Documentation:CHBE Exam Wiki/Midterm Exam 2 2016W/MT2 Question 3

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Catalytic converters are used in cars to oxidize emission compounds in order to make them less toxic. A similar process may be used in a chemical plant to clean up emissions. A waste gas contains 0.5 mol% C₃H₈ (propane), 8.5 mol% CO₂, 4 mol% O₂, 10 mol% H₂O and the remainder inert nitrogen (N₂). In order to try to remove the propane, it undergoes complete combustion with a catalyst and can be modeled as a reversible reaction. This reaction is undertaken at 1 atm.

Question 3a [5 points]

Is there excess oxygen, if so, what is the % excess of oxygen?

[show] Hint
showStep 1 After setting up the balanced combustion equation, determine the extent of reaction by setting a basis (i.e 100 moles) showStep 2 Determine the % excess by first calculating the remaining oxygen using the extent of reaction

[show] Solution

showStep 1 The reaction for complete combustion is $C_{3}H_{8}+5O_{2}\rightarrow 3CO_{2}+4H_{2}O$ Using 100 moles as the basis, we have 0.5 moles of C3H8 and 4 moles of O2 Since all of propane is combusted, the extent of reaction, $\xi =0.5moles$. showStep 2 Using stoichiometry, ${\begin{aligned}n_{excess,O_{2}}&=4-5\cdot \xi \\&=1.5moles\\\end{aligned}}$ The % excess is then, ${\begin{aligned}\%excess&={\frac {n_{excess,O_{2}}}{n_{reacted,O_{2}}}}\times 100\%\\&=60\%\\\end{aligned}}$

Question 3b [15 points]

Write out a formula for the equilibrium constant (K_e) in terms of the pressure (P), reaction extent (E) and molar feed (n°C₃H₈, n°CO₂, n°O₂, n°H₂O, n°N₂) of each compound. Simplify the expression algebraically as much as possible. Note that you should not need to perform any calculations for this.

[show] Hint
showStep 1 Determine the expressions for the equilibrium concentration of each component in the gas showStep 2 Using the expression for equilibrium constant, Ke, rewrite the expression using the expressions determined in step 1

[show] Solution

showStep 1 Find the equilibrium concentration of each component, ${\begin{aligned}n_{C_{3}H_{8}}&=n_{C_{3}H_{8}}^{\circ }-\xi \\n_{O_{2}}&=n_{O_{2}}^{\circ }-5\xi \\n_{CO_{2}}&=n_{CO_{2}}^{\circ }+3\xi \\n_{H_{2}O}&=n_{H_{2}O}^{\circ }+4\xi \\n_{N_{2}}&=n_{N_{2}}^{\circ }(inert)\\n_{total}&=n_{C_{3}H_{8}}^{\circ }+n_{O_{2}}^{\circ }+n_{CO_{2}}^{\circ }+n_{H_{2}O}^{\circ }+n_{N_{2}}^{\circ }+\xi \end{aligned}}$ showStep 2 Since $y_{i}={\frac {n_{i}}{n_{total}}}$ ${\begin{aligned}K_{e}&={\frac {[y_{CO_{2}}\cdot P]^{3}[y_{H_{2}O}\cdot P]^{4}}{[y_{C_{3}H_{8}}\cdot P][y_{O_{2}}\cdot P]^{5}}}\\&={\frac {[n_{CO_{2}}\cdot P/n_{total}]^{3}[n_{H_{2}O}\cdot P/n_{total}]^{4}}{[n_{C_{3}H_{8}}\cdot P/n_{total}][n_{O_{2}}\cdot P/n_{total}]^{5}}}\\&={\frac {[n_{CO_{2}}]^{3}[n_{H_{2}O}]^{4}\cdot P}{[n_{C_{3}H_{8}}][n_{O_{2}}]^{5}\cdot n_{total}}}\\&={\frac {[n_{CO_{2}}^{\circ }+3\xi ]^{3}[n_{H_{2}O}^{\circ }+4\xi ]^{4}\cdot P}{[n_{C_{3}H_{8}}^{\circ }-\xi ][n_{O_{2}}^{\circ }-5\xi ]^{5}[n_{C_{3}H_{8}}^{\circ }+n_{O_{2}}^{\circ }+n_{CO_{2}}^{\circ }+n_{H_{2}O}^{\circ }+n_{N_{2}}^{\circ }+\xi ]}}\\\end{aligned}}$

Question 3c [15 points]

A conversion of 95% of propane from the combustion reaction is achieved. Following this, the remaining propane is removed using an adsorption bed at the same pressure containing 1 tonne activated carbon. Laboratory experiments were undertaken to model the adsorption of propane and gave the results found below, given this how many kilograms of propane can this bed remove?

$X^{*}(g_{C_{3}H_{8}}/100g_{\text{activated carbon}})=K'\cdot p_{C_{3}H_{8}}$

where $K'=67g_{C_{3}H_{8}}/(kPa*100g_{\text{activated carbon}})$

[show] Hint
showStep 1 By setting a basis, determine the total number of moles. Subsequently, determine the partial pressure of propane showStep 2 Use the adsorption equation to determine the mass of propane adsorped by activated carbon

[show] Solution

showStep 1 Using 100 moles as the basis again, ${\begin{aligned}\xi &=0.95\cdot n_{C_{3}H_{8}}^{\circ }\\&=0.95\cdot 0.5moles\\&=0.475moles\\\end{aligned}}$ ${\begin{aligned}n_{total}&=n_{C_{3}H_{8}}^{\circ }+n_{O_{2}}^{\circ }+n_{CO_{2}}^{\circ }+n_{H_{2}O}^{\circ }+n_{N_{2}}^{\circ }+\xi \\&=100moles+0.475moles\\&=100.475moles\\\end{aligned}}$ Determine the partial pressure of propane, ${\begin{aligned}p_{C_{3}H_{8}}&=P\cdot {\frac {n_{C_{3}H_{8}}}{n_{total}}}\\&=1atm\cdot {\frac {0.025}{100.475}}\cdot {\frac {101.325kPa}{1atm}}\\&=2.52\times 10^{-2}kPa\end{aligned}}$ showStep 2 ${\begin{aligned}X^{*}&=K'\cdot p_{C_{3}H_{8}}\\&=67g_{C_{3}H_{8}}/(kPa*100g_{\text{activated carbon}})\cdot 0.0252kPa&=1.69g_{C_{3}H_{8}}/(100g_{\text{activated carbon}})\end{aligned}}$ The mass of propane when 1 tonne of activated carbon is used is, ${\begin{aligned}m_{C_{3}H_{8}}&=m_{\text{activated carbon}}\cdot X^{*}\\&=1tonne_{\text{activated carbon}}\cdot {\frac {10^{6}g_{\text{activated carbon}}}{1tonne_{\text{activated carbon}}}}\times 1.69g_{C_{3}H_{8}}/(100g_{\text{activated carbon}})\\&=16.9kg\\\end{aligned}}$