Coke Deposit in Propane Dehydrogenation: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2013, Q4

Question 4: Coke Deposit in Propane Dehydrogenation (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Basis: 12 kg pure C₃H₈ fed ($M=44.096$ g/mol). Gas effluent $G$ (mol%) and the coke deposit $D$ (96 mass% C, 4 mass% H):

Gas	mol%	C atoms	H atoms
H₂	25.4	0	2
CH₄	3.2	1	4
C₂H₄	0.3	2	4
C₂H₆	5.3	2	6
C₃H₆	21.3	3	6
C₃H₈	44.5	3	8

Find. The deposit mass $D$ as a weight percentage of the 12 kg propane feed.

Figure 4 — Fixed-bed reactor: propane in; a gas effluent $G$ and a C/H coke deposit $D$ out. Only C and H are conserved — two atom balances, two unknowns ($G$, $D$).

Approach. With only carbon and hydrogen present, write a carbon-atom and a hydrogen-atom balance on the reactor; the two equations solve simultaneously for the effluent gas moles $G$ and the deposit mass $D$.

Feed atoms. Moles of propane $=12{,}000/44.096 = 272.13$ mol, giving $$C_{in} = 3(272.13)=816.38\ \text{mol C},\qquad H_{in}=8(272.13)=2177.0\ \text{mol H}.$$
Atoms per mole of effluent gas. Weighting each species' atom count by its mole fraction: $$\bar C_G = \textstyle\sum c_i y_i = 2.118\ \tfrac{\text{mol C}}{\text{mol gas}},\qquad \bar H_G = \textstyle\sum h_i y_i = 5.804\ \tfrac{\text{mol H}}{\text{mol gas}}.$$
Carbon and hydrogen balances. Let $G$ = effluent gas (mol) and $D$ = deposit (g), with C and H in the deposit as $0.96D/12.011$ and $0.04D/1.008$ mol: $$\text{C:}\quad 816.38 = 2.118\,G + \frac{0.96}{12.011}D,\qquad \text{H:}\quad 2177.0 = 5.804\,G + \frac{0.04}{1.008}D.$$
Eliminate $G$ and solve for $D$. Solving the pair simultaneously, $$D = \frac{\bar H_G\,C_{in}-\bar C_G\,H_{in}}{\bar H_G\,b_C-\bar C_G\,b_H}=335.3\ \text{g},\qquad G = 372.8\ \text{mol gas},$$ where $b_C=0.96/12.011$ and $b_H=0.04/1.008$. Hence $$\boxed{D \approx 335\ \text{g of coke}}.$$
Express as weight percent of feed. $$\frac{D}{m_{feed}}=\frac{335.3}{12{,}000}=0.0279=\boxed{2.79\ \text{wt\% of the propane feed}}.$$

Quantity	Result
Effluent gas produced, $G$	372.8 mol (per 12 kg feed)
Coke deposit, $D$	335 g
Deposit as wt% of feed	2.79%

Concept & Theory

Topic. Elemental (atom) balances on a reacting system with an unmeasured solid product.

Key relations. Conservation of each element: $\text{(element in feed)} = \text{(element in gas)} + \text{(element in deposit)}$; atom counts per mole of gas from $\sum(\text{atoms})_i y_i$; deposit split via its mass composition and atomic weights.

Why this works. When several reactions occur at once (cracking, dehydrogenation and coking here), tracking individual reaction extents is hopeless — but atoms are conserved regardless of how many reactions run. Two elements give two independent balances, which is exactly enough to find the two unknowns (gas quantity and deposit mass) without ever writing a stoichiometric equation.

Common pitfall. Forgetting that the deposit contains hydrogen too (4%); treating it as pure carbon over-counts gas hydrogen and skews both balances. Another slip is mixing mass and moles — the gas balances are in moles, the deposit is given in mass, so convert with atomic weights before equating.

Source. Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes §4.7 (combustion / atomic-species balances for multi-reaction systems).

Question 4 of 7: Coke Deposit in Propane Dehydrogenation

Question 4: Coke Deposit in Propane Dehydrogenation (equal value)