Write all the 5 equations as given:
Br2 + M ------> 2 Br + M, Ea,1 = 192.9 kJ/mol (chain initiation) ….(1)
Br + H2 ------> HBr + H, Ea,3 = 73.6 kJ/mol (propagation) ….(2)
HBr + H ------> H2 + Br, Ea,4 = 3.8 kJ/mol (propagation) …..(3)
H + Br2 -----> HBr + Br, Ea,5 = 3.8 kJ/mol (propagation) …..(4)
2 Br + M ------> Br2 + M, Ea,2 = 0 kJ/mol (termination) …..(5)
Multiply (2) by 2 and add all the equations above to obtain
Br2 + M + 2 Br + 2 H2 + HBr + H + H + Br2 + 2 Br + M ------> 2 Br + M + 2 HBr + 2 H + H2 + Br + HBr + Br + Br2 + M
====> 2 Br2 + 2 M + 4 Br + 2 H2 + HBr + 2 H -----> 4 Br + 2 M + 3 HBr + 2 H + H2 + Br2
Cancel out common terms from both sides to write
Br2 + H2 -----> 2 HBr
which is our required equation.
Note that Ea terms are extensive, i.e, they depend on the number of moles; hence we should multiply Ea,3 by 2 and add to get the overall activation energy as
Ea = Ea,1 + 2*Ea,3 + Ea,4 + Ea,5 + Ea,2 = (192.9 kJ/mol) + (2)*(73.6 kJ/mol) + (3.8 kJ/mol) + (3.8 kJ/mol) + (0 kJ/mol) = 347.7 kJ/mol (ans).
Note that k’obs = k4/k5. As per Arrhenius equation, we know that
k4 = A.e^(-Ea,4/RT)
and k5 = A.e^(-Ea,5/RT)
where A is Arrhenius constant and T is the temperature.
Therefore,
k4/k5 = e^(-Ea,4/RT)/e^(-Ea,5/RT)
===> k4/k5 = e^(-Ea,4/RT + Ea,5/RT) = e^(1/RT)*(Ea,5 – Ea,4) = e^(1/RT).(3.8 – 3.8) = e^(1/RT).(0) = e0 = 1
Therefore, k’obs = 1 and hence independent of temperature (ans).
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