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?
Hint
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Step 1
After setting up the balanced combustion equation, determine the extent of reaction by setting a basis (i.e 100 moles)
Step 2
Determine the % excess by first calculating the remaining oxygen using the extent of reaction
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Solution
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Step 1
The reaction for complete combustion is
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, .
Step 2
Using stoichiometry,
![{\displaystyle {\begin{aligned}n_{excess,O_{2}}&=4-5\cdot \xi \\&=1.5moles\\\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d0bb5fcb8ba337748afeea85a59ed6193d9059a5)
The % excess is then,
![{\displaystyle {\begin{aligned}\%excess&={\frac {n_{excess,O_{2}}}{n_{reacted,O_{2}}}}\times 100\%\\&=60\%\\\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/cd4a6b624bf569673290df2aecb13072e553e6b9)
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Question 3b [15 points]
Write out a formula for the equilibrium constant (Ke) 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.
Hint
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Step 1
Determine the expressions for the equilibrium concentration of each component in the gas
Step 2
Using the expression for equilibrium constant, Ke, rewrite the expression using the expressions determined in step 1
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Solution
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Step 1
Find the equilibrium concentration of each component,
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/24ece2c5c74636ce178863a259a4c04e97b34443)
Step 2
Since
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9f9a99ddfd0dc8861752a00a624fcf3bf9d6fcf1)
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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?
![{\displaystyle X^{*}(g_{C_{3}H_{8}}/100g_{\text{activated carbon}})=K'\cdot p_{C_{3}H_{8}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/16129edbc24fa2def63a23fdecca65cd55d054fb)
where
Hint
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Step 1
By setting a basis, determine the total number of moles. Subsequently, determine the partial pressure of propane
Step 2
Use the adsorption equation to determine the mass of propane adsorped by activated carbon
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Solution
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Step 1
Using 100 moles as the basis again,
![{\displaystyle {\begin{aligned}\xi &=0.95\cdot n_{C_{3}H_{8}}^{\circ }\\&=0.95\cdot 0.5moles\\&=0.475moles\\\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/56647f4f8a20722da869efde18dc64ef0ac1aac4)
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/381425724ae0d1f4313dc8cbf87dbfc86c7b1740)
Determine the partial pressure of propane,
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/647b40021ac2593db349417bc6986395a28a974b)
Step 2
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/97cdbb6a31ef7696357da058578328e9d0f7a39e)
The mass of propane when 1 tonne of activated carbon is used is,
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/71af0c9ecbb2705910b6961d7d12cf6115ab5c53)
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