91Ó°ÊÓ

The demand for a particular hydrogenated compound, \(\mathrm{S}\), is \(5.00 \mathrm{kmol} / \mathrm{h}\). This chemical is synthesized in the gas-phase reaction $$A+H_{2}=S$$ The reaction equilibrium constant at the reactor operating temperature is $$K_{p}=\frac{p_{\mathrm{S}}}{p_{\Lambda} p_{\mathrm{H}_{2}}}=0.1 \mathrm{atm}^{-1}$$ 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 \(3 \mathrm{kmol} \mathrm{A} / \mathrm{kmol} \mathrm{H}_{2},\) 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).

Step by step solution

01

Set up the mass balance equations

As per the stoichiometry of the reaction, for every mole of A and H2 reacted, one mole of S is produced. The reaction products are in equilibrium which means the molar flow rate of S in the output stream equals the demand rate of 5 kmol/h. Let's denote the molar flow rates of A and H2 in the feed as \(F_A\) and \(F_{H2}\) respectively, and in the recycle stream as \(R_A\) and \(R_{H2}\). The molar flow rates in the reactor are then given as: For A: \(F_A + R_A - 5 = R_A\) For H2: \(F_{H2} + R_{H2} - 5 = R_{H2}\).

02

Use the equilibrium constant expression

The equilibrium constant expression \(K_p = 0.1 atm^{-1} = p_S/(p_A p_{H2})\) and the total reactor pressure of 10 atm gives us: \((5/10)/((R_A/10)(R_{H2}/10)) = 0.1\). Simplifying this expression gives us \(R_A = 50/R_{H2}\).

03

Solve for the molar flow rates

Substitute \(R_A = 50/R_{H2}\) from step 2 into the mass balance equations from step 1 to get: \(F_A = 5 - 50/R_{H2}\) and \(F_{H2} = 5\). Since the feed mixture contains 3 moles of A for every mole of H2, we get \(F_A = 3F_{H2} = 3*5 = 15 kmol/h\). Substituting this value in the first mass balance equation gives us \(R_{H2} = 10 kmol/h\) and \(R_A = 50/10 = 5 kmol/h\).

04

Convert the recycle stream flow rate to SCMH

The total moles flowing in the recycle stream are \(R_A + R_{H2} = 5 + 10 = 15 kmol/h\). Using the ideal gas law under standard conditions (0 °C and 1 atm), we get \(V = nRT/P = 15*10^3*(0.0821*273)/1 = 328350 m^3/h\) or approximately 328.35 SCMH.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Balance Equations

Understanding mass balance equations is crucial for any chemical engineering calculation. It's based on the principle of conservation of mass, which states that mass cannot be created nor destroyed in a chemical process.

In the context of our exercise, a mass balance equation helps to quantify the amount of each reactant and product moving through the system. Since we are dealing with a synthesis reaction of the form A + H₂ = S, it’s essential to track how much of each component (A, H₂, and S) is present in different stages of the process—feed, reaction, and separation.

Setting up the equations takes into account not only the stoichiometry of the reaction but also the reaction conditions such as the inflows and outflows, as well as the recycling process. In our case, for each kmol of the hydrogenated compound S produced, a kmol of A and H₂ is consumed. Thus, the balanced equation signifies that the flow rates of the reactants must adjust correspondingly to produce the desired 5 kmol/h of S.

Equilibrium Constant Expression

The equilibrium constant expression quantifies the concentrations of reactants and products in a chemical reaction at equilibrium. It is defined by the ratio of the product of the partial pressures of the products to the reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.

In our example, the expression is given by the formula K_p = p_S / (p_A p_H2). Understanding this helps us determine the relation between the partial pressures of A, H₂, and S in the reactor at equilibrium. In practice, we used the given value of K_p and the reactor pressure to establish a relationship between the molar flow rates of A and H₂, once we accounted for the partial pressures being fractions of the total pressure.

Stoichiometry

The cornerstone of most chemical reaction calculations is stoichiometry, which informs us about the relative quantities of reactants and products involved in a chemical reaction. Stoichiometry is grounded in the balanced chemical equation, which stipulates the proportion in which reactants combine and products form.

For our problem, stoichiometry tells us that for each mole of A reacting with each mole of H₂, one mole of S is produced. This simple ratio (1:1:1) is the key to calculating other necessary parameters, such as the feed rates of A and H₂. The stoichiometric coefficients help link the mass balance equations with the equilibrium calculations, allowing us to solve for the unknown flow rates. Notably, when the exercise mentioned a feed mixture containing 3 kmol of A per 1 kmol of H₂, this stoichiometric ratio was vital in finalizing the composition of the feed to the reactor and completing the solution.