium lies far to the right. O The reaction will proceed very slowly. O The reaction contains significant amounts of products and reactants at equilibrium. O The equilibrium lies far to the left.
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Consider the following system at equilibrium at 698 K. 2 HI(g)→ H₂(g) + I₂ (9) Use the References to access importa When some HI(g) is removed from the equilibrium system at constant temperature, the reaction must O run in the forward direction to reestablish equilibrium. O run in the reverse direction to reestablish equilibrium. O remain the same. It is already at equilibrium. The concentration of H₂ will O increase. O decrease. O remain the same.
Consider the equilibrium reaction and its equilibrium constant expression. [ICI]? 1(g) + Cl,(g) 2 2ICI(g) K = L] C] For the reaction 2ICI(g) = 1(g) + Cl,(g) select the equilibrium constant expression. K' О к [ICI]4 2|ICÍ] [L] [CL] L][Cl] [ICI]? [ICi]* K': K' = [Ici]? K' %3D L][CL]
COWLv2 | Online teaching and learning resource from Cengage Le... Consider the reaction: K= [Review Topics] [References] Use the References to access important values if needed for this question. P(s) + 3/2Cl₂ (g) → PCl3 (g) In progress Write the equilibrium constant for this reaction in terms of the equilibrium constants, K₁ and K2, for the reactions below: P(s) + 5/2Cl₂ (g) PC13 (9) + Cl₂ (g) For answers with both a subscript and a superscript, enter the subscript first. For example, enter K if the first equilibrium constant should be squared. Submit Answer PC15 (9) K₁ 55 PC15 (g) K₂ b Home | bartleby Retry Entire Group 9 more group attempts remaining C +
A student ran the following reaction in the laboratory at 1080 K: 2SO3(g) 2S02(g) + O₂(g) When he introduced SO3(g) at a pressure of 0.886 atm Into a 1.00 L evacuated container, he found the equilibrium partial pressure of SO3(g) to be 0.340 atm. Calculate the equilibrium constant, Kp, he obtained for this reaction. Kp = Submit Answer 44 $ T R F V/ % 5 Retry Entire Group 9 more group attempts remaining T Cengage Learning Cengage Technical Support F5 G ^ 6 MacBook Air Y H & 7 RN U * 0 J ► 11 F8 M ( 9 ►► K F9 O 1 F11 + { Save and Exit = F12 11
9 q eq eq req req 1req OWLv2 | Online teaching and learning resource from Cengage Le... Submit Answer [Review Topics] [References] Use the References to access important values if needed for this question. The equilibrium constant, Kp, for the following reaction is 0.497 at 500. K. PC15 (9) PC13 (g) + Cl₂ (g) If an equilibrium mixture of the three gases in a 12.7 L container at 500. K contains PC15 at a pressure of 0.804 atm and PC13 at a pressure of 0.911 atm, the equilibrium partial pressure of Cl₂ is l atm. b Home | bartleby Retry Entire Group 9 more group attempts remaining
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At 777 K the reaction shown below is run with initial concentrations of [N2]=0.120 M and [O2] = 0.120 M.The reaction is allowed to reach equilibrium and has K=1.50x10-3 Calculate the concentration of NO at equilibrium.N2(g)+O2(g)↔2NO(g)
The value of K, for the reaction 2 NO, (g) N204 (g) is 1.52 at 319 K. What is the value of the Kp of the reaction NO2 (g) 1/2 N204 (g) at this temperature: 0.761 O2.31 1.23 1.52 0.811 A Moving to another question will save this response. Esc AI C 中 F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 23 %24
Consider the following equilibrium: N204(g) = 2NO2(g) Suppose two different experiments were conducted on this equilibrium, at two different temperatures. Trial Temperature Initial Moles of N204 Equilibrium Moles of N204 1 25°C 1.00 0.815 55°C 1.00 0.500 What conclusion may be drawn from the above data? O Kc> Kp at all temperatures. The reaction is exothermic. The forward reaction proceeds faster than the reverse reaction. OKp =D1 at 55°C. O The equilibrium constant increases with increasing temperature.
A chemical equilibrium A ⇄ 2B has a forward rate constant, kf = 12 M –1 s–1, and a reverse rate constant, kr = 18 s–1. If the system has a concentration of [A] = 0.15 M at equilibrium, what is the concentration of B at equilibrium? (can't be in science notation)