Chapter 4: Stoichiometry

Writing -r_A Solely as a Function of X. 

Write the rate law for the elementary liquid phase reaction

\[3A + 2B \to 4C\]

solely in terms of conversion. The feed to the batch reactor is equal molar A and B with C_A0 = 2 mol/dm³ and k_A= .01 (dm³/mol)⁴1/s.

Hint 1: What is the rate law?

(a) \( -r_A = k_A C_A C_B^{\frac{2}{3}} \)

(b) \( -r_A = k_A C_A C_B \)

(c) \( -r_A = k_A C_A^3 C_B^2 \)

Hint 2: What is the concentration of A?

(a) \( C_A = C_{A0}(1 - x) \)

(b) \( C_A = C_{A0}(2 - x) \)

(c) \( C_A = C_{A0}(1 - 3x) \)

Hint 3: What is the concentration of B?

(a) \( C_B = C_{A0}\left(1 - \frac{2}{3}x\right) \)

(b) \( C_B = C_{A0}\left(\frac{2}{3} - \frac{2}{3}x\right) \)

(c) \( C_B = C_{A0}(1 - 2x) \)

Full Solution

What is the rate law?

The reaction is elementary so what happens to the coefficients from the reaction?

Back to Problem

What is the concentration of A?

Liquid phase, v = v_o (no volume change)

\( C_A = \frac{N_A}{V} = \frac{N_A}{V_0} = \frac{N_{A0}}{V_0}(1 - X) = C_{A0}(1 - X) \)

Back to Problem





























Hint 3

What is the concentration of B?

\( C_B = \frac{N_B}{v} = \frac{N_B}{v_0} = \frac{N_{A0} \left( \Theta_B - \frac{b}{a} X \right)}{v_0} = C_{A0} \left( \Theta_B - \frac{b}{a} X \right) \)

What is \( \theta_B\ \)?

Equal molar

\( \theta_B = \frac{N_{B0}}{N_{A0}} = \frac{1}{1} = 1 \)

Species A is the limiting reactant because the feed is equal molar in A and B, and two moles of B consumes 3 moles of A.

\( \text{A} + \frac{2}{3}\text{B} \rightarrow \frac{4}{3}\text{C} \)

\( \frac{b}{a} = \frac{\frac{2}{3}}{1} = \frac{2}{3} \)

\( C_B = C_{A0} \left(1 - \frac{2}{3} X\right) \)

Back to Problem

Write the rate law for the elementary liquid phase reaction
\[ 3A + 2B \to 4C \]

solely in terms of conversion. The feed to the batch reactor is equal molar A and B with C_A0 = 2 mol/dm³ and k_A= .01 (dm³/mol)⁴1/s.

1) Rate Law: -r_A=kC³_AC²_B

2) Stoichiometry:

Species A

Liquid phase, v = v_o (no volume change)

\( C_A = \frac{N_A}{V} = \frac{N_A}{V_0} = \frac{N_{A0}}{V_0}(1 - X) = C_{A0}(1 - X) \)

Species B

\( C_B = \frac{N_B}{v} = \frac{N_B}{v_0} = \frac{N_{A0} \left( \Theta_B - \frac{b}{a} X \right)}{v_0} = C_{A0} \left( \Theta_B - \frac{b}{a} X \right) \)

What is \( \theta_B\ \)?

\( \theta_B = \frac{N_{B0}}{N_{A0}} = \frac{1}{1} = 1 \)

Species A is the limiting reactant because the feed is equal molar in A and B, and two moles of B consumes 3 moles of A.

\( \text{A} + \frac{2}{3}\text{B} \rightarrow \frac{4}{3}\text{C} \)

\( \frac{b}{a} = \frac{\frac{2}{3}}{1} = \frac{2}{3} \)

\( C_B = C_{A0} \left(1 - \frac{2}{3} X\right) \)

\( -r_A = k C_{A0}^5 (1 - X)^3 \left(1 - \frac{2}{3} X\right)^2 \)

\( = (0.01) \left(\frac{\text{dm}^3}{\text{mol}}\right)^4 \left(\frac{1}{\text{s}}\right) \left(\frac{2 \, \text{mol}}{\text{dm}^3}\right)^5 (1 - X)^3 \left(1 - \frac{2}{3} X\right)^2 \)

\( = 0.32 (1 - X)^3 \left(1 - \frac{2}{3} X\right)^2 \frac{\text{mol}}{\text{dm}^3 \cdot \text{s}} \)

We now have -r_A=f(X) and can size reactors or determine batch reaction times.

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