2. Consider the following reactions: A -> B (1) B + M -> C (2) Assume M is present in great excess. The concentrations of B and C are zero at t =0. (a) Write out the rate equations for A, B and C A: d[A]/dt = -k1[A], B: d[B]/dt = k1[A]-k2[B][M], C: d[C]/dt = k2 [BM] (b) Assuming the pseudo-steady -state approximation for B, derive a set of equations for the time dependent variation in concentration of A, B and C. These should be derivative rate equations

Using the data in the table, determine the rate constant of the reaction and select the appropriate units. A+2B⟶C+D   Trial [?] (?) [?] (?) Rate (M/s) 1 0.340 0.210 0.0204 2 0.340 0.420 0.0204 3 0.680 0.210 0.0816 ?=

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?

Using the data in the table, determine the rate constant of the reaction and select the appropriate units.   A+2B⟶C+D Trial [?] (?) [?] (?) Rate (M/s) 1 0.360 0.290 0.0144 2 0.360 0.580 0.0144 3 0.720 0.290 0.0576 k=