The reactiorn is first order in both 12 and SiH4, with k = 2.94 L mori,-1 at 264

Question

The reactiorn is first order in both 12 and SiH4, with k = 2.94 L mori,-1 at 264°C. If 1.00 L of 12(g) at a concentration of 2.52x10. 3 M is rapidly mixed with the same volume of SiH(g) also at a concentration of 2.52-10-3 M what is the time (in seconds) required for the I2 concentration to decrease to a value of 1.37 10-M HINT Because the reaction conditions and stoichiometry are such that the concentrations of the two reacting species are always equal, the integrated rate law for the reaction is: c Co where c and co represent concentrations of either of the reactants

Explanation / Answer

Ans. Note that the volume and concertation of both [I2] and [SiH4] taken in the reaction mixture are equal to each other. SO, following 1:1 stoichiometry of reactants, there is NO limiting reactant in the reaction mixture.

# We have-

            Overall order of reaction = 1 (w.r.t.[I]) + 1 (w.r.t [SiH4]) = 2

            Initial [I2], c0 = 2.52 x 10-3 M

            Final [I2], c = 1.37 x 10-4 M

Now, putting the values in second-order kinetic expression-

            1/ (1.37 x 10-4 M) = 1 / (2.52 x 10-3 M) + (2.94 L mol-1 s-1) x t

            Or, 7299.3 M-1 - 396.83 M-1 = (2.94 L mol-1 s-1) x t

            Or, t = 6902.44 M-1 / (2.94 M-1 s-1)

            Hence, t = 2347.77 s

Therefore, required time, t = 2347.77 s

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