Lindemann’s mechanism for a unimolecular react 6. At 500°c and an initial pressu
ID: 703719 • Letter: L
Question
Lindemann’s mechanism for a unimolecular react 6. At 500°c and an initial pressure of 101.325kPa, the thermal decomposition of a hydrocarbon compound has a half-life period of 2 s. If the initial pressure decreases to 10.133kPa, the half-life period increases to 20 s. Determine the reaction order and the rate constant. 7. At 950 K the following data has been measured for the reaction 40 150 80 166.7 t/min P(total)/mmHg 100 At the beginning of the reaction there is only PHa. Calculate the reaction order and the rate constant. 8. The rate equation for catalytic decomposition of HI on Platinum has been determined as, r ki, at high pressures r k2 P(HI), at low pressures At 100°C, k1 - 5.0*10t Pa s;k-50 s Assume the rate of surface reaction is proportional to the quantity of adsorption of Hl on Platinum, calculate P(HIl) when r 2.5*10t Pa s Lindemann's mechanism for a unimolecular reaction: 9. Involves a mediator, M, that creates an excited-state species, A. The mechanism has the following elementary steps. A- Product A+ MAM > Product Derive the rate law for appearance of product at steady state.Explanation / Answer
Ans 9
The given reaction is A - - - - > Product
But in the mechanism
The elementary reaction gives the product
A* - - - - > Product
Rate law for appearance of product = k3[A*]
But A* is an intermediate so final rate law cannot be in terms of A* and M seems to be a catalyst
Rate of A*
d[A*]/dt = k1 [A] - k2 [A*] - k3 [A*]
At steady state
d[A*]/dt =0 = k1 [A] [M] - k2 [A*] [M] - k3 [A*]
k1 [A] - k2 [A*] - k3 [A*] = 0
k1 [A] = k2 [A*] + k3 [A*]
(k1 [A] ) / (k2 + k3) = [A*]
Rate law for appearance of product
= (k3 k1 [A]) / (k2+ k3)
k1, k2 and k3 are the rate constants of elementary reactions 1, 2 and 3 respectively
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