Chapter 4: Stoichiometry

Old Exam Questions

The elementary gas phase reaction

$2\text{A} + \text{B} \longrightarrow 3\text{C}$

takes place isothermally in a flow reactor with no pressure drop. The feed is equal molar in A and B. Write the rate of reaction, -r_A, solely as a function of the conversion, X [i.e. -r_A = f(X)].

$-r_A = \left(\, \right) \frac{\text{kmol}}{\text{m}^3 \cdot \text{s}}$

Hint 1: What is the rate law?

(a) $-r_{A} = k_{A}C_{A}{C_{A}}^{\frac{1}{2}}$

(b) $-r_{A} = k_{A}C_{A}{C_{B}}$

Hint 2: What is e?

Hint 3: What is C_A as a function of X?

(a) $C_{A} = C_{A0}(1 - x)$

(b) $C_{A} = C_{A0}\frac{(1-x)}{(1+3x)}$

Hint 4: What is C_B as a function of X?

(a) $C_{B} = C_{A0}(\frac{1}{2} - \frac{1}{2}x)$

(b) $C_{B} = C_{A0}(1 - \frac{1}{2}x)$

Solution

Hint 1:

Remember that the rate law for gases includes volume change.

$-r_A = \frac{\phantom{.}}{\phantom{.}} \, \frac{\text{kmol}}{\text{m}^3 \cdot \text{s}}$

$k = 2 \, \text{[appropriate MKS units]}$

$K = 10 \, \text{[appropriate MKS units]}$

$C_{A0} = 2 \, \text{kmoles}/\text{m}^3$

$-r_A = kC_A^2 C_B, \, k = 2 \, \frac{\text{dm}^6}{\text{mol}^2 \cdot \text{s}}$

$T = T_0, \, P = P_0$

Back to Problem

Hint 2

$\varepsilon = y_{A0} \delta$

$A + \frac{1}{2}B \longrightarrow \frac{3}{2}C$

$\varepsilon = \frac{1}{2} \left( \frac{3}{2} - \frac{1}{2} - 1 \right) = 0$

Back to Problem

Hint 3

What is C_A as a function of X?

$C_A = \frac{C_{A0}(1 - X)}{(1 + \varepsilon X)} = C_{A0}(1 - X)$

Back to Problem

Hint 4

What is C_B as a function of X?

From the stoichiometry: b/a = 1/2

Equal Molar:

$\theta_B = \frac{F_{B0}}{F_{A0}} = 1$

$C_B = C_{A0} \frac{\left(\theta_B - \frac{1}{2}X\right)}{(1 + \varepsilon X)} = C_{A0}(1 - 0.5X)$

Back to Problem

Solution

We see species A is the limiting reactant as the feed is equal molar in A and B, but one mole of B consumes two moles of A.

$-r_A = k C_A^2 C_B, \, k = 2 \, \frac{\text{dm}^6}{\text{mol}^2 \cdot \text{s}}$

$T = T_0, \, P = P_0$

$\varepsilon = y_{A0} \delta$

Need to put everything in terms of per mole of limiting reactant A.

$A + \frac{1}{2}B \longrightarrow \frac{3}{2}C$

$\varepsilon = \frac{1}{2} \left( \frac{3}{2} - \frac{1}{2} - 1 \right) = 0$

$C_A = \frac{C_{A0}(1 - X)}{(1 + \varepsilon X)} = C_{A0}(1 - X)$

From the stoichiometry: b/a = 1/2

Equal Molar:

$\theta_B = \frac{F_{B0}}{F_{A0}} = 1$

$C_B = C_{A0} \frac{\left(\theta_B - \frac{1}{2}X\right)}{(1 + \varepsilon X)} = C_{A0}(1 - 0.5X)$

$-r_A = kC_{A0}^3(1 - X)^2\left(1 - \frac{X}{2}\right)$

$-r_A = 2 \frac{\text{dm}^6}{\text{mol}^2 \cdot \text{s}} \left(\frac{2 \text{mol}}{\text{dm}^3}\right)^3 (1 - 0.4)^2 (1 - 0.2)$

$-r_A = 4.61 \, \text{mol}/\text{dm}^3 \cdot \text{s}$

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