Give step-by-step solution with explanation and final answer:In stream reforming of ethane the following reaction occurs
C2Hs + 2H20 2CO + 5H

Consider the situation in which gaseous ethane and w

Question

Give step-by-step solution with explanation and final answer:In stream reforming of ethane the following reaction occurs
C2Hs + 2H20 > 2CO + 5H

Consider the situation in which gaseous ethane and water vapor diffuse through a
gas film of thickness L, to the surface of nickel catalyst where they react to form
gaseous CO and H> which diffuse back through the gas film. The mole fraction of
ethane at the catalyst surface is y; and y; on the gas side of the film. The gaseous
concentration is C (ideal gas) and the diffusivity of ethane in the mixture is Dam.
Derive an equation for the flux of ethane, Na.

VIPSolutions · Accepted Answer

Let's break down the problem and derive the equation for the flux of ethane ($ N_A $).

---

## **Step 1: Identify the Diffusing Species**

- **A (C₂H₆, Ethane):** Diffusing toward the catalyst and consumed.
- **B (H₂O):** Also consumed at the surface.
- **C (CO) and D (H₂):** Produced at the surface and diffuse back.

We focus on **ethane (A)**.

---

## **Step 2: Stoichiometry and Fluxes**

Reaction:  
$$
\text{C}_2\text{H}_6 + 2\text{H}_2\text{O} \rightarrow 2\text{CO} + 5\text{H}_2
$$

Let:
- $ N_A $: Molar flux of C₂H₆ **toward** the surface (consumed)
- $ N_C $: Molar flux of CO **away from** the surface (produced)
- $ N_D $: Molar flux of H₂ **away from** the surface (produced)

From stoichiometry:
- 1 mol C₂H₆ consumed → 2 mol CO and 5 mol H₂ produced

So, for every $ N_A $ moles of C₂H₆ that diffuse in,
- $ N_C = -2N_A $ (CO, negative sign because away from the surface)
- $ N_D = -5N_A $ (H₂, negative sign because away from the surface)

The **total molar flux** in the z-direction:
$$
N_{total} = N_A + N_B + N_C + N_D
$$
But only A, C, D are considered here (since B is water vapor and can be a non-diffusing component or in excess).

---

## **Step 3: Stefan-Maxwell Equation (for equimolar counterdiffusion with net flux)**

In this problem, for a gas film of thickness $ L $, the flux of ethane is affected by the counter-diffusion of CO and H₂. The net flux of all gases is **not zero** because more moles are produced than consumed (1 mol C₂H₆ + 2 mol H₂O → 2 mol CO + 5 mol H₂: 3 mol in, 7 mol out).

**Total moles in = 3**, **Total moles out = 7**  
Net increase in moles per mole ethane reacted: $ 7 - 3 = 4 $

---

## **Step 4: Find the Flux Equation**

For **species A** (ethane):

From Fick's Law with a net flux (non-equimolar):
$$
N_A = -D_{A,M} C \frac{dy_A}{dz} + y_A N_{total}
$$

Where:
- $ D_{A,M} $: Diffusivity of A in mixture
- $ C $: Total molar concentration (ideal gas)
- $ y_A $: Mole fraction of A
- $ N_{total} $: Total molar flux (sum of fluxes of all components)

### **Find $ N_{total} $ in Terms of $ N_A $:**

From stoichiometry (per 1 mol C₂H₆ consumed):
$$
\text{C}_2\text{H}_6 + 2\text{H}_2\text{O} \rightarrow 2\text{CO} + 5\text{H}_2
$$
- 1 mol C₂H₆ disappears ($ N_A $)
- 2 mol H₂O disappears ($ N_B $)
- 2 mol CO forms ($ N_C $)
- 5 mol H₂ forms ($ N_D $)

If the flux of A is $ N_A $, the fluxes are proportional to stoichiometric coefficients:
$$
\frac{N_A}{-1} = \frac{N_B}{-2} = \frac{N_C}{2} = \frac{N_D}{5}
$$
So,
- $ N_B = 2 N_A $
- $ N_C = -2 N_A $
- $ N_D = -5 N_A $

