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. 0.0225
• B. 0.0126
• C. 0.0428
• D. 0.0316

Hint:

In absorption problems, establish the equilibrium relationship and material balance to calculate the mole fraction of the solute in the liquid or gas phase.

Answer 1

The Correct Option is B

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 \( 0.0126 \).