Suppose the liquid-phase reaction A → products is second-order, with $( -r_A) = k_Ac_A^2$ , and takes place in a PFR. Show that the SFM gives the same result for $1 – f_A = c_A /c_{Ao}$ as does the integration of equation 15.2-16, the material balance .

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Integration of equation 15.2-16 with the rate law given results in (for a PFR)

[latex] \frac{dc_A}{dt} + ( -r_A) = 0 [/latex] (15.2-16)

[latex] 1 - f_A = \frac{c_A}{c_{Ao}} = \frac {1}{1 + ktc_{Ao}} [/latex] (A)

where t is the residence time (the same result is obtained for a BR, for a constant-density system) .

Using the SFM, equation 13.5-2, with [latex] [c_A(t) /c_{Ao} ]_{BR}[/latex] given by (A), and E(t) by equation 13.4-8, we obtain

[latex]1-f_A = \frac {c_A}{c_{Ao}} = \int_{0}^{\infty}{\left[\frac {c_A(t)}{c_{Ao}} ight]_{BR} E(t)dt} (SFM) [/latex] (13.5-2)

[latex]E(t) = \delta (t - \bar{t}) = \infty ; t = \bar{t} = 0 ; t eq \bar{t} (PF)[/latex] (13.4-8)

[latex]1 - f_A = \frac {c_A}{c_{Ao}} = \int_{0}^{\infty}{\left(\frac{1}{1 + ktc_{Ao}} ight)\delta (t - \bar{t} ) dt} = \frac{1}{1 + k\bar{t} c_{Ao}} [/latex]

which is the same result as in (A) from the material balance, since [latex]\bar{t}[/latex] is the same as the residence time.
The same conclusion would be reached for any form of rate law introduced into equation 15.2-16 on the one hand, and into equation 13.5-2 on the other.

Question 15.5: Suppose the liquid-phase reaction A → products is second-ord...

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