Question

The elementary gas-phase reaction2A → Bis carried out in a constant-volume batch reactor where 50% conversion is achieved in 1 hour. Pure Ais charged to the reactor at an initial concentration of 0.2 mol/dm3. If the same reaction is carried outin a CSTR, what volume would be necessary to achieve 50% conversion for a feed molar flow rate of500 mol/h and an entering concentration of A of 0.2 mol/dm3? (Ans.:V = 5,000 dm3

Answer 1

Answer:

$V=5000dm^3$

Explanation:

Hello,

In this case, for the STR reactor carrying out this chemical reaction, the material balance is:

$\frac{dC_A}{dt}=r_A\\\\\frac{dC_A}{dt}=-kC_A^2$

Thus, with the given information we compute the rate constant as follows:

$C_A_0 \frac{dX_A}{dt}=kC_A_0^2(1-X_A)^2\\\\\int\limits^{0.5}_0 {\frac{dX_A}{(1-X_A)^2} } \,=kC_A_0t\\\\\frac{0.5}{1-0.5}=k*0.2\frac{mol}{dm^3}*1h\\ \\k=\frac{1}{(0.2\frac{mol}{dm^3})*1h} \\\\k=5\frac{dm^3}{mol*h}$

Now, for the CSTR we have the following design equation in terms of conversion for finding the volume:

$V=\frac{F_A_0*X_A}{-r_A}\\\\V=\frac{F_A_0*X_A}{k*C_A_0^2(1-X_A)^2}$

Therefore, it turns out:

$V=\frac{500\frac{mol}{h} *0.5}{5\frac{dm^3}{mol*h} *(0.2\frac{mol}{dm^3} )^2(1-0.5)^2}\\\\V=5000dm^3$

Best regards.