Total molar flux:
$$
N_{total} = N_A + N_B + N_C + N_D = N_A + 2 N_A - 2 N_A - 5 N_A = (N_A + 2N_A) - (2N_A + 5N_A) = 3N_A - 7N_A = -4N_A
$$

But this seems off: Let's check with the sign convention.
- $ N_A $: toward surface, **positive**
- $ N_B $: toward surface, **positive**
- $ N_C $: **away** from surface, **negative**
- $ N_D $: **away** from surface, **negative**

So:
$$
N_{total} = N_A + N_B + N_C + N_D = N_A + N_B + (-N_C) + (-N_D)
$$
But if we use the direction convention (toward the surface **positive**):

If $ N_A $ is **positive** toward the surface, then $ N_C $ and $ N_D $ are **negative** (move away from the surface).

So:
$$
N_{total} = N_A + N_B + N_C + N_D
$$
But, using stoichiometry:
- $ N_A = -\nu_A r $
- $ N_B = -\nu_B r $
- $ N_C = \nu_C r $
- $ N_D = \nu_D r $

Where $ r $ is the rate per area.

So,
$$
N_{total} = N_A + N_B + N_C + N_D = -\nu_A r - \nu_B r + \nu_C r + \nu_D r = (\nu_C + \nu_D - \nu_A - \nu_B) r
$$
$$
= (2 + 5 - 1 - 2) r = (7 - 3) r = 4 r
$$
But $ N_A = - r \to r = -N_A $, so
$$
N_{total} = 4 (-N_A) = -4 N_A
$$

---

## **Step 5: Integrate the Flux Equation**

The general equation:
$$
N_A = -D_{A,M} C \frac{dy_A}{dz} + y_A N_{total}
$$

But $ N_{total} = -4 N_A $ (from above):

$$
N_A = -D_{A,M} C \frac{dy_A}{dz} + y_A (-4 N_A)
$$
$$
N_A + 4 y_A N_A = -D_{A,M} C \frac{dy_A}{dz}
$$
$$
N_A (1 + 4 y_A) = -D_{A,M} C \frac{dy_A}{dz}
$$
$$
N_A = \frac{ -D_{A,M} C }{1 + 4 y_A} \frac{dy_A}{dz}
$$

Integrate both sides from $ z= $ to $ z=L $ (film thickness), with $ y_A = y_1 $ at $ z= $ (gas side), and $ y_A = y_2 $ at $ z=L $ (catalyst surface):

$$
\int_{y_1}^{y_2} (1 + 4 y_A) dy_A = - \int_^L \frac{N_A}{D_{A,M} C} dz
$$
$$
\left[ y_A + 2 y_A^2 \right]_{y_1}^{y_2} = - \frac{N_A L}{D_{A,M} C}
$$
$$
(y_2 + 2 y_2^2) - (y_1 + 2 y_1^2) = - \frac{N_A L}{D_{A,M} C}
$$

Solve for $ N_A $:

$$
N_A = - \frac{D_{A,M} C}{L} \left[ (y_2 + 2 y_2^2) - (y_1 + 2 y_1^2) \right]
$$
or
$$
\boxed{
N_A = \frac{D_{A,M} C}{L} \left[ (y_1 + 2 y_1^2) - (y_2 + 2 y_2^2) \right]
}
$$

---

## **Final Answer**

### **Flux of Ethane ($N_A$)**
$$
\boxed{
N_A = \frac{D_{A,M} C}{L} \left[ (y_1 + 2 y_1^2) - (y_2 + 2 y_2^2) \right]
}
$$

**Where:**
- $ N_A $: Flux of ethane (mol/(area·time))
- $ D_{A,M} $: Diffusivity of ethane in mixture (m²/s)
- $ C $: Total molar concentration of gas (mol/m³)
- $ L $: Film thickness (

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