Question

The mechanism for the chain reaction between gaseous hydrogen and gaseous bromine is where M is a chaperon. An analysis of this mechanism gives the rate equation as: when k_obs and k' _obs are the observed rate constants. What is the value of the activation energy for the overall reaction H_2(g) + Br_2(g) rightarrow 2 HBr(g)? What is the temperature dependence of k'_obs?

Answer 1

Write all the 5 equations as given:

Br₂ + M ------> 2 Br + M, E_a,1 = 192.9 kJ/mol (chain initiation) ….(1)

Br + H₂ ------> HBr + H, E_a,3 = 73.6 kJ/mol (propagation) ….(2)

HBr + H ------> H₂ + Br, E_a,4 = 3.8 kJ/mol (propagation) …..(3)

H + Br₂ -----> HBr + Br, E_a,5 = 3.8 kJ/mol (propagation) …..(4)

2 Br + M ------> Br₂ + M, E_a,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 E_a terms are extensive, i.e, they depend on the number of moles; hence we should multiply E_a,3 by 2 and add to get the overall activation energy as

E_a = E_a,1 + 2*E_a,3 + E_a,4 + E_a,5 + E_a,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^(-E_a,4/RT)

and k₅ = A.e^(-E_a,5/RT)

where A is Arrhenius constant and T is the temperature.

Therefore,

k₄/k₅ = e^(-E_a,4/RT)/e^(-E_a,5/RT)

===> k₄/k₅ = e^(-E_a,4/RT + E_a,5/RT) = e^(1/RT)*(E_a,5 – E_a,4) = e^(1/RT).(3.8 – 3.8) = e^(1/RT).(0) = e⁰ = 1

Therefore, k’_obs = 1 and hence independent of temperature (ans).