The demand for a particular hydrogenated compound, S, is 5.00 kmol/h. This chemical is synthesized in the gas-phase reaction $\mathrm{A}+\mathrm{H}_{2} \rightleftharpoons \mathrm{S}$ The reaction equilibrium constant at the reactor operating temperature is The fresh feed to the process is a mixture of A and hydrogen that is mixed with a recycle stream consisting of the same two species. The resulting mixture, which contains 3 kmol $\mathrm{A} / \mathrm{kmol} \mathrm{H}_{2}$, is fed to the reactor, which operates at an absolute pressure of 10.0 atm. The reaction products are in equilibrium. The effluent from the reactor is sent to a separation unit that recovers all of the S in essentially pure form. The A and hydrogen leaving the separation unit form the recycle that is mixed with fresh feed to the process. Calculate the feed rates of hydrogen and A to the process in kmol/h and the recycle stream flow rate in SCMH (standard cubic meters per hour).

Question

Quizlet Admin · Accepted Answer

The feed rates of hydrogen and A :\
A balance:\

\begin{center}$\mathrm{n_{1}}=$
\begin{tabular}{ c|c }
5 kmol S & 1 kmol A \ 
\hline
1 h& 1 kmol S \ 
\end{tabular}
$=\mathrm{5 \ kmol \ A/h}$ \end{center}

$\mathrm{H_{2}} $ balance:\

\begin{center}$\mathrm{n_{2}}=$
\begin{tabular}{ c|c }
5 kmol S & 1 kmol $\mathrm{H_{2}} $ \ 
\hline
1 h& 1 kmol S \ 
\end{tabular}
$=\mathrm{5 \ kmol \ H_{2} /h}$ \end{center}

Extent of reaction equations:\
$$\mathrm{\dot{n}_{1}=\dot{n}_{i0}+v_{i}\xi}$$
\
The gas-phase reaction
$\mathrm{A+H_{2}\Leftrightarrow S}$

A:$\mathrm{\dot{n}_{4}=3 \dot{n}_{3}-\xi}$\
$\mathrm{H_{2}:\dot{n}_{4}=3 \dot{n}_{3}-\xi}$\
S: $\mathrm{\dot{n}_{S}xi=5}$\
\
By combining the two equations the equations are obtained:\
A:$\mathrm{\dot{n}_{4}=3 \dot{n}_{3}-5}$\
$\mathrm{H_{2}:\dot{n}_{4}=3 \dot{n}_{3}-5}$\
\
The sum of these equations is equal:\
$\mathrm{\dot{n}=4\dot{n}_{3}-5}$

$$\mathrm{p_{A}=y_{A}P=\dfrac{\dot{n}_{4}}{\dot{n}}P}$$
$$\mathrm{p_{A}=\dfrac{3\dot{n}_{3}-5}{4\dot{n}_{3}-5}\cdot 10}$$

$$\mathrm{p_{H_{2}}=y_{H_{2}}P=\dfrac{\dot{n}_{5}}{\dot{n}}P}$$
$$\mathrm{p_{H_{2}}=\dfrac{\dot{n}_{3}-5}{4\dot{n}_{3}-5}\cdot 10}$$

$$\mathrm{p_{S}=y_{S}P}$$
$$\mathrm{p_{S}=\dfrac{5}{4\dot{n}_{3}-5}\cdot 10}$$

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