Question: Conventional equilibrium considerations do not apply when a reaction is being driven by light absorption. Thus the steady-state concentration of products and reactants might differ

Conventional equilibrium considerations do not apply when a reaction is being driven by light absorption. Thus 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 I_a, but that the reverse reaction B→A is bimolecular and second order with a rate k_r[B]². What is the stationary state concentration of B? Why does this ‘photostationary state’ differ from the equilibrium state?

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