Chapter 2: Q62P (page 36)

62.In Section 18.4 we considered the following mechanism for the reaction of Br₂with H₂
${Br}_{2} + M ⇋_{k_{- 1}}^{k_{1}} Br + Br + M Br + H_{2} \overset{k_{2}}{\to} HBr + H {Br}_{2} + H \overset{k_{3}}{\to} HBr + Br$
Although this is adequate for calculating the initial rate of reaction, before product HBr builds up, there is an additional process that can participate as the reaction continues
$HBr + H \overset{k_{4}}{\to} H_{2} + Br$
(a) Write an expression for the rate of change of [H].
(b) Write an expression for the rate of change of [Br].
(c) As hydrogen and bromine atoms are both short-lived species, we can make the steady-state approximation and set the rates from parts (a) and (b) to 0. Express the steady-state concentrations [H] and [Br] in terms of concentrations of H₂, Br₂, HBr, and M. [Hint: Try adding the rate for part (a) to that for part (b).]
(d) Express the rate of production of HBr in terms of concentrations of H₂, Br₂, HBr, and M.

Short Answer

Expert verified

a.Rate expression for change of $[H]$ is $k_{2} [H_{2}] [Br] - k_{3} [H_{2}] [Br] - k_{4} [HBr] [H]$

b.Rate expression for change of $[B r]$ is $k_{1} [{Br}_{2}] [M], k_{3} [{Br}_{2}] [H], k_{4} [HBr] [H] - k_{- 1} {[Br]}^{2} [M] and k_{2} [Br] [H_{2}] . . .$

c. Steady state concendration of $[Br] and [H] is [Br] = \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}}, [H] = k_{2} [H_{2}] \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}}$

d.The the rate of production of is $k_{2} \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}} [H_{2}] + \sqrt{\frac{k_{1} [{Br}_{2}] [M] k_{2} [H_{2}]}{k_{- 1} [M] + k_{4} [HBr]}} (k_{3} [{Br}_{2}] - k_{4} [HBr])$

Step by step solution

Rate expression for change of H:

The reaction mentioned in the question is

${Br}_{2} + M ⇌_{k_{- 1}}^{k_{1}} Br + Br + M Br + H_{2} \overset{k_{2}}{\to} HBr + H {Br}_{2} + H \overset{k_{3}}{\to} HBr + Br HBr + H \overset{k_{4}}{\to} H_{2} + Br$

In the above equation, H is produced at the rate of $k_{2} [H_{2}] [Br]$ and it is disappeared at the rate of $k_{3} [H_{2}] [Br]$ and $k_{4} [HBr] [H]$ ,the overall rate is

$\frac{d [H]}{dt} = k_{2} [H_{2}] [Br] - k_{3} [H] [{Br}_{2}] - k_{4} [HBr] [H] . . . . . . (1)$

Hence the rate expression for the change of $[H]$ is

$k_{2} [H_{2}] [Br] - k_{3} [H_{2}] [Br] - k_{4} [HBr] [H]$

Rate expression for change of Br :

The change in concentration of bromine ie production of bromine occurs at the rate of

$k_{1} [{Br}_{2}] [M], k_{3} [{Br}_{2}] [H], k_{4} [HBr] [H]$ .

While bromine is lost at the rate of $k_{- 1} {[Br]}^{2} [M] and k_{2} [Br] [H_{2}]$ , The net rate is

$\frac{d [Br]}{dt} = k_{1} [{Br}_{2}] [M], k_{3} [{Br}_{2}] [H], k_{4} [HBr] [H] - k_{- 1} {[Br]}^{2} [M] and k_{2} [Br] [H_{2}] . . . . . . . . (2)$

The rate expression for the change of

[B r]

is given as

$k_{1} [{Br}_{2}] [M], k_{3} [{Br}_{2}] [H], k_{4} [HBr] [H] - k_{- 1} {[Br]}^{2} [M] and k_{2} [Br] [H_{2}] . . .$

Steady state concendration of H and Br:

C) Add eqaution 1 and 2

$\frac{d [H]}{dt} + \frac{d [Br]}{dt} = k_{2} [H_{2}] [Br] - k_{3} [H_{2}] [Br] - k_{4} [HBr] [H] + k_{1} [{Br}_{2}] [M] + k_{3} [{Br}_{2}] [H] + k_{4} [HBr] [H] - k_{- 1} {[Br]}^{2} [M] - k_{2} [Br] [H_{2}] . . . . . . . . . . . (3) k_{1} [{Br}_{2}] [M] - k_{- 1} {[Br]}^{2} [M] = 0 (\frac{d [H]}{dt} = \frac{d [Br]}{dt} = 0) - k_{- 1} {[Br]}^{2} [M] = k_{1} [{Br}_{2}] [M] {[Br]}^{2} = \frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]} [Br] = \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}}$

Steady state concendration of H and Br ;

Substituting the value of $[Br]$ in equation-1

$\frac{d [H]}{dt} = k_{2} [H_{2}] [Br] - k_{3} [H] [{Br}_{2}] - k_{4} [HBr] [H] . . . . . . (1) [Br] = \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}} \frac{d [H]}{dt} = k_{2} [H_{2}] \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}} - k_{3} [H] [{Br}_{2}] - k_{4} [HBr] [H] . . . . . . . (4) (\frac{d [H]}{dt} = 0) k_{2} [H_{2}] \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}} = k_{3} [H] [{Br}_{2}] + k_{4} [HBr] [H] . Taking out the commonterms out k_{2} [H_{2}] \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}} = [H] (k_{3} [{Br}_{2}] + k_{4} [HBr]) [H] = k_{2} [H_{2}] \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}} . . . . . . . . . . . . . . . . (5)$

Steady state concendration of hydrogen is $[H] = k_{2} [H_{2}] \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}}$

Steady state concendration of Bromine is $[Br] = \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}}$

Rate of production of HBr:

d) ) HBr is produced and lost at the rate of $k_{2} [Br] [H_{2}], k_{3} [H] [{Br}_{2}]$ and lost at the rate of $k_{4} [HBr] [H]$

So net change in the concentration of $[HBr]$ is

$\frac{d [HBr]}{[dt]} = k_{2} [Br] [H_{2}] + k_{3} [H] [{Br}_{2}] - k_{4} [HBr] [H] . . . . . . . . (6) Taking common terms out \ hfill \frac{d [HBr]}{[dt]} = k_{2} [Br] [H_{2}] + [H] (k_{3} [{Br}_{2}] - k_{4} [HBr]) . . . . . . . . (7)$

$[Br] = \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}} [H] = k_{2} [H_{2}] \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}} Substituting the [Br] and [H] valueof inequation - (7) \frac{d [HBr]}{[dt]} = k_{2} \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}} [H_{2}] + \sqrt{\frac{k_{1} [{Br}_{2}] [M] k_{2} [H_{2}]}{k_{- 1} [M] + k_{4} [HBr]}} (k_{3} [{Br}_{2}] - k_{4} [HBr]) - (8)$

Thus the rate of production of HBr is

$k_{2} \sqrt{\frac{k_{1} [{Br}_{2}] [M]}{k_{- 1} [M]}} [H_{2}] + \sqrt{\frac{k_{1} [{Br}_{2}] [M] k_{2} [H_{2}]}{k_{- 1} [M] + k_{4} [HBr]}} (k_{3} [{Br}_{2}] - k_{4} [HBr])$

Short Answer

Step by step solution

Rate expression for change of H:

Rate expression for change of Br :

Steady state concendration of H and Br:

Steady state concendration of H and Br ;

Rate of production of HBr:

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