Conventional equilibrium considerations do not apply when a reaction is driven by light absorption and the steady-state concentration of products and reactants might differ significantly from equilibrium values. For instance, suppose the reaction A → B is driven by light absorption, and that its rate is Ia, but that the reverse reaction B → A is bimolecular and second order with a rate kr[B]2. What is the stationary state concentration of B? Why does this ‘photostationary state’ differ from the equilibrium state?
Conventional equilibrium considerations do not apply when a reaction is driven by light absorption and the steady-state concentration of products and reactants might differ significantly from equilibrium values. For instance, suppose the reaction A → B is driven by light absorption, and that its rate is Ia, but that the reverse reaction B → A is bimolecular and second order with a rate kr[B]2. What is the stationary state concentration of B? Why does this ‘photostationary state’ differ from the equilibrium state?
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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Conventional equilibrium considerations do not apply when a reaction is driven by light absorption and the steady-state concentration of products and reactants might differ significantly from equilibrium values. For instance, suppose the reaction A → B is driven by light absorption, and that its rate is Ia, but that the reverse reaction B → A is bimolecular and second order with a rate kr[B]2. What is the stationary state concentration of B? Why does this ‘photostationary state’ differ from the equilibrium state?
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