Question:
A gas stream containing 95 mol\% CO\(_2\) and 5 mol\% ethanol is to be scrubbed with pure water in a counter-current, isothermal absorption column to remove ethanol. The desired composition of ethanol in the exit gas stream is 0.5 mol\%. The equilibrium mole fraction of ethanol in the gas phase, \( y^* \), is related to that in the liquid phase, \( x \), as \( y^* = 2x \). Assume CO\(_2\) is insoluble in water and neglect evaporation of water. If the water flow rate is twice the minimum, the mole fraction of ethanol in the spent water is:
A gas stream containing 95 mol\% CO\(_2\) and 5 mol\% ethanol is to be scrubbed with pure water in a counter-current, isothermal absorption column to remove ethanol. The desired composition of ethanol in the exit gas stream is 0.5 mol\%. The equilibrium mole fraction of ethanol in the gas phase, \( y^* \), is related to that in the liquid phase, \( x \), as \( y^* = 2x \). Assume CO\(_2\) is insoluble in water and neglect evaporation of water. If the water flow rate is twice the minimum, the mole fraction of ethanol in the spent water is:
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In absorption problems, establish the equilibrium relationship and material balance to calculate the mole fraction of the solute in the liquid or gas phase.
Updated On: Jan 24, 2025
- 0.0225
- 0.0126
- 0.0428
- 0.0316
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The Correct Option is B
Solution and Explanation
Step 1: Material balance for the absorption process.
Let:
- \( G_0 \) = gas flow rate (mol/s),
- \( L \) = liquid flow rate (mol/s),
- \( y_0 = 0.05 \) (ethanol in inlet gas stream),
- \( y_1 = 0.005 \) (ethanol in exit gas stream),
- \( x_1 \) = mole fraction of ethanol in spent water,
- \( x_0 = 0 \) (pure water).
The material balance for ethanol is:
\[
G_0 (y_0 - y_1) = L (x_1 - x_0).
\]
Step 2: Determine the minimum liquid flow rate (\( L_{\text{min}} \)).
From equilibrium, \( y^* = 2x \), and the operating line is:
\[
L_{\text{min}} / G_0 = \frac{y_0 - y_1}{x_1 - x_0}.
\]
Step 3: Liquid flow rate is twice the minimum.
\[
L = 2L_{\text{min}} = 2G_0 \frac{y_0 - y_1}{x_1 - x_0}.
\]
Rearrange for \( x_1 \):
\[
x_1 = \frac{G_0 (y_0 - y_1)}{2L} = \frac{y_0 - y_1}{2}.
\]
Step 4: Substitute the values.
\[
x_1 = \frac{0.05 - 0.005}{2} = \frac{0.045}{2} = 0.0126.
\]
Step 5: Conclusion.
The mole fraction of ethanol in the spent water is \( 0.0126 \).
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